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# Gen Physics II Noncalc PHYS 2020

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This 24 page Class Notes was uploaded by Iva Cormier on Sunday October 11, 2015. The Class Notes belongs to PHYS 2020 at East Tennessee State University taught by Donald Luttermoser in Fall. Since its upload, it has received 14 views. For similar materials see /class/221410/phys-2020-east-tennessee-state-university in Physics 2 at East Tennessee State University.

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Date Created: 10/11/15

PHYS2020 General Physics II Course Lecture Notes Section XIV Dr Donald G Luttermoser East Tennessee State University Edition 31 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2020 General Physics taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics Enhanced 7th Editizm 2006 textbook by Serway Faughn and Vuille XIV Optical Instruments A Common Optical Tools 1 A camera consists of a lens or series of lenses that focus an image onto a light sensing detector 69 lm CCD etc a The diameter of the lens can be stopped down to smaller 0 C sizes by adjusting the aperture opening i These stops are technically referred to as the f number f number 2 XIV 1 f focal length of lens D diameter of the lens or aperture whichever is smaller ii The larger the f number the less light is allowed into the camera iii Cameras typically have the following f stops f 28 lets the most light in f4 f56 f8 fll f16 lets the least amount of light in The magni cation of the image follows the thin lens for mulae 69 Eq XII 10 with an image size given by h f XIV 2 p The amount of light that falls on the detector is deter mined by the f stop w the shutter speed i For fast moving objects one should use shutter speeds llOOOth or l500th of a second ii For indoor low light levels speeds greater than 1 60th of a second are required gt speeds longer XIV 1 XIV 2 PHYS 2020 General Physics II than 160 sec usually require the camera to be mounted on a tripod to keep the camera steady iii Astronomical photographs typically require the shutter to be open from minutes to hours in order to record the image 2 The organic analogy to the camera is the eye In the case of the human eye we have the following characteristics a The iris is equivalent to the aperture of the camera gt the opening is called the pupil b The cornea is a transparent lens cap for the eye s lens c The retina records the image and sends the signal to the brain via the optic nerve gt analogous to the lm in a camera or a CCD chip The retina is composed of 2 types of receptor cells i Rods are able to detect low light levels in black and white ii Cones come in 3 types that respond respectively to red green and blue light 1 Defects of the eye i Hyperopia Image of a nearby object forms be hind the retina farsighted ii Myopia Maximum focal length of eye lens is shorter than diameter of eye nearsighted iii AstigmatismA point source produces a line im Donald G Luttermoser ETSU XIV 3 age on the retina iv Cataracts Lens andor cornea becomes par tially or totally opaque v Glaucoma Abnormal increase in uid pressure inside the eyeball e Corrective lenses are often used to correct many defects of the eye The power of such a lens is measured in diopters 73diopters XIV 3 1 f In gt f is the lens focal length in meters i The near point is the closest distance for which the lens can accommodate to focus light on the retina This distance changes as the eye ages Age 10 near point 18 cm Age 20 near point 25 cm Age 40 near point 50 cm Age 60 near point 500 cm ii The far point is the farthest distance for which the lens of a relaxed eye can focus light on the retina People with normal Vision can focus ob jects that are very far away 69 the Moon For these people the far point is essentially in nity 3 The Simple Magni er increases the image size of an object with a single lens e magnifying glass with a magni cation of XIV 4 00 gt 6 angle the object subtends with the lens XIV 4 PHYS 2020 General Physics II gt 00 angle the object subtends to the naked eye B Microscopes and Telescopes 1 Both microscopes and refractlng telescopes have a series of 2 or more lenses that are used to magnify an image a The objective lens is the light collector gt the bigger the objective the more light you receive the fainter the object you can see b The eyepiece lens is the main component of magni ca tion c A re ecting telescope has a mirror usually parabolic in shape to collect the light and is called the primary mirror instead of the objective lens 2 These instruments magnify objects following the relations a For compound microscopes L 25 M Mlme 7 lt cm XIV5 f 0 fe M1 magni cation of the objective me magni cation of the eyepiece gt L distance between objective and eyepiece in cm f0 focal length of the objective in cm f6 focal length of the eyepiece in cm b For a telescope m XIV 6 i Note that one can get as high as a magni cation as you want by just getting eyepieces with smaller and smaller focal