ms ms Comparing the light speeds in sapphire and diamond, we obtain ms m

Transcription

1 Chapter 35. The idex of refractio is foud fro Eq. 35-3: c = = v s s = Note that Sell s Law (the law of refractio) leads to θ = θ whe =. The graph idicates that θ = 30 (which is what the proble gives as the value of θ ) occurs at =.5. Thus, =.5, ad the speed with which light propagates i that ediu is 8 c s v = = = s. 3. Coparig the light speeds i sapphire ad diaod, we obtai v= vs vd = c = ( ) = s d (a) The frequecy of yellow sodiu light is s s. f c = = s = Hz. (b) Whe travelig through the glass, its wavelegth is = = = (c) The light speed whe travelig through the glass is ( )( ) v= f l = = Hz s. 5. (a) We take the phases of both waves to be zero at the frot surfaces of the layers. The phase of the first wave at the back surface of the glass is give by φ = k L ωt, where k (= π/ ) is the agular wave uber ad is the wavelegth i glass. Siilarly, the phase of the secod wave at the back surface of the plastic is give by φ = k L ωt, where k (= π/ ) is the agular wave uber ad is the wavelegth i plastic. The agular frequecies are the sae sice the waves have the sae wavelegth i air ad the 37

2 37 CHAPTER 35 frequecy of a wave does ot chage whe the wave eters aother ediu. The phase differece is φ φ = ( k k) L= p L. l l Now, = air /, where air is the wavelegth i air ad is the idex of refractio of the glass. Siilarly, = air /, where is the idex of refractio of the plastic. This eas that the phase differece is π φ φ = ( ) L. The value of L that akes this 5.65 rad is L = b air 9 φ φg = = b g c h. air. b (b) 5.65 rad is less tha π rad = 6.8 rad, the phase differece for copletely costructive iterferece, ad greater tha π rad (= 3.4 rad), the phase differece for copletely destructive iterferece. The iterferece is, therefore, iterediate, either copletely costructive or copletely destructive. It is, however, closer to copletely costructive tha to copletely destructive. 6. I cotrast to the iitial coditios of Proble 35-5, we ow cosider waves W ad W with a iitial effective phase differece (i wavelegths) equal to, ad seek positios of the sliver which cause the wave to costructively iterfere (which correspods to a iteger-valued phase differece i wavelegths). Thus, the extra distace L traveled by W ust aout to 3,, ad so o. We ay write this requireet succictly as + L= where = 0,,, K. 4 (a) Thus, the sallest value of L / that results i the fial waves beig exactly i phase is whe =0, which gives L / = / 4 = 0.5. (b) The secod sallest value of L / that results i the fial waves beig exactly i phase is whe =, which gives L / = 3/ 4 = (c) The third sallest value of L / that results i the fial waves beig exactly i phase is whe =, which gives L / = 5/4= The fact that wave W reflects two additioal ties has o substative effect o the calculatios, sice two reflectios aout to a (/) = phase differece, which is g 6.

3 373 effectively ot a phase differece at all. The substative differece betwee W ad W is the extra distace L traveled by W. (a) For wave W to be a half-wavelegth behid wave W, we require L = /, or L = /4 = (60 )/4 =55 usig the wavelegth value give i the proble. (b) Destructive iterferece will agai appear if W is 3 behid the other wave. I this case, L = 3, ad the differece is 3 60 L L= = = = (a) The tie t it takes for pulse to travel through the plastic is t L L L L 630. L = =. c 55. c 70. c 60. c 45. c Siilarly for pulse : t L L L 633. L = + + =. c 59. c 65. c 50. c Thus, pulse travels through the plastic i less tie. (b) The tie differece (as a ultiple of L/c) is 633. L 630. L 003. L t = t t = =. c c c Thus, the ultiple is (a) Eq. 35- (i absolute value) yields L = 6 c h = 9 b g (b) Siilarly, L = 6 c h = 9 b g (c) I this case, we obtai L = 6 c35. 0 h = 9 b g

