Physics 41 Chapter 38 HW Serway 9 th Edition

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1 Physics 4 Chapter 38 HW Serway 9 th Eition Questions: 3, 6, 8, Problems:, 4, 0,, 5,,, 9, 30, 34, 37, 40, 4, 50, 56, 57 *Q383 Answer () The power of the light coming through the slit ecreases, as you woul expect The central imum increases in with as the with of the slit ecreases n the conition sin for estructive interference on each sie of the central imum, increases as a ecreases *Q386 Answer (c) The ability to resolve light sources epens on iffraction, not on intensity Q388 We apply the equation m for the resolution of a circular aperture, the pupil of your eye D Suppose your ark-aapte eye has pupil iameter D 5 mm An average wavelength for visible light is 550 nm Suppose the healights are m apart an the car is a istance away Then m 4 m 0 so ~0 km The actual istance is less than this because the variabletemperature air between you an the car makes the light refract unpreictably The healights twinkle like stars Q38 Vertical Glare, as usually encountere when riving or boating, is horizontally polarize Reflecte light is polarize in the same plane as the reflecting surface As unpolarize light hits a shiny horizontal surface, the atoms on the surface absorb an then reemit the light energy as a reflection We can moel the surface as containing conuction electrons free to vibrate easily along the surface, but not to move easily out of surface The light emitte from a vibrating electron is partially or completely polarize along the plane of vibration, thus horizontally a Problems y P38 The positions of the first-orer minima are sin ± Thus, the spacing between these two minima is a Δ y a an the wavelength is 3 3 Δ y a 40 0 m m 06 m 547 nm P384 For estructive interference, 500 cm sin m 039 a a 360 cm an 798 y tan gives y ( ) tan 650 m tan m y 9 cm

2 P380 (a) Double-slit interference ima are at angles given by sin m For m 0, 0 0 m ( 80 μm ) sin ( μm ) : ( ) For, sin Similarly, for m, 3, 4, 5 an 6, 0, 3 35, 4 458, an 6 sin ( 07 ) Thus, there are irections for interference ima nonexistent (b) We check for missing orers by looking for single-slit iffraction minima, at asin m m, ( 0700 μm ) sin ( μm ) For an 458 Thus, there is no bright fringe at this angle There are only nine bright fringes, at 0, ± 03, ± 0, ± 35, an ± 636 (c) ( ) sin π sin At 0, At 03, sin an 00 π ( μ ) μ 0700 m sin ra m Similarly, at 0, At 35, At 636, sin ra 900 an 36 ra 35 an 393 ra 5 an The pupil of a cat s eye narrows to a vertical slit of with 0500 mm in aylight What is the angular resolution for horizontally separate mice? Assume that the average wavelength of the light is 500 nm P m 3 sin 00 0 ra 4 a 500 0

3 5 The mpressionist painter Georges Seurat create paintings with an enormous number of ots of pure pigment, each of which was approximately 00 mm in iameter The iea was to have colors such as re an green next to each other to form a scintillating canvas (Fig P387) Outsie what istance woul one be unable to iscern iniviual ots on the canvas? (Assume that 500 nm an that the pupil iameter is 400 mm) SuperStock Figure P385: Sunay Afternoon on the slan of a Grane Jatte, by Georges Seurat P385 By Rayleigh s criterion, two ots separate center-tocenter by 00 mm woul overlap when min D Thus, 3 3 ( 00 0 m )( m ) ( ) D m 3 m 8 A circular raar antenna on a Coast Guar ship has a iameter of 0 m an raiates at a frequency of 50 GHz Two small boats are locate 900 km away from the ship How close together coul the boats be an still be etecte as two objects? P388 c m D f D 0 m 9000 m ( m )( 9000 m ) 05 m 0 m P38The grating spacing is 00 0 m 0 6 m 4500 n the st-orer spectrum, iffraction angles are given by m sin : sin m so that for re 77 an for blue so that m sin m The angular separation is in first-orer, Δ n the secon-orer spectrum, Δ sin sin 3 Again, in the thir orer, 3 3 Δ sin sin 65 Since the re oes not appear in the fourth-orer spectrum, the answer is complete

