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# Week 7 short assignments 1220

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This 18 page Bundle was uploaded by Dragon Note on Thursday March 24, 2016. The Bundle belongs to 1220 at University of Missouri - Columbia taught by Y Zhang in Spring 2016. Since its upload, it has received 44 views. For similar materials see College Physics II in Physics 2 at University of Missouri - Columbia.

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Short Assignment By 3/11/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Short Assignment By 3/11/2016 Due: 11:00am on Friday, March 11, 2016 To understand how points are awarded, read theding Policy for this assignment. Ray Tracing and Image Formation with a Concave Lens A concave lens refracts parallel rays in such a way that they are bent away from the axis of the lens. For this reason, a concave lens is referred to as a diverging lens. Part A Consider the following diagrams, where F represents the focal point of a concave lens. In these diagrams, the image formed by the lens is obtained using the ray tracing technique. Which diagrams are accurate? Type A if you think that only diagram A is correct, type AB if you think that only diagrams A and B are correct, and so on. Hint 1. A ray parallel to the lens axis A ray parallel to the axis of a concave lens is refracted along a line that extends back through the focal point on the same side of the lens. Hint 2. A ray that passes through the focal point A ray that is directed toward the focal point on the other side of the lens is refracted parallel to the lens axis. Hint 3. A ray that passes through the middle of the lens A ray that passes through the middle of a concave lens continues on its original direction with essentially no displacement after passing through the lens. ANSWER: Correct A concave lens always forms an image that is on the same side of the lens as the object. Short Assignment By 3/11/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Part B If the focal length of the concave lens is -7.50 {\rm cm} , at what distance \texttip{d_{\rm o}}{d_o} from the lens should an object be placed so that its image is formed 3.70 {\rm cm} from the lens? Express your answer in centimeters. Hint 1. How to approach the problem To determine the object distance you can use the thin-lens equation, but be careful to assign the correct sign to each quantity involved in the equation. Hint 2. The thin-lens equation The thin-lens equation for a lens with a focal length \texttip{f}{f} is \large{\frac{1}{d_{\rm o}}+\frac{1}{d_{\rm i}}=\frac{1}{f}}, where \texttip{d_{\rm o}}{d_o} and \texttip{d_{\rm i}}{d_i} are the object distance and the image distance, respectively. Hint 3. Find the image distance What is the image distance \texttip{d_{\rm i}}{d_i} for a concave lens that forms an image 3.70 {\rm cm} from the lens? Express your answer in centimeters. Hint 1. Sign convention for image distances Conventionally the image distance has a positive sign when the image is on the opposite side of the lens from the object (a real image), and a negative sign when the image is on the same side of the lens as the object (a virtual image). ANSWER: \texttip{d_{\rm i}}{d_i} = -3.70\rm cm ANSWER: \texttip{d_{\rm o}}{d_o} = 7.30 \rm cm Correct Part C What is the magnification \texttip{m}{m} produced by the concave lens described in Part B? Express your answer numerically. Hint 1. Magnification Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/AMS/Regular/Main.js Short Assignment By 3/11/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... The magnification \texttip{m}{m} produced by a lens is given by the formula \large{m=-\frac{d_{\rm i}}{d_{\rm o}}}, where \texttip{d_{\rm o}}{d_o} and \texttip{d_{\rm i}}{d_i} are the object distance and the image distance, respectively. ANSWER: \texttip{m}{m} = 0.507 Correct Part D Where should the object be moved to have a larger magnification? Hint 1. Magnification and image size. Recall that a larger magnification corresponds to a larger image. You may find the ray diagrams found in Part A helpful in determining how the size of the image varies as the object is moved closer or farther from the lens. ANSWER: The object should be moved closer to the lens. The object should be moved farther from the lens. The object should be moved to the focal point of the lens. The object should not be moved closer to the lens than the focal point. Correct The Magnification Produced by a Lens Part A What can one say about the image produced by a thin lens that produces a positive magnification? Hint 1. How to approach the problem When a lens produces an inverted image, the image lies below the lens axis and its height is taken to be negative. In such a case, the ratio of the image height to the object height is negative. Relate this to the magnification produced by the lens and the image location. Hint 2. Lateral magnification LoadinThe lateral magnification \texttip{m}{m} produced by a lens is the ratio of the image height \texttip{y'}{y'} to the Short Assignment By 3/11/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... object