4 374 CHAPTER 35 (d) Sice their phase differeces were idetical, the brightess should be the sae for (a) ad (b). Now, the phase differece i (c) differs fro a iteger by 0.30, which is also true for (a) ad (b). Thus, their effective phase differeces are equal, ad the brightess i case (c) should be the sae as that i (a) ad (b). 0. (a) We ote that ray travels a extra distace 4L ore tha ray. To get the least possible L which will result i destructive iterferece, we set this extra distace equal to half of a wavelegth: L= L= = = (b) The ext case occurs whe that extra distace is set equal to 3. The result is 3 3(40.0 ) L = = = (a) We wish to set Eq. 35- equal to /, sice a half-wavelegth phase differece is equivalet to a π radias differece. Thus, L i 60 = = b g b g = 550 = 55. µ. (b) Sice a phase differece of 3 (wavelegths) is effectively the sae as what we required i part (a), the L = 3 = 3Li = 355. µ = 465. µ. b g b g. (a) The exitig agle is 50º, the sae as the icidet agle, due to what oe ight call the trasitive ature of Sell s law: siθ = siθ = 3 siθ 3 = (b) Due to the fact that the speed (i a certai ediu) is c/ (where is that ediu s idex of refractio) ad that speed is distace divided by tie (while it s costat), we fid t = L/c = (.45)(5 0 9 )/( /s) = s = 0.4 ps. 3. (a) We choose a horizotal x axis with its origi at the left edge of the plastic. Betwee x = 0 ad x = L the phase differece is that give by Eq. 35- (with L i that equatio replaced with L ). Betwee x = L ad x = L the phase differece is give by a expressio siilar to Eq. 35- but with L replaced with L L ad replaced with (sice the top ray i Fig is ow travelig through air, which has idex of refractio

5 375 approxiately equal to ). Thus, cobiig these phase differeces with = µ, we have L L L 3.50 µ 4.00 µ 3.50 µ µ µ = ( ) + ( ) = (.60.40) + (.40) (b) Sice the aswer i part (a) is closer to a iteger tha to a half-iteger, the iterferece is ore early costructive tha destructive. 4. (a) We use Eq with = 3: F θ = H G I K J si = (b) θ = (0.6) (80 /π) =.4. L N c h O M QP = si 6 d rad. 5. Iterferece axia occur at agles θ such that d si θ =, where is a iteger. Sice d =.0 ad = 0.50, this eas that si θ = 0.5. We wat all values of (positive ad egative) for which 0.5. These are 4, 3,,, 0, +, +, +3, ad +4. For each of these except 4 ad +4, there are two differet values for θ. A sigle value of θ ( 90 ) is associated with = 4 ad a sigle value (+90 ) is associated with = +4. There are sixtee differet agles i all ad, therefore, sixtee axia. 6. (a) For the axiu adjacet to the cetral oe, we set = i Eq ad obtai ( )( l ) l θ = = = si si 0.00 rad. d = 00l (b) Sice y = D ta θ (see Fig. 35-0(a)), we obtai y = (500 ) ta (0.00 rad) = 5.0. The separatio is y = y y 0 = y 0 = The agular positios of the axia of a two-slit iterferece patter are give by dsiθ =, where d is the slit separatio, is the wavelegth, ad is a iteger. If θ is sall, si θ ay be approxiated by θ i radias. The, θ = /d to good approxiatio. The agular separatio of two adjacet axia is θ = /d. Let ' be the wavelegth for which the agular separatio is greater by0.0%. The,.0/d = '/d. or ' =.0 =.0(589 ) = 648.

6 376 CHAPTER I Saple Proble 35-, a experietally useful relatio is derived: y = D/d. Dividig both sides by D, this becoes θ = /d with θ i radias. I the steps that follow, however, we will ed up with a expressio where degrees ay be directly used. Thus, i the preset case, θ 00. θ = = = = = 05.. d d The coditio for a axiu i the two-slit iterferece patter is d si θ =, where d is the slit separatio, is the wavelegth, is a iteger, ad θ is the agle ade by the iterferig rays with the forward directio. If θ is sall, si θ ay be approxiated by θ i radias. The, θ = /d, ad the agular separatio of adjacet axia, oe associated with the iteger ad the other associated with the iteger +, is give by θ = /d. The separatio o a scree a distace D away is give by Thus, y = D θ = D/d. 9 c500 0 hb 540. g y = = = (a) The phase differece (i wavelegths) is φ = d siθ/ = (4.4 µ)si(0 )/(0.500 µ) =.90. (b) Multiplyig this by π gives φ = 8. rad. (c) The result fro part (a) is greater tha 5 (which would idicate the third iiu) ad is less tha 3 (which would correspod to the third side axiu).. Iitially, source A leads source B by 90, which is equivalet to 4 wavelegth. However, source A also lags behid source B sice r A is loger tha r B by 00, which is = 4 wavelegth. So the et phase differece betwee A ad B at the detector is zero.. (a) We use Eq to fid d: d siθ = d = (4)(450 )/si(90 ) = 800. For the third order spectru, the wavelegth that correspods to θ = 90 is = d si(90 )/3 = 600. Ay wavelegth greater tha this will ot be see. Thus, 600 < θ 700 are abset.

