A candle 4.85 cm tall is 39.2 cm to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?
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Textbook Solutions for Sears and Zemansky's University Physics with Modern Physics
Question
Problem 57E
BIO If the person in Exercise 34.54 chooses ordinary glasses over contact lenses, what power lens (in diopters) does she need to correct her vision if the lenses are 2.0 cm in front of the eye?
34.54 . BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?
Solution
Solution 57E
Step 1
Lens formula is,
The power of the lens is,
Here, s is object distance, is image distance and f is focal length.
full solution
BIO If the person in Exercise 34.54 chooses ordinary
Chapter 34 textbook questions
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The image of a tree just covers the length of a plane mirror 4.00 cm tall when the mirror is held 35.0 cm from the eye. The tree is 28.0 m from the mirror. What is its height?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A pencil that is 9.0 cm long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead 12.0 cm from the mirror surface and the end of the eraser 21.0 cm from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A concave mirror has a radius of curvature of 34.0 cm. (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Repeat Exercise 34.5 for the case in which the mirror is convex
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The diameter of Mars is 6794 km, and its minimum distance from the earth is When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of 1.75 m.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An object is 24.0 cm from the center of a silvered spherical glass Christmas tree ornament 6.00 cm in diameter. What are the position and magnification of its image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 cm. Reflection from the surface of the shell forms an image of the 1.5-cm-tall coin that is 6.00 cm behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
You hold a spherical salad bowl 90 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a 35-cm radius of curvature. (a) Where is the image of your 2.0-cm-tall nose located? (b) What are the images size, orientation, and nature (real or virtual)?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
(a) Show that Eq. (34.6) can be written as and hence the lateral magnification given by Eq. (34.7) can be expressed as . (b) A concave spherical mirror has focal length . What is the nonzero distance of the object from the mirror vertex if the image has the same height as the object? In this case, is the image erect or inverted? (c) A convex spherical mirror has . What is the nonzero distance of the object from the mirror vertex if the height of the image is one-half the height of the object?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The thin glass shell shown in Fig. E34.12 has a spherical shape with a radius of curvature of 12.0 cm, and both of its surfaces can act as mirrors. A seed 3.30 mm high is placed 15.0 cm from the center of the mirror along the optic axis, as shown in the figure. (a) Calculate the location and height of the image of this seed. (b) Suppose now that the shell is reversed. Find the location and height of the seeds image.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an erect image with a magnification of 2.00 when the mirror is 1.25 cm from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror? (c) Draw a principal-ray diagram to check your answer in part (b).
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. (a) What is the magnification of a person’s face when it is 12.0 cm to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice (n = 1.309) What is its apparent depth when viewed at normal incidence?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 m above the swimmer. What is the actual height of the diving board above the surface of the water?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. The person looks at a penny that is on the bottom of the pool directly below her. The penny appears to the person to be a distance of 8.00 m from her. What is the depth of the water at this point?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) 12.0 cm; (c) 2.00 cm.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The glass rod of Exercise 34.20 is immersed in oil (n = 1.45). An object placed to the left of the rod on the rod’s axis is to be imaged 1.20 m inside the rod. How far from the left end of the rod must the object be located to form the image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Repeat Exercise 34.22 for the case in which the end of the rod is ground to a concave hemispherical surface with radius 4.00 cm.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The glass rod of Exercise 34.23 is immersed in a liquid. An object 14.0 cm from the vertex of the left end of the rod and on its axis is imaged at a point 9.00 cm from the vertex inside the liquid. What is the index of refraction of the liquid?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A converging meniscus lens (see Fig. 34.32a) with a refractive index of 1.52 has spherical surfaces whose radii are 7.00 cm and 4.00 cm. What is the position of the image if an object is placed 24.0 cm to the left of the lens? What is the magnification?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A converging lens with a focal length of 90.0 cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A converging lens forms an image of an 8.00-mm-tall real object. The image is 12.0 cm to the left of the lens, 3.40 cm tall, and erect. What is the focal length of the lens? Where is the object located?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50 cm. Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 cm from the lens. What is the index of refraction of the lens?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The Lens of the Eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 mm, which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 cm tall is placed 30.0 cm from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The Cornea As a Simple Lens. The cornea behaves as a thin lens of focal length approximately 1.8 cm, although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.0 mm. (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b) The closest distance at which a typical person can focus on an object (called the near point) is about 25 cm, although this varies considerably with age. Where would the cornea focus the image of an 8.0-mm-tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A converging lens with a focal length of 7.00 cm forms an image of a 4.00-mm-tall real object that is to the left of the lens. The image is 1.30 cm tall and erect. Where are the object and image located? Is the image real or virtual?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
For each thin lens shown in Fig. E34.35, calculate the location of the image of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Repeat Exercise 34.36 for the case in which the lens is diverging, with a focal length of
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Combination of Lenses I. A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm. (b) I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Combination of Lenses II. Repeat Problem 34.39 using the same lenses except for the following changes: (a) The second lens is a diverging lens having a focal length of magnitude 60.0 cm. (b) The first lens is a diverging lens having a focal length of magnitude 40.0 cm. (c) Both lenses are diverging lenses having focal lengths of the same magnitudes as in Problem 34.39.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Combination of Lenses III. Two thin lenses with a focal length of magnitude 12.0 cm, the first diverging and the second converging, are located 9.00 cm apart. An object 2.50 mm tall is placed 20.0 cm to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two problems.)
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. (a) If the slide is placed 15.0 cm from the lens, what focal length lens is required? (b) If the dimensions of the picture on a 35-mm color slide are \(24\mathrm{\ mm}\times36\mathrm{\ mm}\), what is the minimum size of the projector screen required to accommodate the image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A camera lens has a focal length of 200 mm. How far from the lens should the subject for the photo be if the lens is 20.4 cm from the film?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
When a camera is focused, the lens is moved away from or toward the film. If you take a picture of your friend, who is standing 3.90 m from the lens, using a camera with a lens with a 85-mm focal length, how far from the film is the lens? Will the whole image of your friend, who is 175 cm tall, fit on film that is
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Figure 34.41 shows photographs of the same scene taken with the same camera with lenses of different focal length. If the object is 200 m from the lens, what is the magnitude of the lateral magnification for a lens of focal length (a) 28 mm; (b) 105 mm; (c) 300 mm?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A photographer takes a photograph of a Boeing 747 airliner (length 70.7 m) when it is flying directly overhead at an altitude of 9.50 km. The lens has a focal length of 5.00 m. How long is the image of the airliner on the film?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Choosing a Camera Lens. The picture size on ordinary 35-mm camera film is Focal lengths of lenses available for 35-mm cameras typically include 28, 35, 50 (the normal lens), 85, 100, 135, 200, and 300 mm, among others. Which of these lenses should be used to photograph the following objects, assuming that the object is to fill most of the picture area? (a) a building 240 m tall and 160 m wide at a distance of 600 m, and (b) a mobile home 9.6 m in length at a distance of 40.0 m.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Zoom Lens. Consider the simple model of the zoom lens shown in Fig. 34.43a. The converging lens has focal length and the diverging lens has focal length The lenses are separated by 4 cm as shown in Fig. 34.43a. (a) For a distant object, where is the image of the converging lens? (b) The image of the converging lens serves as the object for the diverging lens. What is the object distance for the diverging lens? (c) Where is the final image? Compare your answer to Fig. 34.43a. (d) Repeat parts (a), (b), and (c) for the situation shown in Fig. 34.43b, in which the lenses are separated by 8 cm.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A camera lens has a focal length of 180.0 mm and an aperture diameter of 16.36 mm. (a) What is the \(f \text {-number }\) of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{30} \mathrm{~s} \text { at } f / 11\), what is the correct exposure at \(f / 2.8\)? Text Transcription: f-number 1/30 s at f/11 f/2.8
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Recall that the intensity of light reaching film in a camera is proportional to the effective area of the lens. Camera A has a lens with an aperture diameter of 8.00 mm. It photographs an object using the correct exposure time of What exposure time should be used with camera B in photographing the same object with the same film if this camera has a lens with an aperture diameter of 23.1 mm?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Photography. A 35-mm camera has a standard lens with focal length 50 mm and can focus on objects between 45 cm and infinity. (a) Is the lens for such a camera a concave or a convex lens? (b) The camera is focused by rotating the lens, which moves it on the camera body and changes its distance from the film. In what range of distances between the lens and the film plane must the lens move to focus properly over the 45 cm to infinity range?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Curvature of the Cornea. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the corneas vertex is focused on the retina?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
(a) Where is the near point of an eye for which a contact lens with a power of diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of diopters is prescribed for distant vision?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Ordinary Glasses. Ordinary glasses are worn in front of the eye and usually 2.0 cm in front of the eyeball. Suppose that the person in Problem 34.54 prefers ordinary glasses to contact lenses. What focal length lenses are needed to correct his vision, and what is their power in diopters?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