height \texttip{y}{y}: \large{m=\frac{y'}{y}}. This means that when the magnification produced by a certain lens is negative, the image height and the object height have opposite signs. Hint 3. Find another expression for the magnification produced by a thin lens For a lens that produces an image at \texttip{s'}{s'} of an object located at \texttip{s}{s}, which of the following expressions gives the correct magnification produced by the lens? ANSWER: s/s' s'/s -s/s' -s'/s 1/s' + 1/s Hint 4. Sign rule for image distances When the image is on the same side of the lens as the outgoing light, the image distance is positive; otherwise, it is negative. Also, remember that when the outgoing rays pass through an image point, the image is called a real image. However, when the outgoing rays do not pass through an image point and instead appear to diverge as if they had come from an image point, the image is a virtual image. ANSWER: It is real and inverted. It is real and erect. It is virtual and inverted. It is virtual and erect. Correct Note that the magnification produced by a lens is not a constant property of the lens. In fact, it can change depending on the location of the object. That is, the same lens can produce a positive magnification for a certain object distance from the lens, and a negative magnification for a different object distance. Part B If the diameter of a lens is reduced, what happens to the magnification produced by the lens? Hint 1. How to approach the problem The magnification produced by a lens depends on the image formed by the lens, which is determined exclusively by the position of the object relative to the lens and the optical properties of the lens. Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/AMS/Regular/Main.js Short Assignment By 3/11/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... ANSWER: It increases. It decreases. It is unchanged. Correct Score Summary: Your score on this assignment is 91.7%. You received 1.38 out of a possible total of 1.5 points. Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/AMS/Regular/Main.js Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Short Assignment By 3/7/2016 Due: 11:00am on Monday, March 7, 2016 To understand how points are awarded, read theGrading Policy for this assignment. ± Understanding Spherical Mirrors Learning Goal: To be able to calculate locations and heights of images formed by spherical mirrors, as well as to understand the qualities of such an image. In this problem, you will learn to use the spherical mirror equation: . This equation relates three quantities important to the formation of images with a spherical mirror: The object distance is the distance from the mirror to the object, along the axis of the mirror. The image distance is the distance from the mirror to the image, along the axis of the mirror. The focal length is an intrinsic property of the mirror. It is equal to half the radius of curvature (i.e., ). The equation given above allows you to calculate the locations of images and objects. Frequently, you will also be interested in the size of an image or object. The ratio of the size of an image to the size of the object is called the magnification; it is given by , where is the height of the image, and is the height of the object. The second equality allows you to find the size of the image (or object) with the information provided by the spherical mirror equation. All of the quantities in the above equations can take both positive and negative values. Positive distances correspond to real images or objects, while negative distances correspond to virtual images or objects. Positive heights correspond to upright images or objects, while negative heights correspond to inverted images or objects. The following table summarizes this information: Real Virtual Real Virtual Upright Inverted Upright Inverted Focal length can also take positive and negative values. Positive corresponds to a concave mirror, while negative corresponds to a convex mirror. While it is possible for or to be negative, this can happen only in situations with multiple mirrors or mirrors and lenses. In this problem, you will consider only a single mirror, so and will be positive. You will begin with a relatively standard calculation. Consider a concave spherical mirror with a radius of curvature equal to 60.0 centimeters. An object 6.00 centimeters tall is placed along the axis of the mirror, 45.0 centimeters from the mirror. You are to find the location and height of the image. Part A What is the focal length of this mirror? Express your answer in centimeters to three significant figures. ANSWER: Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... = 30.0 Correct Part B Now use the spherical mirror equation to find the image dista.ce Express your answer in centimeters, as a fraction or to three significant figures. ANSWER: = 90.0 Correct Part C Find the magnification , using and . Express your answer numerically, as a fraction or to three significant figures. ANSWER: = -2.00 Correct Part D Finally, use the magnification to find the height of the .mage Express your answer