7 377 (b) The slit separatio d eeds to be decreased. (c) I this case, the 400 wavelegth i the = 4 diffractio is to occur at 90. Thus d ew siθ = d ew = (4)(400 )/si(90 ) = 600. This represets a chage of d = d d ew = 00 = 0.0 µ. 3. Let the distace i questio be x. The path differece (betwee rays origiatig fro S ad S ad arrivig at poits o the x > 0 axis) is F HG I K J d + x x = + where we are requirig destructive iterferece (half-iteger wavelegth phase differeces) ad = 0,,, L. After soe algebraic steps, we solve for the distace i ters of : d + x =. + 4 To obtai the largest value of x, we set = 0: ( 3.00) b g b g d 3 x 0 = = = 8.75 = 8.75(900 ) = = 7.88µ Iagie a y axis idway betwee the two sources i the figure. Thirty poits of destructive iterferece (to be cosidered i the xy plae of the figure) iplies there are = 5 o each side of the y axis. There is o poit of destructive iterferece o the y axis itself sice the sources are i phase ad ay poit o the y axis ust therefore correspod to a zero phase differece (ad correspods to θ = 0 i Eq. 35-4). I other words, there are 7 dark poits i the first quadrat, oe alog the +x axis, ad 7 i the fourth quadrat, costitutig the 5 dark poits o the right-had side of the y axis. Sice the y axis correspods to a iiu phase differece, we ca cout (say, i the first quadrat) the values for the destructive iterferece (i the sese of Eq. 35-6) begiig with the oe closest to the y axis ad goig clockwise util we reach the x axis (at ay poit beyod S ). This leads us to assig = 7 (i the sese of Eq. 35-6) to the poit o the x axis itself (where the path differece for waves coig fro the sources is siply equal to the separatio of the sources, d); this would correspod to θ = 90 i Eq Thus, d = ( 7 + d ) = 7.5 = The axia of a two-slit iterferece patter are at agles θ give by d si θ =, where d is the slit separatio, is the wavelegth, ad is a iteger. If θ is sall, si θ

8 378 CHAPTER 35 ay be replaced by θ i radias. The, dθ =. The agular separatio of two axia associated with differet wavelegths but the sae value of is θ = (/d)( ), ad their separatio o a scree a distace D away is L D y = D D = NM O ta θ θ d QP L30 = M b. g P O =. c h N Q The sall agle approxiatio ta θ θ (i radias) is ade. b 6. (a) We ote that, just as i the usual discussio of the double slit patter, the x = 0 poit o the scree (where that vertical lie of legth D i the picture itersects the scree) is a bright spot with phase differece equal to zero (it would be the iddle frige i the usual double slit patter). We are ot cosiderig x < 0 values here, so that egative phase differeces are ot relevat (ad if we did wish to cosider x < 0 values, we could liit our discussio to absolute values of the phase differece, so that agai egative phase differeces do ot eter it). Thus, the x = 0 poit is the oe with the iiu phase differece. (b) As oted i part (a), the phase differece φ = 0 at x = 0. (c) The path legth differece is greatest at the rightost edge of the scree (which is assued to go o forever), so φ is axiu at x =. (d) I cosiderig x =, we ca treat the rays fro the sources as if they are essetially horizotal. I this way, we see that the differece betwee the path legths is siply the distace (d) betwee the sources. The proble specifies d = 6.00, or d/ = (e) Usig the Pythagorea theore, we have g D + ( x+ d) D + ( x d) φ = =.7 where we have plugged i D = 0, d = 3 ad x = 6. Thus, the phase differece at that poit is.7 wavelegths. (f) We ote that the aswer to part (e) is closer to 3 (destructive iterferece) tha to (costructive iterferece), so that the poit is iterediate but closer to a iiu tha to a axiu.