If the person in Problem 34.56 chooses ordinary glasses over contact lenses, what power lens (in diopters) does she need to correct her vision if the lenses are 2.0 cm in front of the eye?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A thin lens with a focal length of 6.00 cm is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observers near point, 25.0 cm in front of her eye? (b) If the object is 1.00 mm high, what is the height of its image formed by the magnifier?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
You want to view an insect 2.00 mm in length through a magnifier. If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 0.025 radian?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A certain microscope is provided with objectives that have focal lengths of 16 mm, 4 mm, and 1.9 mm and with eyepieces that have angular magnifications of and Each objective forms an image 120 mm beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 mm is 160 mm from its second focal point. The eyepiece has a focal length of 26.0 mm. (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 mm apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The eyepiece of a refracting telescope (see Fig. 34.53) has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.80 m, and the final image is at infinity. What is the angular magnification of the telescope?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A telescope is constructed from two lenses with focal lengths of 95.0 cm and 15.0 cm, the 95.0-cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 m tall, 3.00 km away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Saturn is viewed through the Lick Observatory refracting telescope (objective focal length 18 m). If the diameter of the image of Saturn produced by the objective is 1.7 mm, what angle does Saturn subtend from when viewed from earth?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A reflecting telescope (Fig. E34.67) is to be made by using a spherical mirror with a radius of curvature of 1.30 m and an eyepiece with a focal length of 1.10 cm. The final image is at infinity. (a) What should the distance between the eyepiece and the mirror vertex be if the object is taken to be at infinity? (b) What will the angular magnification be?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Where must you place an object in front of a concave mirror with radius R so that the image is erect and \(2 \frac{1}{2}\) times the size of the object? Where is the image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
If you run away from a plane mirror at at what speed does your image move away from you?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An object is placed between two plane mirrors arranged at right angles to each other at a distance from the surface of one mirror and a distance from the other. (a) How many images are formed? Show the location of the images in a diagram. (b) Draw the paths of rays from the object to the eye of an observer
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
What is the size of the smallest vertical plane mirror in which a woman of height h can see her full-length image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.00 m from the mirror. The filament is 6.00 mm tall, and the image is to be 24.0 cm tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Rear-View Mirror. A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 cm. (a) Another car is behind your car, 9.00 m from the mirror, and this car is viewed in the mirror by your passenger. If this car is 1.5 m tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Suppose the lamp filament shown in Example 34.1 (Section 34.2) is moved to a position 8.0 cm in front of the mirror. (a) Where is the image located now? Is it real or virtual? (b) What is the height of the image? Is it erect or inverted? (c) In Example 34.1, the filament is 10.0 cm in front of the mirror, and an image of the filament is formed on a wall 3.00 m from the mirror. If the filament is 8.0 cm from the mirror, can a wall be placed so that an image is formed on it? If so, where should the wall be placed? If not, why not?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A layer of benzene 4.20 cm deep floats on water that is 6.50 cm deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
CP CALC You are in your car driving on a highway at when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of when the truck is 2.0 m from the mirror, what is the speed of the truck relative to the highway?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Figure P34.78 shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is 2.0 cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram for the following: (a) Using only the ray shown, decide what type of lens (converging or diverging) this is. (b) What is the focal length of the lens? (c) Locate the image by drawing the other two principal rays. (d) Calculate where the image should be, and compare this result with the graphical solution in part (c).
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Pinhole Camera. A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image without a lens. (a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (Hint: Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is 25 cm square and 20.0 cm deep, with the hole in the middle of one of the faces. If this camera is used to photograph a fierce chicken that is 18 cm high and 1.5 m in front of the camera, how large is the image of this bird on the film? What is the magnification of this camera?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A microscope is focused on the upper surface of a glass plate. A second plate is then placed over the first. To focus on the bottom surface of the second plate, the microscope must be raised 0.780 mm. To focus on the upper surface, it must be raised another 2.50 mm. Find the index of refraction of the second plate.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is 6.00 cm, and the radius of curvature at the right end is 12.0 cm. The length of the rod between vertices is 40.0 cm. The object for the surface at the left end is an arrow that lies 23.0 cm to the left of the vertex of this surface. The arrow is 1.50 mm tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The rod in Problem 34.82 is shortened to a distance of 25.0 cm between its vertices; the curvatures of its ends remain the same. As in Problem 34.82, the object for the surface at the left end is an arrow that lies 23.0 cm to the left of the vertex of this surface. The arrow is 1.50 mm tall and at right angles to the axis. (a) What is the object distance for the surface at the right end of the rod? (b) Is the object for this surface real or virtual? (c) What is the position of the final image? (d) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (e) What is the height of the final image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Figure P34.84 shows an object and its image formed by a thin lens. (a) What is the focal length of the lens, and what type of lens (converging or diverging) is it? (b) What is the height of the image? Is it real or virtual?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Figure P34.85 shows an object and its image formed by a thin lens. (a) What is the focal length of the lens, and what type of lens (converging or diverging) is it? (b) What is the height of the image? Is it real or virtual?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A transparent rod 30.0 cm long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 cm at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 cm from the flat end and 15.0 cm from the vertex of the curved end. When viewed from the flat end of the rod, the apparent depth of the object is 9.50 cm from the flat end. What is its apparent depth when viewed from the curved end?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Focus of the Eye. The cornea of the eye has a radius of curvature of approximately 0.50 cm, and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 mm. (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were 25 cm in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about 5.0 mm, where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A transparent rod 50.0 cm long and with a refractive index of 1.60 is cut flat at the right end and rounded to a hemispherical surface with a 15.0-cm radius at the left end. An object is placed on the axis of the rod 12.0 cm to the left of the vertex of the hemispherical end. (a) What is the position of the final image? (b) What is its magnification?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A glass rod with a refractive index of 1.55 is ground and polished at both ends to hemispherical surfaces with radii of 6.00 cm. When an object is placed on the axis of the rod, 25.0 cm to the left of the left-hand end, the final image is formed 65.0 cm to the right of the right-hand end. What is the length of the rod measured between the vertices of the two hemispherical surfaces?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{~cm} \text { and } R_{2}=+28.0 \mathrm{~cm}\). The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An object is placed 18.0 cm from a screen. (a) At what two points between object and screen may a converging lens with a 3.00-cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A convex mirror and a concave mirror are placed on the same optic axis, separated by a distance L = 0.600 m. The radius of curvature of each mirror has a magnitude of 0.360 m. A light source is located a distance x from the concave mirror, as shown in Fig. P34.93. (a) What distance x will result in the rays from the source returning to the source after reflecting first from the convex mirror and then from the concave mirror? (b) Repeat part (a), but now let the rays reflect first from the concave mirror and then from the convex one.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
As shown in Fig. P34.94 the candle is at the center of curvature of the concave mirror, whose focal length is 10.0 cm. The converging lens has a focal length of 32.0 cm and is 85.0 cm to the right of the candle. The candle is viewed looking through the lens from the right. The lens forms two images of the candle. The first is formed by light passing directly through the lens. The second image is formed from the light that goes from the candle to the mirror, is reflected, and then passes through the lens. (a) For each of these two images, draw a principal-ray diagram that locates the image. (b) For each image, answer the following questions: (i) Where is the image? (ii) Is the image real or virtual? (iii) Is the image erect or inverted with respect to the original object?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of 1.55. When a small leaf is placed 20.0 cm in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 cm from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33) but nothing else were changed?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Two Lenses in Contact. (a) Prove that when two thin lenses with focal lengths \(f_1\) and \(f_2\) are placed in contact, the focal length f of the combination is given by the relationship \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) (b) A converging meniscus lens (see Fig. 34.32a) has an index of refraction of 1.55 and radii of curvature for its surfaces of magnitudes 4.50 cm and 9.00 cm. The concave surface is placed upward and filled with carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) which has n = 1.46. What is the focal length of the \(\mathrm{CCl}_{4}\)-glass combination?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Rays from a lens are converging toward a point image located to the right of the lens. What thickness of glass with index of refraction 1.60 must be interposed between the lens and for the image to be formed at located 0.30 cm to the right of The locations of the piece of glass and of points and are shown in Fig. P34.97
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A Lens in a Liquid. A lens obeys Snells law, bending light rays at each surface an amount determined by the index of refraction of the lens and the index of the medium in which the lens is located. (a) Equation (34.19) assumes that the lens is surrounded by air. Consider instead a thin lens immersed in a liquid with refractive index Prove that the focal length is then given by Eq. (34.19) with replaced by (b) A thin lens with index has focal length in vacuum. Use the result of part (a) to show that when this lens is immersed in a liquid of index it will have a new focal length given by