in centimeters, as a fraction or to three significant figures. ANSWER: = -12.0 Correct Part E Look at the signs of your answers to determine which of the following describes the image formed by this mirror: ANSWER: Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... real and upright real and inverted virtual and upright virtual and inverted Correct Next, you will manipulate the spherical mirror equation to explore the qualities (real/virtual and upright/inverted) of some images. Part F Solve the spherical mirror equation for. Express your answer in terms of and . ANSWER: = Correct Part G What is the magnification ? Use your answer from Part F. Express your answer in terms of and . ANSWER: = Correct Part H If you are working with a convex mirror ( ), which of the following describes the image? Hint 1. Is it real? Consider the effect of having a negative focal length on the equation for image distance, . Recall that is positive in this problem. With that in mind, what is the sign of the numerator? What about the denominator? Finally, what will the sign ofbe when you perform the division? Hint 2. Upright or inverted? Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... To determine whether the image is upright or inverted, look at the sign of the magnification in the equation . What is the sign of the numerator? Keeping in mind that is positive, what is the sign of the denominator? Does the sign of either depend on the size of \texttip{s}{s}? Now, since y' = m y, and \texttip{y}{y} is always positive for a single mirror, the sign of \texttip{y'}{y'} is the same as the sign of \texttip{m}{m}. ANSWER: real and upright real and inverted virtual and upright virtual and inverted depends on the object distance Correct You know now that the magnification of a single convex mirror is always positive. Part I What can one say about the size of the magnification of a single convex mirror? ANSWER: It is greater than one. It is equal to one. It is less than one. Correct Because they create an upright image that is smaller than the object, convex mirrors are used in many different places. You will see them in convenience stores for surveillance, because they allow one person to see most of the store at a glance. They are also used in hospitals, parking garages, and winding roads to provide a limited view around a corner to aid in avoiding collisions. As a spherical mirror becomes flatter, the radius of curvature \texttip{R}{R} gets larger. Notice that as \texttip{R}{R} goes to infinity, so does \texttip{f}{f}, because f={R}/{2}. Thus, as \texttip{R}{R} gets larger, {1}/{f} gets smaller. In the limit where you allow \texttip{R}{R} to go to infinity, {1}/{f} becomes zero. Therefore, if you could construct a mirror with an infinitely large radius of curvature, it would obey the equation {1}/{s} + {1}/{s'} = 0. Part J What is the value of \texttip{s'}{s'} obtained from this new equation? Express your answer in terms of \texttip{s}{s}. ANSWER: Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... \texttip{s'}{s'} = -s Correct The introduction to this part noted that as the mirror gets flatter, \texttip{R}{R} gets larger. Recall that the earth is approximately a sphere. It only appears flat to us because of its large radius of curvature. From this, you should be able to see that you can in fact make a mirror with infinite radius of curvature. It would be a flat mirror! Your answer to Part J is the equation relating \texttip{s}{s} and \texttip{s'}{s'} for a plane mirror. Is Light Reflected or Refracted? When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at the interface separating the two materials. For example, consider a ray of light that travels from air into the water of a lake. As the ray strikes the air-water interface (the surface of the lake), it is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation. These effects on light propagation occur because light travels at different speeds depending on the medium. The index of refraction of a material, denoted by \texttip{n}{n}, gives an indication of the speed of light in the material. It is defined as the ratio of the speed of light \texttip{c}{c} in vacuum to the speed \texttip{v}{v} in the material, or \large{n=\frac{c}{v}}. Part A When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the speed of the light Hint 1. Index of refraction The index of refraction \texttip{n}{n} of a material is defined as the ratio of the speed of light \texttip{c}{c} in vacuum to the speed \texttip{v}{v} in that particular material, or \large{n=\frac{c}{v}}. Since it is the ratio of two positive quantities that have the same units, the index of refraction is a pure (positive) number. Note that the speed of light \texttip{v}{v} in a certain material is inversely proportional to the index of refraction of that material. ANSWER: increases. decreases. remains constant. Correct Part B Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... What is the minimum value that the index of refraction can have? Hint 1. How to approach the problem Remember that the speed of light \texttip{v}{v} in a certain material is inversely proportional to the