Chapter L - Problems

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1 Chpter L - Problems Blinn College - Physics 46 - Terry Honn Problem L.1 Young's ouble slit experiment is performe by shooting He-Ne lser bem (l 63.8 nm) through two slits seprte by 0.15 mm onto screen 8 m wy. () Wht is the istnce between bright fringes on the screen? (b) Wht is the istnce between rk fringes? (c) Wht is the istnce from the center of the centrl mximum to the thir rk fringe? () Wht is the smllest istnce from the centrl mximum where the intensity is 3/4 its mximum? Solution to L.1 l 63.8µ10-9 m, L 8 m n 0.15µ10-3 m () sin q m l Assuming the smll ngle pproximtion sin q y L justifie if the y vlue is much less thn L.) we get y m l. (The smll ngle pproximtion will be L The bright fringes re t integer vlues of m so between bright fringes we hve Dm 1. y m l L ï Dy Dm l L cm Note tht m ` 5 m. This justifies the smll ngle pproximtion. (b) The istnce between rk bns is the sme s the istnce between bright fringes. DJm + 1 N Dm 1 ï Dy 3.37 cm (c) Still ssume the smll ngle pproximtion. For estructive interference (interference minimum) we get: sin q Jm + 1 N l ï y L Jm + 1 N l. Note tht the vlues of Jm + 1 N re 0.5, 1.5,.5,..., so the thir minimum is Jm + 1 N.5. () The intensity formul is Using the smll ngle pproximtion we get y Jm + 1 N l L.5 l L I I 0 cos K p sin q O l I I 0 cos K p y l L O 8.43 cm 3/4 the mximum intensity mens tht I I

2 Chpter L - Problems cos -1 I I 0 p y l L ï y 5.6 mm Mke sure your clcultor is in the rins moe for this clcultion. Problem L. 700 nm light shines through single slit of with 1.µ10-5 m n shines on screen 55 cm wy. () Wht is the istnce from the center to the thir mximum? (Do not use the smll ngle pproximtion here? (b) Repet prt () using the smll ngle pproximtion. Solution to L. 1.µ10-5 m, l 700µ10-9 m n L 55 cm () The centrl mximum is t y 0. For constructive interference, using the smll ngle pproximtion, we get: sin q m l n m 3 ï q sin -1 m l tn q y L ï y L tn q 0.55 tn q m (b) With the smll ngle pproximtion y L m l n m 3 ï y m l L m Problem L.3 When 63.8 nm light shines through iffrction grting the first bright line is observe t eflection ngle of 15. Wht is the spcing between the slits in the grting? Solution to L.3 For iffrction grting the conition for constructive interference is the sme s for the single slit, sin q m l but there is estructive interference everywhere else. The first bright fringe correspons to m 1. sin q m l, m 1, q 15 n l 63.8µ10-9 m ï.44µ10-6 m Problem L.4 The visible spectrum vries between l violet 400 nm n l re 700 nm. Consier iffrction grting with n rbitrry slit seprtion. Show tht for white light the m 1 mximum gives pttern tht oesn't overlp the m mximum. Also show tht the m n m 3 ptterns lwys overlp. Solution to L.4 First consier the m 1 n m mxim.

3 Chpter L - Problems 3 sin q m l ï sin q m l It is sufficient to consier the extremes only. sin q 1,re 1 l re 700 nm n sin q,violet l violet 800 nm Thus for ny we hve sin q 1,re < sin q,violet. Since sine is n incresing function (in the first qurnt) we cn conclue tht Now consier the m n m 3 mxim. q 1,re < q,violet. sin q,re l re 1400 nm n sin q 3,violet 3 l violet 100 nm Thus for ny we hve sin q 3,violet < sin q,re, giving q 3,violet < q,re. Problem L.5 A iffrction grting hs 300 linesêmm. Wht is the highest orer violet (400 nm) line seen? Wht is the highest orer re line (700 nm). Solution to L.5 If there re 50 lines/mm, then the slit spcing is mm µ µ10-6 m nm Here the smll ngle pproximtion cnnot be use. The lrgest ngle is 90 so we hve sin q 1. The conition for seeing the m th bright fringe is: sin q m l n sin q 1 ï sin q m l The highest orer fringe is the lrgest m (n integer) tht stisfies this inequlity. l 400 nm ï m 1 ï m l nm 400 nm ï m mx 8 (Note tht since êl is rel number the istinction between ' ' n '<' in m l shoul not be importnt. In this problem it mtters; it chnges the nswer. Whether the nswer shoul be 10 or 9 is mtter of interprettion n the problem is mbiguous.) l 700 nm ï m nm 700 nm 4.76 ï m mx 4 Problem L nm light psses through single verticl slit of with 0.16 mm. Wht is the with of the centrl mximum observe on screen 1. m wy?