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A glass plate 3.50 cm thick, with an index of refraction of 1.55 and plane parallel faces, is held with its faces horizontal and its lower face 6.00 cm above a printed page. Find the position of the image of the page formed by rays making a small angle with the normal to the plate.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A symmetric, double-convex, thin lens made of glass with index of refraction 1.52 has a focal length in air of 40.0 cm. The lens is sealed into an opening in the left-hand end of a tank filled with water. At the right-hand end of the tank, opposite the lens, is a plane mirror 90.0 cm from the lens. The index of refraction of the water is (a) Find the position of the image formed by the lenswatermirror system of a small object outside the tank on the lens axis and 70.0 cm to the left of the lens. (b) Is the image real or virtual? (c) Is it erect or inverted? (d) If the object has a height of 4.00 mm, what is the height of the image?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
You have a camera with a 35.0-mm-focal-length lens and 36.0-mm-wide film. You wish to take a picture of a 12.0-m-long sailboat but find that the image of the boat fills only \(\frac{1}{4}\) of the width of the film. (a) How far are you from the boat? (b) How much closer must the boat be to you for its image to fill the width of the film?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
What Is the Smallest Thing We Can See? The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, it's image cannot be any smaller than a sin- gle retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the cen- ter of the eye. We shall model the eye as a sphere 2.50 cm in diam- eter with a single thin lens at the front and the retina at the rear, with light receptor cells \(5.0\ \mu\mathrm{m}\) in diameter. (a) What is the small- est object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm {\ \min}\right)\) and compare it with the typical experi- mental value of about 1.0 min. (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.) Text Transcription: (mu m) 5.0 mu m (1^circ = 60 min)
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A camera with a 90-mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 m from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The derivation of the expression for angular magnification, Eq. (34.22), assumed a near point of 25 cm. In fact, the near point changes with age as shown in Table 34.1. In order to achieve an angular magnification of \(2.0 \times\), what focal length should be used by a person of (a) age 10; (b) age 30; (c) age 60? (d) If the lens that gives M = 2.0 for a 10-year-old is used by a 60-year-old, what angular magnification will the older viewer obtain? (e) Does your answer in part (d) mean that older viewers are able to see more highly magnified images than younger viewers? Explain.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Angular Magnification. In deriving Eq. (34.22) for the angular magnification of a magnifier, we assumed that the object is placed at the focal point of the magnifier so that the virtual image is formed at infinity. Suppose instead that the object is placed so that the virtual image appears at an average viewer’s near point of 25 cm, the closest point at which the viewer can bring an object into focus. (a) Where should the object be placed to achieve this? Give your answer in terms of the magnifier focal length (b) What angle \(\theta^{\prime}\) will an object of height subtend at the position found in part (a)? (c) Find the angular magnification with the object at the position found in part (a). The angle \(\theta^{\prime}\) is the same as in Fig. 34.51a, since it refers to viewing the object without the magnifier. (d) For a convex lens with \(f=+10.0 \mathrm{~cm}\), what is the value of with the object at the position found in part (a)? How many times greater is in this case than in the case where the image is formed at infinity? (e) In the description of a compound microscope in Section 34.8, it is stated that in a properly designed instrument, the real image formed by the objective lies just inside the first focal point \(F_{1}^{\prime}\) of the eyepiece. What advantages are gained by having the image formed by the objective be just inside \(F_{1}^{\prime}\) as opposed to precisely at \(F_{1}^{\prime}\)? What happens if the image formed by the objective is just outside \(F_{1}^{\prime}\)?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
BIO In one form of cataract surgery the persons natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the persons eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of 24 cm?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye described in Exercise 34.52. If the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described in Exercise 34.52, what is the distance from the cornea vertex to the retina? What does this tell you about the shape of the nearsighted eye?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
The Galilean Telescope. Figure P34.112 is a diagram of a Galilean telescope, or opera glass, with both the object and its final image at infinity. The image serves as a virtual object for the eyepiece. The final image is virtual and erect. (a) Prove that the angular magnification is (b) A Galilean telescope is to be constructed with the same objective lens as in Exercise 34.65. What focal length should the eyepiece have if this telescope is to have the same magnitude of angular magnification as the one in Exercise 34.65? (c) Compare the lengths of the telescopes.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_1\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right|\). The two lenses are separated by a variable distance d that is always less than \(f_1\). Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right)\). To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_0\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_0\) at some point Q. The distance from the final image \(I^{\prime}\) to the point Q is the effective focal length f of the lens combination; if the combination were replaced by a single lens of focal length f placed at Q, parallel rays would still be brought to a focus at \(I^{\prime}\). Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right)\). (d) If \(f_{1}=12.0 \mathrm{\ cm}\), \(f_{2}=-18.0 \mathrm{\ cm}\) and the separation d is adjustable between 0 and 4.0 cm, find the maximum and minimum focal lengths of the combination. What value of d gives f = 30.0 cm?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A certain reflecting telescope, constructed as shown in Fig. E34.67, has a spherical mirror with a radius of curvature of 96.0 cm and an eyepiece with a focal length of 1.20 cm. If the angular magnification has a magnitude of 36 and the object is at infinity, find the position of the eyepiece and the position and nature (real or virtual) of the final image. (Note: |M| is not equal to \(\left|f_{1}\right| /\left|f_{2}\right|\), so the image formed by the eyepiece is not at infinity.)
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
A microscope with an objective of focal length 8.00 mm and an eyepiece of focal length 7.50 cm is used to project an image on a screen 2.00 m from the eyepiece. Let the image distance of the objective be 18.0 cm. (a) What is the lateral magnification of the image? (b) What is the distance between the objective and the eyepiece?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Spherical aberration is a blurring of the image formed by a spherical mirror. It occurs because parallel rays striking the mirror far from the optic axis are focused at a different point than are rays near the axis. This problem is usually minimized by using only the center of a spherical mirror. (a) Show that for a spherical concave mirror, the focus moves toward the mirror as the parallel rays move toward the outer edge of the mirror. (Hint: Derive an analytic expression for the distance from the vertex to the focus of the ray for a particular parallel ray. This expression should be in terms of (i) the radius of curvature R of the mirror and (ii) the angle \(\theta\) between the incident ray and a line connecting the center of curvature of the mirror with the point where the ray strikes the mirror.) (b) What value of \(\theta\) produces a 2% change in the location of the focus, compared to the location for \(\theta\) very close to zero?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
CALC (a) For a lens with focal length find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
An Object at an Angle. A 16.0-cm-long pencil is placed at a \(45.0^{\circ}\) angle, with its center 15.0 cm above the optic axis and 45.0 cm from a lens with a 20.0-cm focal length as shown in Fig. P34.118. (Note that the figure is not drawn to scale.) Assume that the diameter of the lens is large enough for the paraxial approximation to be valid. (a) Where is the image of the pencil? (Give the location of the images of the points A, B, and C on the object, which are located at the eraser, point, and center of the pencil, respectively.) (b) What is the length of the image (that is, the distance between the images of points A and B)? (c) Show the orientation of the image in a sketch.
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Chapter 34: Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
People with normal vision cannot focus their eyes underwater if they aren’t wearing a face mask or goggles and there is water in contact with their eyes (see Discussion Question Q34.23). (a) Why not? (b) With the simplified model of the eye described in Exercise 34.52, what corrective lens (specified by focal length as measured in air) would be needed to enable a person underwater to focus an infinitely distant object? (Be careful—the focal length of a lens underwater is not the same as in air! See Problem 34.98. Assume that the corrective lens has a refractive index of 1.62 and that the lens is used in eyeglasses, not goggles, so there is water on both sides of the lens. Assume that the eyeglasses are 2.00 cm in front of the eye.)
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Chapter : Problem 80 Sears and Zemansky's University Physics with Modern Physics 13
Problem 80P A microscope is focused on the upper surface of a glass plate. A second plate is then placed over the first. To locus on the bottom surface of the second plate, the microscope must be raised 0.780 mm. To focus on the upper surface, it must be raised another 2.50 mm. Find the index of refraction of the second plate.
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Chapter : Problem 81 Sears and Zemansky's University Physics with Modern Physics 13
Problem 81P What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence?
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Chapter : Problem 117 Sears and Zemansky's University Physics with Modern Physics 13
CALC (a) For a lens with focal length f, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?
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Chapter : Problem 1 Sears and Zemansky's University Physics with Modern Physics 13
Problem 1DQ A spherical mirror is cut in half horizontally. Will an image be formed by the bottom half of the mirror? If so, where will the image be formed?
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Chapter : Problem 1 Sears and Zemansky's University Physics with Modern Physics 13
Problem 1E A candle 4.85 cm tall is 39.2 cm to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?
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Chapter : Problem 2 Sears and Zemansky's University Physics with Modern Physics 13
For the situation shown in Fig. 34.3, is the image distance \(s^{i}\) positive or negative? Is the image real or virtual? Explain your answers. Equation transcription: Text transcription: s^{i}
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Chapter : Problem 2 Sears and Zemansky's University Physics with Modern Physics 13
The image of a tree just covers the length of a plane mirror 4.00 cm tall when the mirror is held 35.0 cm from the eye. The tree is 28.0 m from the mirror. What is its height?
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Chapter : Problem 3 Sears and Zemansky's University Physics with Modern Physics 13
Problem 3DQ The laws of optics also apply to electromagnetic waves invisible to the eye. A satellite TV dish is used to detect radio waves coming from orbiting satellites. Why is a curved reflecting surface (a “dish”) used? The dish is always concave, never convex; why? The actual radio receiver is placed on an arm and suspended in front of the dish. How far in front of the dish should it be placed?