9 Cosider the two waves, oe fro each slit, that produce the seveth bright frige i the absece of the ica. They are i phase at the slits ad travel differet distaces to the seveth bright frige, where they have a phase differece of π = 4π. Now a piece of ica with thickess x is placed i frot of oe of the slits, ad a additioal phase differece betwee the waves develops. Specifically, their phases at the slits differ by x x x = b g where is the wavelegth i the ica ad is the idex of refractio of the ica. The relatioship = / is used to substitute for. Sice the waves are ow i phase at the scree, x bg = 4 or 9 7 7c550 0 h 6 x = = = The proble asks for the greatest value of x exactly out of phase which is to be iterpreted as the value of x where the curve show i the figure passes through a phase value of π radias. This happes as soe poit P o the x axis, which is, of course, a distace x fro the top source ad (usig Pythagoras theore) a distace d + x fro the botto source. The differece (i oral legth uits) is therefore d + x x, or (expressed i radias) is π ( d + x x). We ote (lookig at the leftost poit i the graph) that at x = 0, this latter quatity equals 6π, which eas d = 3. Usig this value for d, we ow ust solve the coditio π ( d ) x x + = π. Straightforward algebra the lead to x = (35/4), ad usig = 400 we fid x = 3500, or 3.5 µ. 9. The phasor diagra is show below. Here E =.00, E =.00, ad φ = 60. The resultat aplitude E is give by the trigooetric law of cosies: E = E + E E E cos 80 φ. b g Thus, b g b g b gb g E = cos 0 =. 65.

10 380 CHAPTER I addig these with the phasor ethod (as opposed to, say, trig idetities), we ay set t = 0 (see Saple Proble 35-4) ad add the as vectors: y y h v = 0cos cos 30 = 6. 9 = 0si si 30 = 4. 0 so that Thus, y = y + y = 7. 4 R h v β yv ta y = F H G I K J = y = y+ y = yr si ωt+ β = 7. 4 si ωt+ 33. h b g b g. Quotig the aswer to two sigificat figures, we have y 7si ( ωt 3 ) With phasor techiques, this aouts to a vector additio proble R ρ = A ρ + B ρ + C ρ where (i agitude-agle otatio) A ρ = 0 0, B ρ = 5 45, ad C ρ = 5 45, where the agitudes are uderstood to be i µv/. We obtai the resultat (especially efficiet o a vector-capable calculator i polar ode): ρ R = = 7. 0 b g b g b g b gb gb gb g which leads to b g bg ER = 7. µ Vsi ωt where ω = rad/s. 3. (a) We ca use phasor techiques or use trig idetities. Here we show the latter approach. Sice si a + si(a + b) = cos(b/)si(a + b/),

11 38 we fid E + E = E cos( φ / )si( ωt+ φ/ ) 0 where E o =.00 µv/, ω = rad/s, ad φ = 39.6 rad. This shows that the electric field aplitude of the resultat wave is E = E cos( φ / ) = (.00 µ V/) cos(9. rad) =.33 µ V/. 0 (b) Eq. 35- leads to at poit P, ad I = 4I cos ( φ / ) =.35 I 0 0 I = 4I cos (0) = 4I ceter 0 0 at the ceter. Thus, I / I ceter =.35/ 4 = (c) The phase differece φ (i wavelegths) is gotte fro φ i radias by dividig by π. Thus, φ = 39.6/π = 6.3 wavelegths. Thus, poit P is betwee the sixth side axiu (at which φ = 6 wavelegths) ad the seveth iiu (at which φ = 6 wavelegths). (d) The rate is give by ω = rad/s. (e) The agle betwee the phasors is φ = 39.6 rad = 70 (which would look like about 0 whe draw i the usual way). 33. I addig these with the phasor ethod (as opposed to, say, trig idetities), we ay set t = 0 (see Saple Proble 35-4) ad add the as vectors: y y h v b g b g = 0cos0 + 5cos cos 45 = 65. = 0si 0 + 5si si 45 = 40. so that y = y + y = R h v y β = = yh v ta 8.5. Thus, y y y y y ( ωt β) ( ωt ) = = R si + = 7si (a) Referrig to Figure 35-0(a) akes clear that θ = ta (y/d) = ta (0.05/4) =.93.