4 4 Chpter L - Problems Solution to L.6 l 550µ10-9 m, L 1. m n 0.16µ10-3 m The conition for estructive interference (rk fringes) for single slit is sin q m l, where m ±1, ±, ±3,... Note tht m 0 is not rk fringe; it is the center of the centrl bright fringe. The the centrl bright fringe is from m -1 to m +1. Use the smll ngle pproximtion. sin q m l ï y L m l ï y m l L The with of the centrl mximum is y 8.5 mm. 1µ550µ10-7 µ µ µ10-3 Problem L.7 When He-Ne lser bem (l 63.8 nm) psses through nrrow verticl slit iffrction pttern is observe on screen 80 cm wy. If the istnce from the first rk fringe to the thir is.5 mm then wht is the with of the slit? Solution to L.7 The conition for estructive interference with the smll ngle pproximtion becomes sin q m l ï y L m l y m l L ï Dy Dm l L ï H3-1L 63.8µ10-9 µ0.80 ï mm Problem L.8 When light of wvelength l psses through nrrow verticl slit the highest orer rk fringe observe (over ll possible ngles) is 5. (The fifth fringe cn be seen n the sixth cnnot.) Wht cn one conclue bout the with of the slit? Solution to L.8 The smll ngle pproximtion cnnot be use. sin q m l. To o this use the fct tht sin q 1 for m 5 n sin q 1 for m 6. sin q m l ï 5 l 1 n 6 l 1 ï 5 l 6 l Problem L.9 A sop film in ir hs thickness of 10 nm. Wht visible wvelengths re strongly reflecte by this film? Tke the inex of the sop film to be the sme s for wter, The visible spectrum is between 400 nm n 700 nm. Solution to L.9

5 Chpter L - Problems 5 Strongly reflecte implies constructive interference. The thickness of the film is t 10 nm. The sop is between ir lyers on both sies so n > n 1. n > n n const. int. ï t Jm + 1 N l n ï l n t µ1.33µ10 nm nm Jm+ 1 N Jm+ 1 N Jm+ 1 N For vlues of Jm + 1 N 0.5, 1.5,.5,... we get list of possible wvelengths. l 638 nm, 13 nm, 18 nm,... Since the visible spectrum is between 400 nm n 700 nm, the only visible wvelength in the list is 638 nm. Problem L.10 Re light (l 650 nm) is normlly incient on thin oil lyer (n 1.5) tht sits on pule of wter. () Wht is the smllest nonzero thickness tht will strongly reflect the re light? (b) Wht is the smllest nonzero thickness tht will minimlly reflect the re light? Solution to L.10 l 650 nm. The inex of the film is n 1.5, which is lrger thn tht of wter In 1.33M, so n < n () Strongly reflect implies constructive interference n t m l ï t m l mµ60 nm n The vlues of m re 0, 1,,... so the smllest nonzero thickness is when m 1. (b) Minimlly reflect implies estructive interference t min 60 nm n t Jm + 1 N l ï t Jm + 1 N l n Jm + 1 Nµ60 nm Since the vlues of m + 1 re 0.5, 1.5,.5,.... It follows the the minimum is when m t min 130 nm Problem L.11 You re ske to esign n ntiglre coting for computer monitor. Sitting on the glss (n 1.50) screen is thin coting of mgnesium fluorie with n inex of refrction of To be n ntiglre coting it must stisfy two conitions. It must give estructive interference for light in the mile of the visible spectrum, t 550 nm. Seconly, there must be no constructive interference for ny wvelength in the visible spectrum, from 400 nm to 700 nm. First fin the ifferent thicknesses tht give the estructive interference n then use the lck of consructive interference to select the thickness uniquely. Solution to L.11 Here we hve n 1.38 for the coting n the glss is n For estructive interference t 550 nm:

6 6 Chpter L - Problems n < n n est. int. ï n t Jm + 1 N l. ï t Jm + 1 N l n Jm + 1 N nm Jm + 1 N 1, 3, 5, 7,... ï t nm, nm, nm,... Thus the minimum thickness is t nm n the others re 3 t 0, 5 t 0, 7 t 0,... Now consier the constructive interference conition for the smllest thickness t nm n < n n est. int. ï n t 0 m l ï l n t 0 m 75 nm m m 1,, 3,..... ï l 76 nm, 138 nm, 9 nm,.... The visible spectrum is between 400 nm n 700 nm, so there re no visible wvelengths tht re intensifie. It shoul lso be cler tht if we choose other thicknesses tht give estructive interference 3 t 0, 5 t 0, 7 t 0,..., then we will multiply the bove wvelength list by 3, 5, 7, 9,... This will lwys give constructive interference t some visible wvelength nm The unique thickness is thus Problem L.1 Unpolrize light with n intensity of 4000 Wëm psses through three polrizing filters, the first is t 15 ngle from verticl, the secon is 40 from verticl n the thir is horizontl. Wht re the intensities between the first n secon filters, between the secon n thir filters n fter the thir filter? Solution to L.1 When polrize light with intensity I 0 psses through polrizing filter it leves with hlf the intensity I I 0 ë n is polrize long the xis of the filter. It follows tht the intensity between the first n secon filters is: I 1 I W m, n this is polrize t q When polrize light with intensity I 0 psses through filter it leves polrize long the xis of the filter n with intensity I I 0 cos q, where q is the ngle between the polriztion of the light n the xis of the filter. After the secon filter t q 40 the intensity is I I cos Hq - q 1 L 000 cos H40-15 L W m, The sme nlysis gives the intensity fter the thir filter. I I cos Iq 3 - q M cos H90-40 L 679 W m, Problem L.13 When the sun is wht ngle bove the horizon will sunlight reflecte off still pon be totlly polrize? Solution to L.13

7 Chpter L - Problems 7 The polriztion ngle for light from ir reflecting off wter is q p. tn q p n 1.33 n 1 1 ï q p 53.1 This ngle is the incient ngle q 1 which is is mesure from the norml. The ngle bove the horizon is its complement

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