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Chapter : Problem 3 Sears and Zemansky's University Physics with Modern Physics 13
Problem 3E A pencil that is 9.0 cm long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead 12.0 cm from the mirror surface and the end of the eraser 21.0 cm from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?
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Chapter : Problem 4 Sears and Zemansky's University Physics with Modern Physics 13
Problem 4DQ Explain why the focal length of a plane mirror is infinite, and explain what it means for the focal point to be at infinity.
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Chapter : Problem 4 Sears and Zemansky's University Physics with Modern Physics 13
Problem 4E A concave mirror has a radius of curvature of 34.0 cm. (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?
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Chapter : Problem 5 Sears and Zemansky's University Physics with Modern Physics 13
Problem 5DQ If a spherical mirror is immersed in water, does its focal length change? Explain.
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Chapter : Problem 5 Sears and Zemansky's University Physics with Modern Physics 13
Problem 5E An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.
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Chapter : Problem 6 Sears and Zemansky's University Physics with Modern Physics 13
Problem 6DQ For what range of object positions does a concave spherical mirror form a real image? What about a convex spherical mirror?
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Chapter : Problem 6 Sears and Zemansky's University Physics with Modern Physics 13
Problem 6E Repeat Exercise 34.5 for the case in which the mirror is convex. 34.5 . An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.
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Chapter : Problem 7 Sears and Zemansky's University Physics with Modern Physics 13
Problem 7DQ When a room has mirrors on two opposite walls, an infinite series of reflections can be seen. Discuss this phenomenon in terms of images. Why do the distant images appear fainter?
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Chapter : Problem 7 Sears and Zemansky's University Physics with Modern Physics 13
Problem 7E The diameter of Mars is 6794 km, and its minimum distance from the earth is 5.58 × 107 km. When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of 1.75 m.
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Chapter : Problem 8 Sears and Zemansky's University Physics with Modern Physics 13
For a spherical mirror, if \(s=f\), then \(s^{i}=\infty\), and the lateral magnification is infinite. Does this make sense? If 80, what does it mean? Equation transcription: Texts=f transcription: s=f s^{i}=\infty
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Chapter : Problem 8 Sears and Zemansky's University Physics with Modern Physics 13
Problem 8E An object is 24.0 cm from the center of a silvered spherical glass Christmas tree ornament 6.00 cm in diameter. What are the position and magnification of its image?
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Chapter : Problem 9 Sears and Zemansky's University Physics with Modern Physics 13
Problem 9DQ You may have noticed a small convex mirror next to your bank’s ATM. Why is this mirror convex, as opposed to flat or concave? What considerations determine its radius of curvature?
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Chapter : Problem 9 Sears and Zemansky's University Physics with Modern Physics 13
Problem 9E A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 cm. Reflection from the surface of the shell forms an image of the 1.5-cm-tall coin that is 6.00 cm behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.
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Chapter : Problem 10 Sears and Zemansky's University Physics with Modern Physics 13
Problem 10DQ A student claims that she can start a fire on a sunny day using just the sun’s rays and a concave mirror. How is this done? Is the concept of image relevant? Can she do the same thing with a convex mirror? Explain.
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Chapter : Problem 10 Sears and Zemansky's University Physics with Modern Physics 13
Problem 10E You hold a spherical salad Bowl 90 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a 35-cm radius of curvature. (a) What is the image of your 2.0-cm-tall nose located? (b) What are the image’s size, orientation, and nature (real or virtual)?
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Chapter : Problem 11 Sears and Zemansky's University Physics with Modern Physics 13
Problem 11DQ A person looks at his reflection in the concave side of a shiny spoon. Is it right side up or inverted? Does it matter how far his face is from the spoon? What if he looks in the convex side? (Try this yourself!)
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Chapter : Problem 11 Sears and Zemansky's University Physics with Modern Physics 13
Show that Eq. (34.6) can be written as \(s^{\prime}=s f /(s-f)\) and hence the lateral magnification given by Eq. (34.7) can be expressed as m = f / (f - 5). (b) A concave spherical mirror has focal length f = +14.0 cm. What is the nonzero distance of the object from the mirror vertex if the image has the same height as the object? In this case, is the image erect or inverted? (c) A convex spherical mirror has f = -8.00 cm. What is the nonzero distance of the object from the mirror vertex if the height of the image is one-half the height of the object?
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Chapter : Problem 12 Sears and Zemansky's University Physics with Modern Physics 13
In Example (Section ), there appears to be an ambiguity for the case \(s=10 \mathrm{~cm}\) as to whether \(s^{i}\) is \(+\infty 0 r-\infty\) and whether the image is erect or inverted. How is this resolved? Or is it? Equation transcription: Text transcription: s=10 \mathrm{~cm} s^{i} +\infty 0 r-\infty
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Chapter : Problem 12 Sears and Zemansky's University Physics with Modern Physics 13
The thin glass shell Figure E34.12 shown in has a spherical shape with a radius of curvature of , and both of its surfaces can act as mirrors. A seed high is placed from the center of the mirror along the optic axis, as shown in the figure. a) Calculate the location and height of the image of this seed. (b) Suppose now that the shell is reversed. Find the location and height of the seed's image.
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Chapter : Problem 13 Sears and Zemansky's University Physics with Modern Physics 13
Suppose that in the situation of Example of Section (scc Fig. ) a vertical arrow tall is painted on the side of the pool beneath the water line. According to the calculations in the example, this arrow would appear to the person shown in Fig. to be long. But the discussion following Eq. ( ) states that the magnification for a plane refracting surface is , which suggests that the arrow would appear to the person to be long. How can you resolve this apparent contradiction?
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Chapter : Problem 13 Sears and Zemansky's University Physics with Modern Physics 13
Problem 13E Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an erect image with a magnification of 2.00 when the mirror is 1.25 cm from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror? (c) Draw a principal-ray diagram to check your answer in part (b).
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Chapter : Problem 14 Sears and Zemansky's University Physics with Modern Physics 13
The bottom of the passenger-side mirror on your car notes, “Objects in the mirror are closer than they appear.” Is this true? Why?
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Chapter : Problem 14 Sears and Zemansky's University Physics with Modern Physics 13
Problem 14E A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. (a) What is the magnification of a person’s face when it is 12.0 cm to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.
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Chapter : Problem 15 Sears and Zemansky's University Physics with Modern Physics 13
Problem 15DQ How could you very quickly make an approximate measurement of the focal length of a converging lens? Could the same method be applied if you wished to use a diverging lens? Explain.
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Chapter : Problem 15 Sears and Zemansky's University Physics with Modern Physics 13
Problem 15E A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice (n = 1.309). What is its apparent depth when viewed at normal incidence?
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Chapter : Problem 16 Sears and Zemansky's University Physics with Modern Physics 13
Problem 16DQ The focal length of a simple lens depends on the color (wavelength) of light passing through it. Why? Is it possible for a lens to have a positive focal length for some colors and negative for others? Explain.
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Chapter : Problem 16 Sears and Zemansky's University Physics with Modern Physics 13
Problem 16E A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?
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Chapter : Problem 17 Sears and Zemansky's University Physics with Modern Physics 13
Problem 17DQ When a converging lens is immersed in water, does its focal length increase or decrease in comparison with the value in air? Explain.
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Chapter : Problem 17 Sears and Zemansky's University Physics with Modern Physics 13
Problem 17E A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 m above the swimmer. What is the actual height of the diving board above the surface of the water?
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Chapter : Problem 18 Sears and Zemansky's University Physics with Modern Physics 13
Problem 18DQ A spherical air bubble in water can function as a lens. Is it a converging or diverging lens? How is its focal length related to its radius?
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Chapter : Problem 18 Sears and Zemansky's University Physics with Modern Physics 13
A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. The person looks at a penny that is on the bottom of the pool directly below her. The penny appears to the person to be a distance of 8.00 m from her. What is the depth of the water at this point?
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Chapter : Problem 19 Sears and Zemansky's University Physics with Modern Physics 13
Problem 19DQ Can an image formed by one reflecting or refracting surface serve as an object for a second reflection or refraction? Does it matter whether the first image is real or virtual? Explain.
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Chapter : Problem 19 Sears and Zemansky's University Physics with Modern Physics 13
Problem 19E A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?
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Chapter : Problem 20 Sears and Zemansky's University Physics with Modern Physics 13
Problem 20DQ If a piece of photographic film is placed at the location of a real image, the film will record the image. Can this be done with a virtual image? How might one record a virtual image?
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Chapter : Problem 20 Sears and Zemansky's University Physics with Modern Physics 13
Problem 20E The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) 12.0 cm; (c) 2.00 cm.
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Chapter : Problem 21 Sears and Zemansky's University Physics with Modern Physics 13
Problem 21DQ According to the discussion in Section 34.2, light rays are reversible. Are the formulas in the table in this chapter’s Summary still valid if object and image are interchanged? What does reversibility imply with respect to the forms of the various formulas?