lengths Donald G Luttermoser ETSU XIV 5 ii The higher the magni cation the fainter the im age gt rule of thumb for useful magni cations 50 power for each inch of the objective gt a 24 telescope s maximum useful power is 120l Example XIVil Problems 2529 Page 840 from the Serway amp Faughn textbook Astronomers often take photographs with the objective lens or mirror of a telescope alone without an eye piece a Show that the image size M for this telescope is given by h fhf p where h is the object size f the objective focal length and p the object distance b Simplify the expression in part a if the object distance is much greater than the objective focal length c The quotwingspanquot of the International Space Station is 1086 m the overall width of its solar panel configuration When it is orbiting at an altitude of 407 km find the width of the image formed by the telescope objective of focal length 400 m Solution a Using the thin lens equation Eq XII 10 we get gig I9 qif lll q fp 1 i q pf pf qpf p f lnserting this into the magni cation equation Eq Xll ll we get XIV 6 PHYS 2020 General Physics II Therefore the image size will be fh fh h Mh 7 7 p f f p Solution b prgtgtfthenf pm pand hm 17 Solution c Suppose the telescope observes the space station while at the zenith e the point directly overhead then M iW rv p 4 407 x 103In C Spatial Resolution 1 The ability of an optical system to distinguish between closely spaced objects is limits due to the wave nature of light a Light gets diffracted when passing through apertures see Figures 2513 and 2514 in the textbook b When the central maximum of one image falls on the rst minimum of another image the images are said to just be resolved gt Rayleigh s criterion 2 The limiting angle which an object is viewed depends both on the size of the aperture and the wavelength of light Donald G Luttermoser ETSU XIV 7 a For a slit 0m g XIV 7 where a is the width of the slit opening Both the wave length x and a must be measured in the same units then 0m the resolving angle is in radians b For a circular aperture A m 1227 XIV8 0 D where D is the circular aperture diameter Again A and D must be in the same units to give 0m in radians 3 Turbulence in the Earth s atmosphere limits the spatial resolu tion of Earth based telescopes to about 1 arcsecondl Example XIV72 Problems 2533 Page 840 from the Serway amp Faughn textbook A converging lens with a diameter of 300 cm forms an image of a satellite passing overhead The satellite has two green lights wavelength 500 nm spaced 100 m apart If the lights can just be resolved according to the Rayleigh criterion what is the altitude of the satellite Solution Since the lens corresponds to a circular aperture we can just use Eq XIV 8 to determine the angular separation 0 0m 122 122 203 x 106 rad The altitude can then be determined from the arc length equation e d h639 where d is the size of the arc length and h the radius of the circle in the form d 100 m h i 492 105 0 203 x 106 rad X m 492 km XIV 8 PHYS 2020 General Physics II D Spectral Resolution 1 Earlier we mentioned that prisms disperse light through refrac tion Diffraction gratings also can disperse light through diffrac tion see Xlll of these notes a Diffraction gratings have parallel lines etched scratched on the surface of a glass or some other transparent ma terial plate b These etched lines act like a series of parallel slits 2 Diffraction is caused by interference of many light waves By us ing the mathematics of interference patterns the grating equa tion results dsin639m m012 XIV 9 a d is the slit spacing lN where N is the number of slits or lines per cm b 6 is the grating angle or angle of deviation c m is the order number 3 Spectral resolution Aka is typically measured in A or nm Spectral resolution measures how close together 2 spectral lines can be separated in a spectrograph 4 The resolving power R is related to the spectral resolution Via A A R277 Ag Al AAG XIV10 Donald G Luttermoser ETSU XIV 9 a The higher the resolving power the ner the detail seen in a spectrum the better the spectral resolution b For a grating the resolving power a unitless number is directly proportional to the line density on the grating R Nm X1 cm XIV 11 where N is measured in inverse centimeters Example XIV73 Problems 2439 Page 815 from the Serway amp Faughn textbook A light source emits two major spec tral lines an orange line of wavelength 610 nm and a blue green line of wavelength 480 nm If the spectrum is resolved by a diffraction grating having 5000 linescm and viewed on a screen 200 m from the grating what is the distance in cm between the two spectral lines in the second order spectrum Solution The grating spacing is d lN 1 cm5000 200gtlt 10 4 cm 200 X 10 6 m and Eq XIV 9 gives the angular position of the second order m 2 spectral line as 2A 2A sin6 F or 6 sin 1 For the given wavelengths the angular positions are 2610 x 109 m 1 o 376 61 8111 200 x 106 m l 2480 x 109 m 0 1 287 2 m 200 x 106 m l lf L is the distance from the grating to the