12 38 CHAPTER 35 Thus, the phase differece at poit P is φ =dsiθ/ = wavelegths, which eas it is betwee the cetral axiu (zero wavelegth differece) ad the first iiu ( wavelegth differece). Note that the above coputatio could have bee siplified soewhat by avoidig the explicit use of the taget ad sie fuctios ad akig use of the sall-agle approxiatio (taθ siθ). (b) Fro Eq. 35-, we get (with φ = (0.397)(π) =.495 rad) I = 4I cos ( φ / ) = I 0 0 at poit P ad I = 4I cos (0) = 4I ceter 0 0 at the ceter. Thus, I / I ceter = / 4 = Light reflected fro the frot surface of the coatig suffers a phase chage of π rad while light reflected fro the back surface does ot chage phase. If L is the thickess of the coatig, light reflected fro the back surface travels a distace L farther tha light reflected fro the frot surface. The differece i phase of the two waves is L(π/ c ) π, where c is the wavelegth i the coatig. If is the wavelegth i vacuu, the c = /, where is the idex of refractio of the coatig. Thus, the phase differece is L(π/) π. For fully costructive iterferece, this should be a ultiple of π. We solve F π L π π HG I K J = for L. Here is a iteger. The solutio is b + L = g. 4 To fid the sallest coatig thickess, we take = 0. The, L = = bg 8 = (a) We are dealig with a thi fil (aterial ) i a situatio where > > 3, lookig for strog reflectios; the appropriate coditio is the oe expressed by Eq Therefore, with legths i ad L = 500 ad =.7, we have

13 383 = L = 700 for = 850 for = 567 for = 3 45 for = 4 fro which we see the latter two values are i the give rage. The loger wavelegth (=3) is = 567. (b) The shorter wavelegth ( = 4) is = 45. (c) We assue the teperature depedece of the refractio idex is egligible. Fro the proportioality evidet i the part (a) equatio, loger L eas loger. 37. For costructive iterferece, we use Eq : L = b + g. For the sallest value of L, let = 0: 64 L0 = = = 7 = 0.7 µ ( ) (b) For the secod sallest value, we set = ad obtai ( ) + 3 L = = = 3L = µ = 0.35µ. 0 ( ) 38. (a) O both sides of the soap is a ediu with lower idex (air) ad we are exaiig the reflected light, so the coditio for strog reflectio is Eq With legths i, = L + = 3360 for = 0 0 for = 67 for = 480 for = for = for = 5 fro which we see the latter four values are i the give rage. (b) We ow tur to Eq ad obtai

14 384 CHAPTER 35 = L = 680 for = 840 for = 560 for = 3 40 for = for = 5 fro which we see the latter three values are i the give rage. 39. For coplete destructive iterferece, we wat the waves reflected fro the frot ad back of the coatig to differ i phase by a odd ultiple of π rad. Each wave is icidet o a ediu of higher idex of refractio fro a ediu of lower idex, so both suffer phase chages of π rad o reflectio. If L is the thickess of the coatig, the wave reflected fro the back surface travels a distace L farther tha the wave reflected fro the frot. The phase differece is L(π/ c ), where c is the wavelegth i the coatig. If is the idex of refractio of the coatig, c = /, where is the wavelegth i vacuu, ad the phase differece is L(π/). We solve L F HG I K J = b + g for L. Here is a iteger. The result is b + L = g. 4 To fid the least thickess for which destructive iterferece occurs, we take = 0. The, l L = = = ( ) 40. The situatio is aalogous to that treated i Saple Proble 35-6, i the sese that the icidet light is i a low idex ediu, the thi fil of acetoe has soewhat higher =, ad the last layer (the glass plate) has the highest refractive idex. To see very little or o reflectio, accordig to the Saple Proble, the coditio L= ( + ) l where = 0,,, K ust hold. This is the sae as Eq which was developed for the opposite situatio (costructive iterferece) regardig a thi fil surrouded o both sides by air (a very differet cotext tha the oe i this proble). By aalogy, we expect Eq to apply i this proble to reflectio axia. A ore careful aalysis such as that give i 35-7 bears this out. Thus, usig Eq with =.5 ad = 700 yields