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Chapter : Problem 21 Sears and Zemansky's University Physics with Modern Physics 13
Problem 21E The glass rod of Exercise 34.22 is immersed in oil (n = 1.45). An object placed to the left of the rod on the rod’s axis is to be imaged 1.20 m inside the rod. How far from the left end of the rod must the object be located to form the image? The glass rod of Exercise 34.22 is immersed in oil (n = 1.45). An object placed to the left of the rod on the rod’s axis is to be imaged 1.20 m inside the rod. How far from the left end of the rod must the object be located to form the image? 34.22 . The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) 12.0 cm; (c) 2.00 cm.
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Chapter : Problem 22 Sears and Zemansky's University Physics with Modern Physics 13
Problem 22DQ You’ve entered a survival contest that will include building a crude telescope. You are given a large box of lenses. Which two lenses do you pick? How do you quickly identify them?
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Chapter : Problem 22 Sears and Zemansky's University Physics with Modern Physics 13
Problem 22E The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?
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Chapter : Problem 23 Sears and Zemansky's University Physics with Modern Physics 13
Problem 23DQ BIO You can’t see clearly underwater with the naked eye, but you can if you wear a face mask or goggles (with air between your eyes and the mask or goggles). Why is there a difference? Could you instead wear eyeglasses (with water between your eyes and the eyeglasses) in order to see underwater? If so, should the lenses be converging or diverging? Explain.
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Chapter : Problem 23 Sears and Zemansky's University Physics with Modern Physics 13
Problem 23E Repeat Exercise 34.24 for the case in which the end of the rod is ground to a concave hemispherical surface with radius 4.00 cm. 34.24 .. The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?
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Chapter : Problem 24 Sears and Zemansky's University Physics with Modern Physics 13
Problem 24DQ You take a lens and mask it so that light can pass through only the bottom half of the lens. How does the image formed by the masked lens compare to the image formed before masking?
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Chapter : Problem 24 Sears and Zemansky's University Physics with Modern Physics 13
Problem 24E The glass rod of Exercise 34.25 is immersed in a liquid. An object 14.0 cm from the vertex of the left end of the rod and on its axis is imaged at a point 9.00 cm from the vertex inside the liquid. What is the index of refraction of the liquid? 34.25 .. Repeat Exercise 34.24 for the case in which the end of the rod is ground to a concave hemispherical surface with radius 4.00 cm. 34.24 .. The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?
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Chapter : Problem 25 Sears and Zemansky's University Physics with Modern Physics 13
Problem 25E An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.
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Chapter : Problem 26 Sears and Zemansky's University Physics with Modern Physics 13
A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.
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Chapter : Problem 27 Sears and Zemansky's University Physics with Modern Physics 13
A converging meniscus lens (see Fig. 34.32a) with a refractive index of 1.52 has spherical surfaces whose radii are \(7.00 \mathrm{~cm}\) and \(4.00 \mathrm{~cm}\). What is the position of the image if an object is placed 24.0 cm to the left of the lens? What is magnification? Equation transcription: Text transcription: 7.00{~cm} 4.00{~cm}
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Chapter : Problem 28 Sears and Zemansky's University Physics with Modern Physics 13
A converging lens with a focal length of 90.0 cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?
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Chapter : Problem 29 Sears and Zemansky's University Physics with Modern Physics 13
Problem 29E A converging lens forms an image of an 8.00-mm-tall real object. The image is 12.0 cm to the left of the lens, 3.40 cm tall, and erect. What is the focal length of the lens? Where is the object located?
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Chapter : Problem 30 Sears and Zemansky's University Physics with Modern Physics 13
Problem 30E A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?
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Chapter : Problem 31 Sears and Zemansky's University Physics with Modern Physics 13
Problem 31E A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50 cm. Using this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 cm from the lens. What is the index of refraction of the lens?
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Chapter : Problem 32 Sears and Zemansky's University Physics with Modern Physics 13
Problem 32E BIO The Lens of the Eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 mm, which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 cm tall is placed 30.0 cm from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)
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Chapter : Problem 33 Sears and Zemansky's University Physics with Modern Physics 13
Problem 33E BIO The Cornea As a Simple Lens. The cornea behaves as a thin lens of focal length approximately 1.8 cm, although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.0 mm. (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b) The closest distance at which a typical person can focus on an object (called the near point) is about 25 cm, although this varies considerably with age. Where would the cornea focus the image of an 8.0-mm-tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)
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Chapter : Problem 34 Sears and Zemansky's University Physics with Modern Physics 13
Problem 34E A converging lens with a focal length of 7.00 cm forms an image of a 4.00-mm-tall real object that is to the left of the lens. The image is 1.30 cm tall and erect. Where are the object and image located? Is the image real or virtual?
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Chapter : Problem 35 Sears and Zemansky's University Physics with Modern Physics 13
For each thin lens shown in Fig. E34.35, calculate the location of the image of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes. \(R=10.0 \mathrm{~cm}\) \(R=15.0 \mathrm{~cm}\) \(\text { - Flat }\) Equation transcription: Text transcription: R=10.0{~cm} R=15.0{~cm} text { - Flat }
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Chapter : Problem 36 Sears and Zemansky's University Physics with Modern Physics 13
Problem 36E A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.
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Chapter : Problem 37 Sears and Zemansky's University Physics with Modern Physics 13
Problem 37E Repeat Exercise 34.38 for the case in which the lens is diverging, with a focal length of -48.0 cm. 34.38 . A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.
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Chapter : Problem 38 Sears and Zemansky's University Physics with Modern Physics 13
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.
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Chapter : Problem 39 Sears and Zemansky's University Physics with Modern Physics 13
Problem 39E Combination of Lenses I. A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm. (b) I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.
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Chapter : Problem 40 Sears and Zemansky's University Physics with Modern Physics 13
Combination of Lenses II. Repeat Problem 34.39 using the same lenses except for the following changes: (a) The second lens is a diverging lens having a focal length of magnitude 60.0 cm. (b) The first lens is a diverging lens having a focal length of magnitude 40.0 cm. (c) Both lenses are diverging lenses having focal lengths of the same magnitudes as in Problem 34.39.
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Chapter : Problem 41 Sears and Zemansky's University Physics with Modern Physics 13
Problem 41E Combination of Lenses III. Two thin lenses with a focal length of magnitude 12.0 cm, the first diverging and the second converging, are located 9.00 cm apart. An object 2.50 mm tall is placed 20.0 cm to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two problems.) 34.41 .. Combination of Lenses I. A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm. (b) I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses. 34.42 .. Combination of Lenses II. Repeat Exercise 34.41 using the same lenses except for the following changes: (a) The second lens is a diverging lens having a focal length of magnitude 60.0 cm. (b) The first lens is a diverging lens having a focal length of magnitude 40.0 cm. (c) Both lenses are diverging lenses having focal lengths of the same magnitudes as in Exercise 34.41.
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Chapter : Problem 42 Sears and Zemansky's University Physics with Modern Physics 13
Problem 42E You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. (a) If the slide is placed 15.0 cm from the lens, what focal length lens is required? (b) If the dimensions of the picture on a 35-mm color slide are 24 mm × 36 mm, what is the minimum size of the projector screen required to accommodate the image?
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Chapter : Problem 43 Sears and Zemansky's University Physics with Modern Physics 13
Problem 43E A camera lens has a focal length of 200 mm. How far from the lens should the subject for the photo be if the lens is 20.4 cm from the film?
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Chapter : Problem 44 Sears and Zemansky's University Physics with Modern Physics 13
Problem 44E When a camera is focused, the lens is moved away from or toward the digital image sensor. If you take a picture of your friend, who is standing 3.90 m from the lens, using a camera with a lens with an 85-mm focal length, how far from the sensor is the lens? Will the whole image of your friend, who is 175 cm tall, fit on a sensor that is 24 mm × 36 mm?
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Chapter : Problem 45 Sears and Zemansky's University Physics with Modern Physics 13
Figure 34.41 shows photographs of the same scene taken with the same camera with lenses of different focal length. If the object is 200 m from the lens, what is the magnitude of the lateral magnification for a lens of focal length (a) 28 mm; (b) 105 mm; (c) 300 mm?
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Chapter : Problem 46 Sears and Zemansky's University Physics with Modern Physics 13
Problem 46E A photographer takes a photograph of a Boeing 747 airliner (length 70.7 m) when it is flying directly overhead at an altitude of 9.50 km. The lens has a focal length of 5.00 m. How long is the image of the airliner on the film?
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Chapter : Problem 47 Sears and Zemansky's University Physics with Modern Physics 13
Problem 47E Choosing a Camera Lens. The picture size on ordinary 35-mm camera film is 24 mm × 36 mm. Focal lengths of lenses available for 35-mm cameras typically include 28, 35, 50 (the “normal” lens), 85, 100, 135, 200, and 300 mm, among others. Which of these lenses should be used to photograph the following objects, assuming that the object is to fill most of the picture area? (a) a building 240 m tall and 160 m wide at a distance of 600 m, and (b) a mobile home 9.6 m in length at a distance of 40.0 m.