index of refraction of that material. Thus, the minimum value of the index of refraction is calculated for the medium where the speed of light is maximum. That occurs in vacuum where v=c. ANSWER: \;\;\;\;0 +1 -1 between 0 and 1 Correct The index of refraction of a material is always a positive number greater than 1 that tells us how fast the light travels in the material. The greater the index of refraction of a material, the more slowly light travels in the material. An example of reflection and refraction of light is shown in the figure. An incident ray of light traveling in the upper material strikes the interface with the lower material. The reflected ray travels back in the upper material, while the refracted ray passes into the lower material. Experimental studies have shown that the incident, reflected, and refracted rays and the normal to the interface all lie in the same plane. Moreover, the angle that the reflected ray makes with the normal to the interface, called the angle of reflection, is always equal to the angle of incidence. (Both of these angles are measured between the light ray and the normal to the interface separating the two materials.) This is known as the law of reflection. The direction of propagation of the refracted ray, instead, is given by the angle that the refracted ray makes with the normal to the interface, which is called the angle of refraction. The angle of refraction depends on the angle of incidence and the indices of refraction of the two materials. In particular, if we let \texttip{n_{\rm 1}}{n_1} be the index of refraction of the upper material and \texttip{n_{\rm 2}}{n_2} the index of refraction of the lower material, then the angle of incidence, \texttip{\theta _{\rm 1}}{theta_1}, and the angle of refraction, \texttip{\theta _{\rm 2}}{theta_2}, satisfy the relation n_1\sin\theta_1=n_2\sin\theta_2. This is the law of refraction, also known as Snell's law. Part C Now consider a ray of light that propagates from water (n=1.33) to air (n=1). If the incident ray strikes the water-air interface at an angle \theta_1\neq 0, which of the following relations regarding the angle of refraction, \texttip{\theta _{\rm 2}}{theta_2}, is correct? Typesetting math: 24%n expression for the ratio of the sines of \texttip{\theta _{\rm 1}}{theta_1} and Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... \texttip{\theta _{\rm 2}}{theta_2} Let the index of refraction of water be \texttip{n_{\rm 1}}{n_1} and that of air be \texttip{n_{\rm 2}}{n_2}. Use Snell's law to find an expression for the ratio of the sine of \texttip{\theta _{\rm 2}}{theta_2} to the sine of \texttip{\theta _{\rm 1}}{theta_1}. Express your answer in terms of some or all of the variables \texttip{n_{\rm 1}}{n_1}, \texttip{n_{\rm 2}}{n_2}, and \texttip{\theta _{\rm 1}}{theta_1}. ANSWER: \large{\frac{\sin\theta_2}{\sin\theta_1}} = \large{\frac{n_{1}}{n_{2}}} ANSWER: \theta_2 > \theta_1 \theta_2<\theta_1 \theta_2=\theta_1 Correct When light propagates from a certain material to another one that has a smaller index of refraction, that is, n_1> n_2, the speed of propagation of the light rays increases and the angle of refraction is always greater than the angle of incidence. This means that the rays are always bent away from the normal to the interface separating the two media. Part D Consider a ray of light that propagates from water (n=1.33) to glass (n=1.52). If the incident ray strikes the water-glass interface at an angle \theta_1\neq 0, which of the following relations regarding the angle of refraction \texttip{\theta _{\rm 2}}{theta_2} is correct? Hint 1. Find an expression for the ratio of the sines of \texttip{\theta _{\rm 1}}{theta_1} and \texttip{\theta _{\rm 2}}{theta_2} Let the index of refraction of water be \texttip{n_{\rm 1}}{n_1} and that of glass be \texttip{n_{\rm 2}}{n_2}. Use Snell's law to find an expression for the ratio of the sine of \texttip{\theta _{\rm 2}}{theta_2} to the sine of \texttip{\theta _{\rm 1}}{theta_1}. Express your answer in terms of some or all of the variables \texttip{n_{\rm 1}}{n_1}, \texttip{n_{\rm 2}}{n_2}, and \texttip{\theta _{\rm 1}}{theta_1}. ANSWER: \large{\frac{\sin\theta_2}{\sin\theta_1}} = \large{\frac{n_{1}}{n_{2}}} ANSWER: Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... \theta_2 > \theta_1 \theta_2 < \theta_1 \theta_2 = \theta_1 Correct When light propagates from a certain material to another one that has a greater index of refraction, that is, n_1< n_2, the speed of propagation of the light rays decreases and the angle of refraction is always smaller than the angle of incidence. This means that the rays are always bent toward the normal to the interface separating the two media. Part E Consider a ray of light that propagates from air (n=1) to any one of the materials listed below. Assuming that the ray strikes the interface with any of the listed materials always at the same angle \texttip{\theta _{\rm 1}}{theta_1}, in which material will the direction of propagation of the ray change the most due to refraction? Hint 1. How to approach the problem The direction of propagation of the ray of light will change the most when