screen the distance on the screen from the central maximum to a second order bright line is y Ltan6 Therefore for the two given wavelengths the XIV 10 PHYS 2020 General Physics II separation of the lines on the screen is Ay Ltan61 tan6392 200 m tan3760 tan287o 0445 m PHYS2020 General Physics II Course Lecture Notes Section XIV Dr Donald G Luttermoser East Tennessee State University Edition 31 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2020 General Physics taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics Enhanced 7th Editizm 2006 textbook by Serway Faughn and Vuille XIV Optical Instruments A Common Optical Tools 1 A camera consists of a lens or series of lenses that focus an image onto a light sensing detector 69 lm CCD etc a The diameter of the lens can be stopped down to smaller 0 C sizes by adjusting the aperture opening i These stops are technically referred to as the f number f number 2 XIV 1 f focal length of lens D diameter of the lens or aperture whichever is smaller ii The larger the f number the less light is allowed into the camera iii Cameras typically have the following f stops f 28 lets the most light in f4 f56 f8 fll f16 lets the least amount of light in The magni cation of the image follows the thin lens for mulae 69 Eq XII 10 with an image size given by h f XIV 2 p The amount of light that falls on the detector is deter mined by the f stop w the shutter speed i For fast moving objects one should use shutter speeds llOOOth or l500th of a second ii For indoor low light levels speeds greater than 1 60th of a second are required gt speeds longer XIV 1 XIV 2 PHYS 2020 General Physics II than 160 sec usually require the camera to be mounted on a tripod to keep the camera steady iii Astronomical photographs typically require the shutter to be open from minutes to hours in order to record the image 2 The organic analogy to the camera is the eye In the case of the human eye we have the following characteristics a The iris is equivalent to the aperture of the camera gt the opening is called the pupil b The cornea is a transparent lens cap for the eye s lens c The retina records the image and sends the signal to the brain via the optic nerve gt analogous to the lm in a camera or a CCD chip The retina is composed of 2 types of receptor cells i Rods are able to detect low light levels in black and white ii Cones come in 3 types that respond respectively to red green and blue light 1 Defects of the eye i Hyperopia Image of a nearby object forms be hind the retina farsighted ii Myopia Maximum focal length of eye lens is shorter than diameter of eye nearsighted iii AstigmatismA point source produces a line im Donald G Luttermoser ETSU XIV 3 age on the retina iv Cataracts Lens andor cornea becomes par tially or totally opaque v Glaucoma Abnormal increase in uid pressure inside the eyeball e Corrective lenses are often used to correct many defects of the eye The power of such a lens is measured in diopters 73diopters XIV 3 1 f In gt f is the lens focal length in meters i The near point is the closest distance for which the lens can accommodate to focus light on the retina This distance changes as the eye ages Age 10 near point 18 cm Age 20 near point 25 cm Age 40 near point 50 cm Age 60 near point 500 cm ii The far point is the farthest distance for which the lens of a relaxed eye can focus light on the retina People with normal Vision can focus ob jects that are very far away 69 the Moon For these people the far point is essentially in nity 3 The Simple Magni er increases the image size of an object with a single lens e magnifying glass with a magni cation of XIV 4 00 gt 6 angle the object subtends with the lens XIV 4 PHYS 2020 General Physics II gt 00 angle the object subtends to the naked eye B Microscopes and Telescopes 1 Both microscopes and refractlng telescopes have a series of 2 or more lenses that are used to magnify an image a The objective lens is the light collector gt the bigger the objective the more light you receive the fainter the object you can see b The eyepiece lens is the main component of magni ca tion c A re ecting telescope has a mirror usually parabolic in shape to collect the light and is called the primary mirror instead of the objective lens 2 These instruments magnify objects following the relations a For compound microscopes L 25 M Mlme 7 lt cm XIV5 f 0 fe M1 magni cation of the objective me magni cation of the eyepiece gt L distance between objective and eyepiece in cm f0 focal length of the objective in cm f6 focal length of the eyepiece in cm b For a telescope m XIV 6 i Note that one can get as high as a magni cation as you want by just getting eyepieces with smaller and smaller focal lengths Donald G Luttermoser ETSU XIV 5 ii The higher the magni cation the fainter the im age gt rule of thumb for useful magni cations 50 power for each inch of the objective gt a 24 telescope s maximum useful power is 120l Example XIVil Problems 2529 Page 840 from the Serway amp Faughn textbook Astronomers often take