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Chapter : Problem 48 Sears and Zemansky's University Physics with Modern Physics 13
Zoom Lens. Consider the simple model of the zoom lens shown in Fig. 34.43a. The converging lens has focal length \(f_{1}=12 \mathrm{~cm}\), and the diverging lens has focal length \(f_{2}=-12 \mathrm{~cm}\). The lenses are separated by as shown in Fig. 34.43a. (a) For a distant object, where is the image of the converging lens? (b) The image of the converging lens serves as the object for the diverging lens. What is the object distance for the diverging lens? (c) Where is the final image? Compare your answer to Fig. 34.43a. (d) Repeat parts (a), (b), and (c) for the situation shown in Fig. 34.43b, in which the lenses are separated by . Equation transcription: Text transcription: f_{1}=12{~cm} f_{2}=-12{~cm}
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Chapter : Problem 49 Sears and Zemansky's University Physics with Modern Physics 13
A camera lens has a focal length of and an aperture diameter of . (a) What is the -number of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{30} s\) at \(f / 11\), what is the correct exposure at \(f / 2.8 ?\) Equation transcription: Text transcription: frac{1}{30} s f / 11 f / 2.8 ?
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Chapter : Problem 50 Sears and Zemansky's University Physics with Modern Physics 13
Recall that the intensity of light reaching film in a camera is proportional to the effective area of the lens. Camera A has a lens with an aperture diameter of 8.00 mm. It photographs an object using the correct exposure time of \(\frac{1}{30}\) s. What exposure time should be used with camera B in photographing the same object with the same film if this camera has a lens with an aperture diameter of 23.1 mm? Text Transcription: 1/30
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Chapter : Problem 51 Sears and Zemansky's University Physics with Modern Physics 13
Photography. A 35-mm camera has a standard lens with local length 50 mm and can focus on objects between 45 cm and infinity. (a) Is the lens for such a camera a concave or a convex lens? (b) The camera is focused by rotating the lens, which moves it on the camera body and changes its distance from the film. In what range of distances between the lens and the film plane must the lens move to focus properly over the 45 cm to infinity range?
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Chapter : Problem 52 Sears and Zemansky's University Physics with Modern Physics 13
Problem 52E BIO Curvature of the Cornea. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the bending occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea’s vertex is focused on the retina?
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Chapter : Problem 53 Sears and Zemansky's University Physics with Modern Physics 13
(a) Where is the near point of an eye for which a contact lens with a power of +2.75 diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of -1.30 diopters is prescribed for distant vision?
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Chapter : Problem 54 Sears and Zemansky's University Physics with Modern Physics 13
Problem 54E BIO Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?
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Chapter : Problem 55 Sears and Zemansky's University Physics with Modern Physics 13
Problem 55E BIO Ordinary Glasses. Ordinary glasses are worn in front of the eye and usually 2.0 cm in front of the eyeball. Suppose that the person in Exercise 34.52 prefers ordinary glasses to contact lenses. What focal length lenses are needed to correct his vision, and what is their power in diopters? 34.52 . BIO Contact Lenses. Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?
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Chapter : Problem 56 Sears and Zemansky's University Physics with Modern Physics 13
BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?
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Chapter : Problem 57 Sears and Zemansky's University Physics with Modern Physics 13
Problem 57E BIO If the person in Exercise 34.54 chooses ordinary glasses over contact lenses, what power lens (in diopters) does she need to correct her vision if the lenses are 2.0 cm in front of the eye? 34.54 . BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?
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Chapter : Problem 58 Sears and Zemansky's University Physics with Modern Physics 13
Problem 58E A thin lens with a focal length of 6.00 cm is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.
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Chapter : Problem 59 Sears and Zemansky's University Physics with Modern Physics 13
Problem 59E The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer’s near point, 25.0 cm in front of her eye? (b) If the object is 1.00 mm high, what is the height of its image formed by the magnifier?
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Chapter : Problem 60 Sears and Zemansky's University Physics with Modern Physics 13
Problem 60E You want to view an insect 2.00 mm in length through a magnifier. If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 0.025 radian?
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Chapter : Problem 61 Sears and Zemansky's University Physics with Modern Physics 13
Problem 61E A certain microscope is provided with objectives that have focal lengths of 16 mm, 4 mm, and 1.9 mm and with eyepieces that have angular magnifications of 5× and 10×. Each objective forms an image 120 mm beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.
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Chapter : Problem 62 Sears and Zemansky's University Physics with Modern Physics 13
Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 mm is 160 mm from its second focal point. The eyepiece has a focal length of 26.0 mm. (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 mm apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?
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Chapter : Problem 63 Sears and Zemansky's University Physics with Modern Physics 13
Problem 63E The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?
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Chapter : Problem 64 Sears and Zemansky's University Physics with Modern Physics 13
The eyepiece of a refracting telescope (see Fig. 34.53) has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.80 m, and the final image is at infinity. What is the angular magnification of the telescope?
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Chapter : Problem 65 Sears and Zemansky's University Physics with Modern Physics 13
Problem 65E A telescope is constructed from two lenses with focal lengths of 95.0 cm and 15.0 cm, the 95.0-cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 m tall, 3.00 km away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?
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Chapter : Problem 66 Sears and Zemansky's University Physics with Modern Physics 13
Problem 66E Saturn is viewed through the Lick Observatory refracting telescope (objective focal length 18 m). If the diameter of the image of Saturn produced by the objective is 1.7 mm. what angle does Saturn subtend from when viewed from earth?
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Chapter : Problem 67 Sears and Zemansky's University Physics with Modern Physics 13
A reflecting telescope (Fig. E34.67) is to be made by using a spherical mirror with a radius of curvature of 1.30 m and an eyepiece with a focal length of 1.10 cm. The final image is at infinity. (a) What should the distance between the eyepiece and the mirror vertex be if the object is taken to be at infinity? (b) What will the angular magnification be?
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Chapter : Problem 68 Sears and Zemansky's University Physics with Modern Physics 13
Where must you place an object in front of a concave mirror with radius so that the image is erect and \(2 \frac{1}{2}\) times the size of the object? Where is the image? Equation transcription: Text transcription: 2 frac{1}{2}
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Chapter : Problem 69 Sears and Zemansky's University Physics with Modern Physics 13
Problem 69P If you run away from a plane mirror at 3.60 m/s, at what speed does your image move away from you?
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Chapter : Problem 70 Sears and Zemansky's University Physics with Modern Physics 13
Problem 70P An object is placed between two plane mirrors arranged at right angles to each other at a distance d1 from the surface of one mirror and a distance d2 from the other. (a) How many images are formed? Show the location of the images in a diagram. (b) Draw the paths of rays from the object to the eye of an observer.
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Chapter : Problem 71 Sears and Zemansky's University Physics with Modern Physics 13
Problem 71P What is the size of the smallest vertical plane mirror in which a woman of height h can see her full-length image?
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Chapter : Problem 72 Sears and Zemansky's University Physics with Modern Physics 13
A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?
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Chapter : Problem 73 Sears and Zemansky's University Physics with Modern Physics 13
Problem 73P A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.00 m from the mirror. The filament is 6.00 mm tall, and the image is to be 24.0 cm tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?
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Chapter : Problem 74 Sears and Zemansky's University Physics with Modern Physics 13
Problem 74P Rear-View Mirror. A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 cm. (a) Another car is behind your car, 9.00 m from the mirror, and this car is viewed in the mirror by your passenger. If this car is 1.5 m tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?
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Chapter : Problem 75 Sears and Zemansky's University Physics with Modern Physics 13
Suppose the lamp filament shown in Example 34.1 (Section 34.2) is moved to a position 8.0 cm in front of the mirror. (a) Where is the image located now? Is it real or virtual? (b) What is the height of the image? Is it erect or inverted? (c) In Example 34.1, the filament is 10.0 cm in front of the mirror, and an image of the filament is formed on a wall 3.00 m from the mirror. If the filament is 8.0 cm from the mirror, can a wall be placed so that an image is formed on it? If so, where should the wall be placed? If not, why not?
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Chapter : Problem 76 Sears and Zemansky's University Physics with Modern Physics 13
Problem 76P A layer of benzene (n = 1.50) 4.20 cm deep floats on water (n = 1.33) that is 6.50 cm deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence?
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Chapter : Problem 77 Sears and Zemansky's University Physics with Modern Physics 13
You are in your car driving on a highway at 25 m/s when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 m/s when the truck is 2.0 m from the mirror, what is the speed of the truck relative to the highway?
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Chapter : Problem 78 Sears and Zemansky's University Physics with Modern Physics 13
Figure P34.78 shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is 2.0 cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram for the following: (a) Using only the ray shown, decide what type of lens (converging or diverging) this is. (b) What is the focal length of the lens? (c) Locate the image by drawing the other two principal rays. (d) Calculate where the image should be, and compare this result with the graphical solution in part (c).
