rgsdle (zdr8) HW13 ditmire (58335) 1 This print-out should hve 1 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. 001 (prt 1 of ) 10.0 points Consider the setup of single slit experiment. The wvelength of the incident light is λ 0 nm. The slit width nd the distnce between the slit nd the is specified in the figure. β 1 π nd the pth length difference is δ 1 λ. The smll ngle pproximtion gives us y 1 tn 1 1 sin 1 δ 1, or y 1 δ 1 λ 5.775 mm. 00 (prt of ) 10.0 points Denote the intensity on the t y 3.075 mm by I nd the intensity on the t y 0 by I 0. 60 µm S 8. m y 1 60 µm S 8. m 3.075 mm Find the position y y 1 of the first intensity minimum. Use smll ngle pproximtion sin tn. Correct nswer: 5.775 mm. Find the intensity rtio I I 0. Correct nswer: 0.681. I I 0 [ ] sin(β/). β/ et : λ 0 nm, 8. m, 60 µm. nd 1 0 π π π or 360 β S [ y δ sin ] y 1 For single slit diffrction, destructive interference occurs when, sin λ, or simply when, δ sin λ. Thus, between the two end rys which correspond to the first minimum, the phse ngle difference is 0 1 et: δ λ 0 90 180 70 360 β π δ λ sin λ R y 3.075 mm y 1 5.775 mm λ y (1) 0.59, ()
where y 1 is the position of the first intensity minimum. From Eq. (1), rgsdle (zdr8) HW13 ditmire (58335) β π y [ λ ] π (60 µm) (3.075 mm) 0 nm (8. m) 3.70708 rd 1., in greement with the bove digrm Following is the lterntive method. From Eq. (1) nd (), we hve R β β 1 δ δ 1 y y 1. β Rβ 1 (0.59) π 3.70708 rd (since β 1 π), the intensity rtio t ny point on the is I I 0 sin ( β β ) [ ] sin(rπ) Rπ [ sin(0.59π) 0.59π 0.681, in greement with the intensity digrm in the question. 003 10.0 points Hint: Use smll ngle pproximtion; e.g., sin tn nd cos 1. Consider the setup of double-slit experiment in the schemtic drwing below. Note: As cn be seen in the figure below, one of the double-slit interference mxim is locted t the first single-slit diffrction minimum. ] d S y Determine the rtio d ; i.e., the slit seprtion d compred to the slit width. 1. d correct. d 6 3. d 5. d 7 5. d 9 6. d 11 7. d 5 8. d 9. d 13 10. d 3 At y there is minimum for single-slit diffrction nd mxim for double-slit interference, s noted in the question. The first minimum for single-slit diffrction occurs when sin λ, (1)
rgsdle (zdr8) HW13 ditmire (58335) 3 nd the mxim for double-slit interference occur when sin (m) λ d. () The first diffrction minimum for single-slit diffrction nd the fourth double-slit interference mximum (m ) occur t the sme position y, s seen in the figure below. Figure: The dshed curve on the left of the is due to single slit interference. The dshed curve on the right of the is due to double slit interference. The position is zero mplitude nd the positive direction is reflected on either side of the. Since the single-slit diffrction minimum msks the fourth double-slit interference mxim, one must estimte where the fourth double-slit mxim occurs using the spcing between the double-slit interference pttern shown on the right-hnd side of the, s seen in the figure bove. Using Eq. 1 nd, we hve keywords: sin λ 1 (m) 1 d d (). 0 m /* If you use ny of these, fix the comment symbols. 00 10.0 points The lines in grting re uniformly spced t 1530 nm. Clculte the ngulr seprtion of the second order bright fringes between light of wvelength 600 nm nd 603.11 nm. Correct nswer: 6.58059 mrd. The eqution for the n th bright fringe is d sin n λ, for n 0, ±1, ±,, where is the ngulr displcement describing the n th fringe. Hence the ngulr difference between the second bright fringe (n ) of two different wvelengths (λ nd λ 1 ) of light is ( ) ( ) λ λ1 1 rcsin rcsin d d [ (6.0311 10 7 ] m) rcsin 1.53 10 6 m [ (6 10 7 ] m) rcsin 1.53 10 6 m 0.00658059 rd 6.58059 mrd. 005 10.0 points ight of wvelength 56 nm from mercury rc flls on diffrction grting ruled with 7700 lines/in. Wht is the ngulr seprtion between the first-order imges on either side of the centrl mximum? (Cution: do not use smll ngle pproximtion here.) Correct nswer: 73.088. For double-slit,.5 cm/in d 7700 lines/in 916.968 nm sin 1 λ d (56 nm) (916.968 nm) 0.5951,
rgsdle (zdr8) HW13 ditmire (58335) so 1 36.51. Since we re sked for the ngulr seprtion between first-order imges to either side of the centrl mximum, it should be 1 73.088. 006 10.0 points A bem of light is diffrcted by single slit. The distnce between the positions of zero intensity (m ±1) is.1 mm. 0.538 mm S.119 m.05 mm Estimte the wvelength of the lser light. Use smll ngle pproximtion sin tn. Correct nswer: 50.81 nm. tn m y m. In the smll ngle pproximtion, when is in rdins sin tn y m mλ y m mλ. The spcing between the minim m 1 nd m 1 is equl to b y 1 λ λ b 1 0.001 m (0.000538 m).119 m 109 nm m 50.81 nm. 007 10.0 points An unpolrized light bem with intensity of I 0 psses through polrizers shown in the picture. Unpolrized light Polrizer et :.119 m, y.05 mm, 0.538 mm. nd Trnsmission xis E0 Polrized lihgt Anlyzer E0cos S y The distnce y m from the m th minimum to the center of the first-order mximum cn be clculted from the formule sin m mλ, If 30,wht is the bem intensity fter the second polrizer? 1. I 5 8 I 0. I 9 3. I 5. I 1 8 I 0 5. I 3 8 I 0 correct