the difference between the angle of refraction and the angle of incidence is maximum. Since we are studying a situation where light propagates from air to a material that has a greater index of refraction, we can make use of the results obtained in Part D. We know that, in this case, the angle of refraction is always smaller than the angle of incidence. Thus, the difference between the angle of refraction and the angle of incidence is maximum when the angle of refraction is smallest. Hint 2. Find an expression for the sine of the angle of refraction Let the index of refraction of the unknown material be \texttip{n_{\rm 2}}{n_2}. Use Snell's law to find an expression for the sine of the angle of refraction, \texttip{\theta _{\rm 2}}{theta_2}. Express your answer in terms of some or all of the variables \texttip{n_{\rm 2}}{n_2} and \texttip{\theta _{\rm 1}}{theta_1}. ANSWER: \sin \theta_2 = \large{\frac{{\sin}\left({\theta}_{1}\right)}{n_{2}}} ANSWER: ice (n=1.309) water (n=1.333) turpentine (n=1.472) glass (n=1.523) diamond (n=2.417) Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Correct The greater the change in index of refraction, the greater the change in the direction of propagation of light. To avoid or minimize undesired bending of the light rays, light should travel through materials with matching indices of refraction. Is light always both reflected and refracted at the interface separating two different materials? To answer this question, let's consider the case of light propagating from a certain material to another material with a smaller index of refraction (i.e., n_1>n_2). Part F In the case of n_1>n_2, if the incidence angle is increased, the angle of refraction Hint 1. How to approach the question Recall that, according to Snell's law, the sine of the angle of refraction is directly proportional to the sine of the angle of incidence. Thus, as the angle of incidence is increased, the angle of refraction changes accordingly. Moreover, since the angle of refraction is greater than the angle of incidence, as you found in Part C, the angle of refraction can reach its maximum value sooner than the angle of incidence. ANSWER: decreases. increases. increases up to a maximum value of 90 degrees. remains constant. Correct Since the light is propagating into a material with a smaller index of refraction, the angle of refraction, \texttip{\theta _{\rm 2}}{theta_2}, is always greater than the angle of incidence, \texttip{\theta _{\rm 1}}{theta_1}. Therefore, as \texttip{\theta _{\rm 1}}{theta_1} is increased, at some point \texttip{\theta _{\rm 2}}{theta_2} will reach its maximum value of 90^\circ and the refracted ray will travel along the interface. The angle of incidence for which \theta_2=90^\circ is called the critical angle \texttip{\theta _{\rm crit}}{theta_crit}. For any angle of incidence greater than \texttip{\theta _{\rm crit}}{theta_crit}, no refraction occurs. The ray no longer passes into the second material. Instead, it is completely reflected back into the original material. This phenomenon is called total internal reflection and occurs only when light encounters an interface with a second material with a smaller index of refraction than the original material. Part G What is the critical angle \texttip{\theta _{\rm crit}}{theta_crit} for light propagating from a material with index of refraction of 1.50 to a material with index of refraction of 1.00? Express your answer in radians. Hint 1. Find an expression for the sine of the angle of incidence Typesetting math: 24%w to find a general expression for the sine of the angle of incidence, \texttip{\theta _{\rm 1}}{theta_1} Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... , for a ray of light that travels from a material with index of refraction \texttip{n_{\rm 1}}{n_1} to a material with index of refraction \texttip{n_{\rm 2}}{n_2}. Express your answer in terms of \texttip{\theta _{\rm 2}}{theta_2}, \texttip{n_{\rm 1}}{n_1}, and \texttip{n_{\rm 2}}{n_2}. ANSWER: \sin\theta_1 = \large{\frac{n_{2} {\sin}\left({\theta}_{2}\right)}{n_{1}}} ANSWER: \texttip{\theta _{\rm crit}}{theta_crit} = 0.730\rm radians Correct In conclusion, light is always both reflected and refracted, except in the special situation when the conditions for total internal reflection occur. In that case, there is no refracted ray and the incident ray is completely reflected. ± Understanding the Propagation of Light Learning Goal: To understand ray diagrams, as well as basic reflection and refraction problems. There are two ways of indicating, in a diagram, the path that light follows. One way is by using wavefronts (shown in blue); the other is by using rays (shown in red) . Wavefronts represent, in a schematic way, the successive peaks of the electromagnetic wave at a specific time. Light is a transverse wave; it moves perpendicular to the wavefronts. Rays are used to indicate the direction of motion of the light. Ray diagrams are typically used in problems where the wave nature of light is not important, as will be the case in geometric optics. Notice in the diagram that the wavefronts get closer together inside of the