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Chapter : Problem 79 Sears and Zemansky's University Physics with Modern Physics 13
Problem 79P Pinhole Camera. A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image without a lens. (a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (Hint: Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is 25 cm square and 20.0 cm deep, with the hole in the middle of one of the 25 cm × 25 cm faces. If this camera is used to photograph a fierce chicken that is 18 cm high and 1.5 m in front of the camera, how large is the image of this bird on the film? What is the lateral magnification of this camera?
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Chapter : Problem 82 Sears and Zemansky's University Physics with Modern Physics 13
A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is 6.00 cm, and the radius of curvature at the right end is 12.0 cm. The length of the rod between vertices is 40.0 cm. The object for the surface at the left end is an arrow that lies 23.0 cm to the left of the vertex of this surface. The arrow is 1.50 mm tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?
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Chapter : Problem 83 Sears and Zemansky's University Physics with Modern Physics 13
Problem 83P The rod in Problem is shortened to a distance of is 25.0 cm between its vertices: the curvatures of its ends remain the same. As in Problem the object for the surface at the left end is an arrow that lies 23.0 cm to the left of the vertex of this surface. The arrow is 1.50 mm tall and at right angles to the axis. (a) What is the object distance for the surface at the right end of the rod? (b) Is the object for this surface real or virtual? (c) What is the position of the final image? (d) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (e) What is the height of the final image? Probelm: A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is 6.00 cm, and the radius of curvature at the right end is 12.0 cm. The length of the rod between vertices is 40.0 cm. The object for the surface at the left end is an arrow that lies 23.0 cm to the left of the vertex of this surface. The arrow is 1.50 cm tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?
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Chapter : Problem 84 Sears and Zemansky's University Physics with Modern Physics 13
Figure P34.84 shows an object and its image formed by a thin lens. (a) What is the focal length of the lens, and what type of lens (converging or diverging) is it? (b) What is the height of the image? Is it real or virtual?
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Chapter : Problem 85 Sears and Zemansky's University Physics with Modern Physics 13
Figure P34.85 shows an object and its image formed by a thin lens. (a) What is the focal length of the lens, and what type of lens (converging or diverging) is it? (b) What is the height of the image? Is it real or virtual?
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Chapter : Problem 86 Sears and Zemansky's University Physics with Modern Physics 13
Problem 86P A transparent rod 30.0 cm long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 cm at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 cm from the flat end and 15.0 cm from the vertex of the curved end. When viewed from the fiat end of the rod, the apparent depth of the object is 9.50 cm from the flat end. What is its apparent depth when viewed from the curved end?
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Chapter : Problem 87 Sears and Zemansky's University Physics with Modern Physics 13
BIO Focus of the Eye. The cornea of the eye has a radius of curvature of approximately 0.50 cm, and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 mm. (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were 25 cm in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about 5.0 mm, where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?
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Chapter : Problem 88 Sears and Zemansky's University Physics with Modern Physics 13
Problem 88P A transparent rod 50.0 cm long and with a refractive index of 1.60 is cut flat at the right end and rounded to a hemispherical surface with a 15.0-cm radius at the left end. An object is placed on the axis of the rod 12.0 cm to the left of the vertex of the hemispherical end. (a) What is the position of the final image? (b) What is its magnification?
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Chapter : Problem 89 Sears and Zemansky's University Physics with Modern Physics 13
Problem 89P A glass rod with a refractive index of 1.55 is ground and polished at both ends to hemispherical surfaces with radii of 6.0 cm. When an object is placed on the axis of the rod, 25.0 cm to the left of the left-hand end, the final image is formed 65.0 cm to the right of the right-hand end. What is the length of the rod measured between the vertices of the two hemispherical surfaces?
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Chapter : Problem 90 Sears and Zemansky's University Physics with Modern Physics 13
Problem 90P The radii of curvature of the surfaces of a thin converging meniscus lens are R1 = +12.0 cm and R2 = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.
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Chapter : Problem 91 Sears and Zemansky's University Physics with Modern Physics 13
Problem 91P An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens.
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Chapter : Problem 92 Sears and Zemansky's University Physics with Modern Physics 13
Problem 92P An object is placed 18.0 cm from a screen. (a) At what two points between object and screen may a converging lens with a 3.00-cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?
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Chapter : Problem 93 Sears and Zemansky's University Physics with Modern Physics 13
A convex mirror and a concave mirror are placed on the same optic axis, separated by a distance \(L=0.600 \mathrm{~m}\).The radius of curvature of each mirror has a magnitude of 0.360 m. A light source is located a distance from the concave mirror, as shown in Fig. P34.93. (a) What distance will result in the rays from the source returning to the source after reflecting first from the convex mirror and then from the concave mirror? (b) Repeat part (a), but now let the rays reflect first from the concave mirror and then from the convex one. Equation transcription: Text transcription: L=0.600 {~m}
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Chapter : Problem 94 Sears and Zemansky's University Physics with Modern Physics 13
As shown in Fig. P34.94 the candle is at the center of curvature of the concave mirror, whose focal length is 10.0 cm. The converging lens has a focal length of 32.0 cm and is 85.0 cm to the right of the candle. The candle is viewed looking through the lens from the right. The lens forms two images of the candle. The first is formed by light passing directly through the lens. The second image is formed from the light that goes from the candle to the mirror, is reflected, and then passes through the lens. (a) For each of these two images, draw a principal-ray diagram that locates the image. (b) For each image, answer the following questions: (i) Where is the image? (ii) Is the image real or virtual? (iii) Is the image erect or inverted with respect to the original object?
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Chapter : Problem 95 Sears and Zemansky's University Physics with Modern Physics 13
One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of 1.55. When a small leaf is placed 20.0 cm in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 cm from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33) but nothing else were changed?
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Chapter : Problem 96 Sears and Zemansky's University Physics with Modern Physics 13
Two Lenses in Contact. (a) Prove that when two thin lenses with focal lengths and are placed in contact, the focal length of the combination is given by the relationship \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) (b) A converging meniscus lens (see Fig. 34.32a) has an index of refraction of and radii of curvature for its surfaces of magnitudes and . The concave surface is placed upward and filled with carbon tetrachloride \(\left(C C I_{4}\right)\), which has . What is the focal length of the \(C C I_{4}\) - glass combination? Equation transcription: Text transcription: frac{1}{f}=\frac{1}{f{1}}+\frac{1}{f{2}} (C C I{4}) C C I{4}
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Chapter : Problem 97 Sears and Zemansky's University Physics with Modern Physics 13
Rays from a lens are converging toward a point image located to the right of the lens. What thickness of glass with index of refraction must be interposed between the lens and for the image to be formed at , located to the right of The locations of the piece of glass and of points and are shown in Fig. P34.97.
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Chapter : Problem 98 Sears and Zemansky's University Physics with Modern Physics 13
A Lens in a Liquid. A lens obeys Snell's law, bending light rays at each surface an amount determined by the index of refraction of the lens and the index of the medium in which the lens is located. (a) Equation ( ) assumes that the lens is surrounded by air. Consider instead a thin lens immersed in a liquid with refractive index \(n_{l i q}\). Prove that the focal length is then given by Eq. ( ) with replaced by \(n / n_{l i q}\) (b) A thin lens with index has focal length in vacuum. Use the result of part (a) to show that when this lens is immersed in a liquid of index \(n_{l i q}\), it will have a new focal length given by \(f^{i}=\left[\frac{n_{k q}(n-1)}{n-n_{l i q}}\right] f\) Equation transcription: Text transcription: n_{l i q} n / n_{l i q} f^{i}=\left[\frac{n_{k q}(n-1)}{n-n_{l i q}}\right] f
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Chapter : Problem 99 Sears and Zemansky's University Physics with Modern Physics 13
Problem 99P When a n object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens?
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Chapter : Problem 100 Sears and Zemansky's University Physics with Modern Physics 13
Problem 100P A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?
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Chapter : Problem 101 Sears and Zemansky's University Physics with Modern Physics 13
Problem 101P A glass plate 3.50 cm thick, with an index of refraction of 1.55 and plane parallel faces, is held with its faces horizontal and its lower face 6.00 cm above a printed page. Find the position of the image of the page formed by rays making a small angle with the normal to the plate.
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Chapter : Problem 102 Sears and Zemansky's University Physics with Modern Physics 13
A symmetric, double-convex, thin lens made of glass with index of refraction 1.52 has a focal length in air of 40.0 cm. The lens is sealed into an opening in the left-hand end of a tank filled with water. At the right-hand end of the tank, opposite the lens, is a plane mirror 90.0 cm from the lens. The index of refraction of the water is 4/3 (a) Find the position of the image formed by the lens-water-mirror system of a small object outside the tank on the lens axis and 70.0 cm to the left of the lens. (b) Is the image real or virtual? (c) Is it erect or inverted? (d) If the object has a height of 4.00 mm, what is-the height of the image?