rgsdle (zdr8) HW13 ditmire (58335) 5 6. I 3 7. I 1 I 0 8. I 7 9. I 1 10. I 1 I 0 The bem intensity fter the first polrizer is I 1 I 0. We use the formul for the intensity of the trnsmitted (polrized) light. Thus the bem intensity fter the second polrizer is I I 1 cos I 0 cos (30 ) 3 I 0 8 008 (prt 1 of ) 10.0 points Consider 3 polrizers #1, #, nd #3 ordered sequentilly. The incident light is unpolrized with intensity I 0. The intensities fter the light psses through the subsequent polrizers re lbeled s I 1, I, nd I 3, respectively (see the sketch). I 0 I 1 I I 3 #1 # #3 Polrizers #1 nd #3 re crossed such tht their trnsmission xes re perpendiculr to ech other. Polrizer # is plced between the polrizers #1 nd #3 with its trnsmission xis t 60 with respect to the trnsmission xis of the polrizer #1 (see the sketch). #1 60 # #3 After pssing through polrizer # the intensity I (in terms of the intermedite intensity I 1 ) is 1. I I 1 correct. None of these. 3. I I 1 ( ) 3. I I 1 5. I I 1 6. I I 1 8 7. I I 1 3 When the light psses through the polrizer #1 it is polrized verticlly. Thus the ngle between its polriztion nd the orienttion of polrizer # is 60. Thus the trnsmitted intensity is I I 1 cos I 1 cos (60 ) 1 I 1. When polrized light psses through polrizer, the trnsmitted intensity is I I 1 cos, where is the ngle between the polriztion of the light (of I 1 ) nd the orienttion of the polrizer #. Thus I I 1. 009 (prt of ) 10.0 points Wht is the finl intensity I 3? 1. I 3 3
rgsdle (zdr8) HW13 ditmire (58335) 6. I 3 5 3 I 0 3. I 3 1 I 0. I 3 1 I 0 et : λ.55 10 7 m d 3.8 10 8 m. Applying Ryleigh s criterion, nd 5. I 3 1 6. I 3 0 7. I 3 3 3 I 0 correct 8. I 3 1 8 I 0 After the polrizer #1 I 1 I 0. After the polrizer # I I 1 cos (60 ) After the polrizer #3 I 1. I 3 I cos (90 60 ) ( ) 3 I ( ) ( ) 3 1 I 1 ( ) ( ) ( ) 3 1 1 I 0 3 3 I 0. 010 10.0 points If we were to send ruby lser bem (wvelength 55 nm) outwrd from the brrel of telescope whose dimeter is 3.6 m, wht would be the dimeter of the big red spot when the bem hit the Moon 3.8 10 5 km wy? Neglect tmospheric dispersion. Correct nswer: 13.13 m. min r D 1. λ d, where r is the rdius of the spot. So, for the dimeter x of the spot we obtin x r 1. λ D d (1.) (.55 10 7 m)(3.8 10 8 m) (3.6 m) 13.13 m. 011 10.0 points On the night of April 18, 1775, signl ws to be sent from the Old North Church steeple to Pul Revere, who ws.3 mi wy: One if by lnd, two if by se. Assume tht Pul Revere s pupils hd dimeter of.93 mm t night, nd tht the lntern light hd predominnt wvelength of 685 nm. At wht minimum seprtion did the sexton hve to set the lnterns so tht Revere could receive the correct messge? One mile is pproximtely equl to 1.609 km. Correct nswer: 1.067 m. The ngle of resolution for the Pul Revere s pupils is Therefore min 1. λ D d. d 1. λ D (685 nm) (.3 mi) 1..93 mm 1. (6.85 10 7 m) (373.88 m) 0.0093 m 1.067 m.
rgsdle (zdr8) HW13 ditmire (58335) 7 01 10.0 points Consider the setup of single slit experiment. Hint: Use smll ngle pproximtion; e.g., sin tn. S 10 y3 Determine the height y 3, where the third minimum occurs. 1. y 3 λ. y 3 5 λ 3. y 3 3 λ. y 3 λ 5. y 3 5 λ 6. y 3 9 λ 7. y 3 λ 8. y 3 λ 9. y 3 3 λ correct 10. y 3 7 λ The third minimum occurs t β 6 π, which corresponds to pth difference between two end rys: b 3 β k ( 6 π ) π λ 3 λ b 3 y 3 y 3 b 3 3 λ. 013 10.0 points A converging lens with dimeter of 9.9 cm forms n imge of stellite pssing overhed. The stellite hs two green lights (wvelength 505 nm) spced 1 m prt. If the lights cn just be resolved ccording to the Ryleigh criterion, wht is the ltitude of the stellite? Correct nswer: 809.933 km. Given : d 1 m, D 9.9 cm 0.99 m, nd λ 505 nm 5.05 10 7 m. The ngulr resolution is m 1. λ D Thus the ltitude is h d 1. 5.05 10 7 m 0.99 m 1.367 10 6 rd. 1 m 1.367 10 6 rd 809.933 km. 1 km 1 10 3 m 01 10.0 points A binry str system in the constelltion Orion hs n ngulr seprtion between the two strs of 1.3 10 5 rd. If the wvelength is 70 nm, wht is the smllest dimeter telescope cn hve nd
just resolve the two strs? Correct nswer: 0.06555 cm. et : λ 70 nm. rgsdle (zdr8) HW13 ditmire (58335) 8 1. λ D, D 1. λ 70 nm 1. 1.3 10 5 rd 10 cm 10 9 nm 0.06555 cm.