glass. This is because the speed of light in glass is less than that in air. The frequency of a wave does not change when it propagates through different media, even though its speed may change. (Waves can be neither created nor destroyed at the boundary between different media; hence, the number of waves that strike the boundary per unit time must equal the number of waves that leave the boundary per unit time.) Let \texttip{v}{v} be the wave's speed, \texttip{\lambda }{lambda} its wavelength, and \texttip{f}{f} its frequency. These quantities are related via the equation v = \lambda f. Note that, if the wave speed decreases, the wavelength must also decrease for the frequency to remain constant. Part A What is the wavelength \texttip{\lambda }{lambda} of light in glass, if its wavelength in air is \texttip{\lambda _{\rm 0}}{lambda_0}, its speed in air is \texttip{c}{c}, and its speed in the glass is \texttip{v}{v}? Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Express your answer in terms of \texttip{\lambda _{\rm 0}}{lambda_0}, \texttip{c}{c}, and \texttip{v}{v}. Hint 1. How to approach the problem Because the frequency stays the same, you can solve the equation relating \texttip{v}{v}, \texttip{\lambda }{lambda}, and \texttip{f}{f} for \texttip{f}{f}. Then, set the expression for the frequency of the light in air equal to its frequency in glass. Solve this equation for \texttip{\lambda }{lambda}. ANSWER: \texttip{\lambda }{lambda} = \large{\frac{v{\lambda}_{0}}{c}} Correct The letter \texttip{c}{c} is used to denote the speed of light in vacuum. Although the speed of light in air should have been used for this part, the speeds of light in air and vacuum are so similar that they are often used interchangeably. You should always keep this in mind, as problems that specify many significant figures may require you to use the correct speed in air. For the rest of this problem, however, assume that the difference between the speeds of light in air and vacuum is too small to affect your results. Part B The index of refraction \texttip{n}{n} for a material is defined to be n=c/v. Rewrite your answer from Part A in terms of the index of refraction. Express your answer in terms of \texttip{\lambda _{\rm 0}}{lambda_0} and \texttip{n}{n}. ANSWER: \texttip{\lambda }{lambda} = \large{\frac{{\lambda}_{0}}{n}} Correct Part C If red light of wavelength 700 {\rm nm} in air enters glass with index of refraction 1.5, what is the wavelength \texttip{\lambda }{lambda} of the light in the glass? Express your answer in nanometers to three significant figures. ANSWER: \texttip{\lambda }{lambda} = 467 {\rm nm} Correct Two important things happen to light when it strikes a transparent boundary: It gets reflected and it gets refracted. When you see your reflection in glass, you are seeing the result of reflection from a transparent boundary. In the figure , the ray moving toward the air/glass interface is called the incident ray. The ray leaving the boundary in air is called the reflected ray. The ray Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... leaving the boundary inside the glass is called the refracted ray. Reflection from a mirror and reflection from a transparent boundary both obey the law of reflection: \theta_{\rm a} = \theta_{\rm r}, where \texttip{\theta _{\rm a}}{theta_a} is the angle of incidence (the angle between the incoming ray and the normal to the surface), and \texttip{\theta _{\rm r}}{theta_r} is the angle of reflection (the angle between the normal line and the reflected ray) . Part D If light strikes the air/glass interface at an angle 32.0 {\rm degrees} to the normal, what is the angle of reflection, \texttip{\theta _{\rm r}}{theta_r}? Express your answer in degrees to three significant figures. ANSWER: \texttip{\theta _{\rm r}}{theta_r} = 32 {\rm degrees} Correct The second important effect of light striking a transparent boundary is refraction. Refraction is the bending of light caused by the difference in the speed of light between materials. When light moves into a medium with a higher index of refraction (i.e., lower speed of light), the refracted ray has a smaller angle to the normal than the incident ray. Snell's law gives this angle of refraction, \texttip{\theta _{\rm b}}{theta_b}: n_{\rm a} \sin(\theta_{\rm a}) = n_{\rm b} \sin(\theta_{\rm b}) . Since we are assuming that the speed of light in air is very close to the speed of light in vacuum, you will use n_{\rm a}=1.00 in this problem. Typesetting math: 24% Short Assignment By 3/7/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Part E If light strikes the air/glass interface at an incidence angle of 32.0 {\rm degrees}, what is the angle of refraction, \texttip{\theta _{\rm b}}{theta_b}? Use 1.50 for the index of refraction of glass. Express your answer in degrees to three significant figures. ANSWER: \texttip{\theta _{\rm b}}{theta_b} = 20.7{\rm degrees} Correct Score Summary: Your score on this assignment is 100%. You received 4 out of a possible total of 4 points. Typesetting math: 24%

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#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.