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Chapter : Problem 103 Sears and Zemansky's University Physics with Modern Physics 13
Problem 103P You have a camera with a 35.0-mm-focal-length lens and 36.0-mm-wide film. You wish to take a picture of a 12.0-m-long sailboat but find that the image of the boat fills only ¼ of the width of the film. (a) How far are you from the boat? (b) How much closer must the boat be to you for its image to fill the width of the film?
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Chapter : Problem 104 Sears and Zemansky's University Physics with Modern Physics 13
Problem 104P BIO What Is the Smallest Thing We Can See? The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns (?m) is typical near the center of the eye. We shall model the eye as a sphere 2.50 cm in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0 mm in diameter. (a) What is the smallest object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes (1o = 60 min), and compare it with the typical experimental value of about 1.0 min. (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)
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Chapter : Problem 105 Sears and Zemansky's University Physics with Modern Physics 13
Problem 105P Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens.
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Chapter : Problem 106 Sears and Zemansky's University Physics with Modern Physics 13
Problem 106P A camera with a 90-mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 m from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?
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Chapter : Problem 107 Sears and Zemansky's University Physics with Modern Physics 13
The derivation of the expression for angular magnification, Eq. (34.22), assumed a near point of 25 cm. In fact, the near point changes with age as shown in Table 34.1. In order to achieve an angular magnification of \(2.0 \times\), what focal length should be used by a person of (a) age 10; (b) age 30; (c) age 60? (d) If the lens that gives M = 2.0 for a 10-year-old is used by a 60-year-old, what angular magnification will the older viewer obtain? (e) Does your answer in part (d) mean that older viewers are able to see more highly magnified images than younger viewers? Explain.
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Chapter : Problem 108 Sears and Zemansky's University Physics with Modern Physics 13
Angular Magnification. In deriving Eq. (34.22) for the angular magnification of a magnifier, we assumed that the object is placed at the focal point of the magnifier so that the virtual image is formed at infinity. Suppose instead that the object is placed so that the virtual image appears at an average viewer's near point of , the closest point at which the viewer can bring an object into focus. (a) Where should the object be placed to achieve this? Give your answer in terms of the magnifier focal length (b) What angle will an object of height subtend at the position found in part (a)? (c) Find the angular magnification with the object at the position found in part (a). The angle is the same as in Fig. , since it refers to viewing the object without the magnifier. (d) For a convex lens with , what is the value of with the object at the position found in part (a)? How many times greater is in this case than in the case where the image is formed at infinity? (e) In the description of a compound microscope in Section , it is stated that in a properly designed instrument, the real image formed by the objective lies just inside the first focal point \(f_{1}^{1}\) of the eyepiece. What advantages are gained by having the image formed by the objective be just inside \(f_{1}^{1}\), as opposed to precisely at \(f_{1}^{1}\) ? What happens if the image formed by the objective is just outside \(f_{1}^{1}\) ? Equation transcription: Text transcription: f{1}^{1}
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Chapter : Problem 109 Sears and Zemansky's University Physics with Modern Physics 13
Problem 109P BIO In one form of cataract surgery the person’s natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the person’s eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of 24 cm?
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Chapter : Problem 110 Sears and Zemansky's University Physics with Modern Physics 13
Problem 110P A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye described in Exercise. If the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described in Exercise, what is the distance from the cornea vertex to the retina? What does this tell you about the shape of the nearsighted eye? Exercise: Curvature of the Cornea. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40 and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea’s vertex is focused on the retina?
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Chapter : Problem 111 Sears and Zemansky's University Physics with Modern Physics 13
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?
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Chapter : Problem 112 Sears and Zemansky's University Physics with Modern Physics 13
The Galilean Telescope. Figure P34.112 is a diagram of a Galilean telescope, or opera glass, with both the object and its final image at infinity. The image serves as a virtual object for the eyepiece. The final image is virtual and erect. (a) Prove that the angular magnification is \(M=-f_{1} / f_{2}\). (b) A Galilean telescope is to be constructed with the same objective lens as in Exercise . What focal length should the eyepiece have if this telescope is to have the same magnitude of angular magnification as the one in Exercise (c) Compare the lengths of the telescopes. Equation transcription: Text transcription: M=-f{1} / f{2}
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Chapter : Problem 113 Sears and Zemansky's University Physics with Modern Physics 13
Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1}\), and the diverging lens has focal length \(f_{2=-}\left|f_{2}\right|\). The two lenses are separated by a variable distance that is always less than . Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right)\). To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{1}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image is formed a distance \(s_{2}^{1}=\left|f_{2}\right|\left(f_{1}-d\right)\left(f_{2} \mid-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius at some point The distance from the final image to the point is the effective focal length of the lens combination; if the combination were replaced by a single lens of focal length placed at , parallel rays would still be brought to a focus at . Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right|\left(\left|f_{2}\right|-f_{1}+d\right)\) (d) If \(f_{1}=12.0 \mathrm{~cm}\), \(f_{2}=-18.0 \mathrm{~cm}\), and the separation is adjustable between 0 and , find the maximum and minimum focal lengths of the combination. What value of gives Equation transcription: Text transcription: F_{1} f_{2=-}\left|f_{2}\right| \left|f_{2}\right|>\left(f_{1}-d\right) r_{0}^{1}=r_{0}\left(f_{1}-d\right) / f_{1} s_{2}^{1}=\left|f_{2}\right|\left(f_{1}-d\right)\left(f_{2} \mid-f_{1}+d\right) f_{2}=-18.0 \mathrm{~cm} f=f_{1}\left|f_{2}\right|\left(\left|f_{2}\right|-f_{1}+d\right) f_{1}=12.0 \mathrm{~cm}
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Chapter : Problem 114 Sears and Zemansky's University Physics with Modern Physics 13
A certain reflecting telescope, constructed as shown in Fig. E34.67, has a spherical mirror with a radius of curvature of and an eyepiece with a focal length of . If the angular magnification has a magnitude of 36 and the object is at infinity, find the position of the eyepiece and the position and nature (real or virtual) of the final image. (Note: is not equal to \(\left|f_{1}\right| /\left|f_{2}\right|\), so the image formed by the eyepiece is not at infinity.) Equation transcription: Text transcription: \left|f{1}right| /\left|f{2}\right|
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Chapter : Problem 115 Sears and Zemansky's University Physics with Modern Physics 13
Problem 115P A microscope with an objective of local length 8.00 mm and an eyepiece of focal length 7.50 cm is used to project an image on a screen 2.00 m from the eyepiece. Let the image distance of the objective be 18.0 cm. (a) What is the lateral magnification of the image? (b) What is the distance between the objective and the eyepiece?
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Chapter : Problem 116 Sears and Zemansky's University Physics with Modern Physics 13
Spherical aberration is a blurring of the image formed by a spherical mirror. It occurs because parallel rays striking the mirror far from the optic axis are focused at a different point than are rays near the axis. This problem is usually minimized by using only the center of a spherical mirror. (a) Show that for a spherical concave mirror, the focus moves toward the mirror as the parallel rays move toward the outer edge of the mirror. (Hint: Derive an analytic expression for the distance from the vertex to the focus of the ray for a particular parallel ray. This expression should be in terms of (i) the radius of curvature R of the mirror and (ii) the angle \(\theta\) between the incident ray and a line connecting the center of curvature of the mirror with the point where the ray strikes the mirror.) (b) What value of \(\theta\) produces a 2% change in the location of the focus, compared to the location for \(\theta\) very close to zero?
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Chapter : Problem 118 Sears and Zemansky's University Physics with Modern Physics 13
An Object at an Angle. A 16.0-cm-long pencil is placed at a 45.0° angle, with its center 15.0 cm above the optic axis and 45.0 cm from a lens with a 20.0-cm focal length as shown in Fig. P34.118. (Note that the figure is not drawn to scale.) Assume that the diameter of the lens is large enough for the paraxial approximation to be valid. (a) Where is the image of the pencil? (Give the location of the images of the points , ,and on the object, which are located at the eraser, point, and center of the pencil, respectively.) (b) What is the length of the image (that is, the distance between the images of points and )? (c) Show the orientation of the image in a sketch.
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Chapter : Problem 119 Sears and Zemansky's University Physics with Modern Physics 13
Problem 119CP BIO People with normal vision cannot focus their eyes underwater if they aren’t wearing a face mask or goggles and there is water in contact with their eyes (see Discussion Question Q34.23). (a) Why not? (b) With the simplified model of the eye described in Exercise 34.50, what corrective lens (specified by focal length as measured in air) would be needed to enable a person underwater to focus an infinitely distant object? (Be careful—the focal length of a lens underwater is not the same as in air! See Problem 34.92. Assume that the corrective lens has a refractive index of 1.62 and that the lens is used in eyeglasses, not goggles, so there is water on both sides of the lens. Assume that the eyeglasses are 2.00 cm in front of the eye.) Q34.23. BIO You can’t see clearly underwater with the naked eye, but you can if you wear a face mask or goggles (with air between your eyes and the mask or goggles). Why is there a difference? Could you instead wear eyeglasses (with water between your eyes and the eyeglasses) in order to see underwater? If so, should the lenses be converging or diverging? Explain.
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