photographs with the objective lens or mirror of a telescope alone without an eye piece a Show that the image size M for this telescope is given by h fhf p where h is the object size f the objective focal length and p the object distance b Simplify the expression in part a if the object distance is much greater than the objective focal length c The quotwingspanquot of the International Space Station is 1086 m the overall width of its solar panel configuration When it is orbiting at an altitude of 407 km find the width of the image formed by the telescope objective of focal length 400 m Solution a Using the thin lens equation Eq XII 10 we get gig I9 qif lll q fp 1 i q pf pf qpf p f lnserting this into the magni cation equation Eq Xll ll we get XIV 6 PHYS 2020 General Physics II Therefore the image size will be fh fh h Mh 7 7 p f f p Solution b prgtgtfthenf pm pand hm 17 Solution c Suppose the telescope observes the space station while at the zenith e the point directly overhead then M iW rv p 4 407 x 103In C Spatial Resolution 1 The ability of an optical system to distinguish between closely spaced objects is limits due to the wave nature of light a Light gets diffracted when passing through apertures see Figures 2513 and 2514 in the textbook b When the central maximum of one image falls on the rst minimum of another image the images are said to just be resolved gt Rayleigh s criterion 2 The limiting angle which an object is viewed depends both on the size of the aperture and the wavelength of light Donald G Luttermoser ETSU XIV 7 a For a slit 0m g XIV 7 where a is the width of the slit opening Both the wave length x and a must be measured in the same units then 0m the resolving angle is in radians b For a circular aperture A m 1227 XIV8 0 D where D is the circular aperture diameter Again A and D must be in the same units to give 0m in radians 3 Turbulence in the Earth s atmosphere limits the spatial resolu tion of Earth based telescopes to about 1 arcsecondl Example XIV72 Problems 2533 Page 840 from the Serway amp Faughn textbook A converging lens with a diameter of 300 cm forms an image of a satellite passing overhead The satellite has two green lights wavelength 500 nm spaced 100 m apart If the lights can just be resolved according to the Rayleigh criterion what is the altitude of the satellite Solution Since the lens corresponds to a circular aperture we can just use Eq XIV 8 to determine the angular separation 0 0m 122 122 203 x 106 rad The altitude can then be determined from the arc length equation e d h639 where d is the size of the arc length and h the radius of the circle in the form d 100 m h i 492 105 0 203 x 106 rad X m 492 km XIV 8 PHYS 2020 General Physics II D Spectral Resolution 1 Earlier we mentioned that prisms disperse light through refrac tion Diffraction gratings also can disperse light through diffrac tion see Xlll of these notes a Diffraction gratings have parallel lines etched scratched on the surface of a glass or some other transparent ma terial plate b These etched lines act like a series of parallel slits 2 Diffraction is caused by interference of many light waves By us ing the mathematics of interference patterns the grating equa tion results dsin639m m012 XIV 9 a d is the slit spacing lN where N is the number of slits or lines per cm b 6 is the grating angle or angle of deviation c m is the order number 3 Spectral resolution Aka is typically measured in A or nm Spectral resolution measures how close together 2 spectral lines can be separated in a spectrograph 4 The resolving power R is related to the spectral resolution Via A A R277 Ag Al AAG XIV10 Donald G Luttermoser ETSU XIV 9 a The higher the resolving power the ner the detail seen in a spectrum the better the spectral resolution b For a grating the resolving power a unitless number is directly proportional to the line density on the grating R Nm X1 cm XIV 11 where N is measured in inverse centimeters Example XIV73 Problems 2439 Page 815 from the Serway amp Faughn textbook A light source emits two major spec tral lines an orange line of wavelength 610 nm and a blue green line of wavelength 480 nm If the spectrum is resolved by a diffraction grating having 5000 linescm and viewed on a screen 200 m from the grating what is the distance in cm between the two spectral lines in the second order spectrum Solution The grating spacing is d lN 1 cm5000 200gtlt 10 4 cm 200 X 10 6 m and Eq XIV 9 gives the angular position of the second order m 2 spectral line as 2A 2A sin6 F or 6 sin 1 For the given wavelengths the angular positions are 2610 x 109 m 1 o 376 61 8111 200 x 106 m l 2480 x 109 m 0 1 287 2 m 200 x 106 m l lf L is the distance from the grating to the screen the distance on the screen from the central maximum to a second order bright line is y Ltan6 Therefore for the two given wavelengths the XIV 10 PHYS 2020 General Physics II separation of the lines on the screen is Ay Ltan61 tan6392 200 m tan3760 tan287o 0445 m

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