4 A helium neon laser ( 638 nm) is use to calibrate a iffraction grating f the first orer imum occurs at 05, what is the spacing between ajacent grooves in the grating? P38 sin 0350 : ine spacing 8 μm 638 nm nm sin 0350 P m 380 nm 400 cm sin m or m y tan tan sin m sin an Thus, m m Δ y tan sin tan sin For m, Δ y ( 00 m ) tan sin tan sin 0554 mm For m, ( ) ( ) Δ y ( 00 m ) tan sin tan sin 54 mm For m 3, ( ) ( ) Δ y ( 00 m ) tan sin tan sin 504 mm Thus, the observe orer must be m P3830 sin m : ( ) sin m sin m nm m P3834 The average value of the cosine-square function is one-half, so the first polarizer transmits the light The secon 3 transmits cos f 3 3 i i The critical angle for total internal reflection for sapphire surroune by air is 344 Calculate the polarizing angle for sapphire P3837 sin c or n n 77 sin sin344 c

5 Also, tan p n Thus, ( ) p n ( ) tan tan P3840 For incient unpolarize light of intensity : The average value of the cosine-square function is one half, so the intensity after transmission by the first isk is After transmitting n isk: cos where the angle between the first an secon isk is ω t After transmitting 3r isk: cos cos ( 90 ) Using trigonometric ientities cos ( + cos ) an cos ( 90 ) sin ( cos) ( + cos ) ( cos ) we have ( cos ) ( cos4 ) 8 8 Since ω t, the intensity of the emerging beam is given by ( 4ω t) 6 *P384 (a) et 0 represent the intensity of unpolarize light incient on the first polarizer n Malus's law the average value of the cosine-square function is /, so the first filter lets through / of the incient intensity Of the light reaching them, the secon filter passes cos 45 / an the thir filter also cos 45 / The transmitte intensity is then 0 (/)(/)(/) 05 0 The reuction in intensity is by a factor of 0875 of the incient intensity (b) By the same logic as in part (a) we have transmitte 0 (/)(cos 30 )(cos 30 )(cos 30 ) 0 (/)(cos 30 ) Then the fraction absorbe is 0789 (c) Yet again we compute transmission 0 (/)(cos 5 ) An the fraction absorbe is 0670 () We can get more an more of the incient light through the stack of ieal filters, approaching 50%, by reucing the angle between the transmission axes of each one an the next

6 50 The Very arge Array (VA) is a set of 7 raio telescope ishes in Caton an Socorro Counties, New Mexico (Fig P3854) The antennas can be move apart on railroa tracks, an their combine signals give the resolving power of a synthetic aperture 360 km in iameter (a) f the etectors are tune to a frequency of 40 GHz, what is the angular resolution of the VA? (b) Clous of hyrogen raiate at this frequency What must be the separation istance of two clous at the center of the galaxy, lightyears away, if they are to be resolve? (c) What f? As the telescope looks up, a circling hawk looks own Fin the angular resolution of the hawk s eye Assume that that the hawk is most sensitive to green light having a wavelength of 500 nm an that it has a pupil of iameter 0 mm () A mouse is on the groun 300 m below By what istance must the mouse s whiskers be separate if the hawk can resolve them? P3850 (a) v : f m s 9 04 m 40 0 s (b) D 04 m 76 μra m min : min 4 min s 76 μra 50 arc secons π 6 min : ( )( ) min 76 0 ra ly 089 ly 508 μra ( 05 secons ofarc ) (c) min D m 0 0 m min 3 () ( )( ) 6 3 min ra 300 m 5 0 m 5 mm P3856 (a) We require min raius of iffraction isk D D Then D 44 (b) D ( )( ) m 050 m 48 μm

7 57 An American stanar television picture is compose of about 485 horizontal lines of varying light intensity Assume that your ability to resolve the lines is limite only by the Rayleigh criterion an that the pupils of your eyes are 500 mm in iameter Calculate the ratio of minimum viewing istance to the vertical imension of the picture such that you will not be able to resolve the lines Assume that the average wavelength of the light coming from the screen is 550 nm P3857 The limiting resolution between lines ( m ) ( m ) ra D min 3 Assuming a picture screen with vertical imension l, the minimum viewing istance for no visible lines is foun from 485 min l The esire ratio is then l ra min 4 ( ) 54