15 385 L = 0, 80, 560, 840,0, K for the first several values. Ad the equatio show above (equivalet to Eq ) gives, with = 600, L = 0,360,600,840,080,Κ for the first several values. The lowest uber these lists have i coo is L = I this setup, we have < ad < 3, ad the coditio for destructive iterferece is L L= =, = 0,,,... Thus, we have L = (380 )(.34) = 08 ( = ) =. L = (380 )(.34) = 509 ( = ) For the wavelegth to be i the visible rage, we choose = with = I this setup, we have > ad > 3, ad the coditio for costructive iterferece is 4L L= + =, = 0,,,... + Thus, we have 4L = 4(35 )(.75) = 75 ( = 0) = 4 L / 3 = 4(35 )(.75) / 3 = 758 ( = ). 4 L / 5 = 4(35 )(.75) / 5 = 455 ( = ) For the wavelegth to be i the visible rage, we choose = with = I this setup, we have > ad > 3, ad the coditio for costructive iterferece is L, 0,,,... L = + = + = The third least thickess is (=)

16 386 CHAPTER 35 L 6 = + = 478. (.60) 44. I this setup, we have > ad > 3, ad the coditio for costructive iterferece is 4L L= + =, = 0,,,... + Thus, we get 4L = 4(85 )(.60) = 84 ( = 0) =. 4 L / 3 = 4(85 )(.60) / 3 = 608 ( = ) For the wavelegth to be i the visible rage, we choose = with = Whe a thi fil of thickess L ad idex of refractio is placed betwee aterials ad 3 such that > ad 3 > where ad 3 are the idexes of refractio of the aterials, the geeral coditio for destructive iterferece for a thi fil is L = = = L, 0,,,... where is the wavelegth of light as easured i air. Thus, we have, for = = L = (00 )(.40) = I this setup, we have > ad < 3, ad the coditio for destructive iterferece is L, 0,,,... L = + = + = The secod least thickess is (=) L 48 = + = 48. (.46) 47. I this setup, we have > ad < 3, ad the coditio for destructive iterferece is 4L = + = = + L, 0,,,...

17 387 Thus, 4L = 4(0 )(.46) = 6 ( = 0) =. 4 L / 3 = 4(0 )(.46) / 3 = 409 ( = ) For the wavelegth to be i the visible rage, we choose = with = I this setup, we have < ad < 3, ad the coditio for costructive iterferece is L, 0,,,... L = + = + = The secod least thickess is (=) L 587 = + = 39. (.34) 49. I this setup, we have < ad > 3, ad the coditio for destructive iterferece is L, 0,,,... L = + = + = The secod least thickess is (=) L 34 = + = 6. (.59) 50. I this setup, we have < ad > 3, ad the coditio for destructive iterferece is Therefore, 4L = + = = + L, 0,,,... 4L = 4(45 )(.59) = 639 ( = 0) = 4 L / 3 = 4(45 )(.59) / 3 = 880 ( = ). 4 L / 5 = 4(45 )(.59) / 5 = 58 ( = ) For the wavelegth to be i the visible rage, we choose =3 with = 58.

18 388 CHAPTER I this setup, we have > ad > 3, ad the coditio for costructive iterferece is L, 0,,,... L = + = + = The third least thickess is (=) L 38 = + = 73. (.75) 5. I this setup, we have < ad < 3, ad the coditio for costructive iterferece is L, 0,,,... L = + = + = The secod least thickess is (=) L 63 = + = 339. (.40) 53. The situatio is aalogous to that treated i Saple Proble 35-6, i the sese that the icidet light is i a low idex ediu, the thi fil has soewhat higher =, ad the last layer has the highest refractive idex. To see very little or o reflectio, accordig to the Saple Proble, the coditio l L= + where = 0,,,... ust hold. The value of L which correspods to o reflectio correspods, reasoably eough, to the value which gives axiu trasissio of light (ito the highest idex ediu which i this proble is the water). b g (Eq ) gives zero reflectio i this type of syste, the we (a) If L= + ight reasoably expect that its couterpart, Eq , gives axiu reflectio here. A ore careful aalysis such as that give i 35-7 bears this out. We disregard the = 0 value (correspodig to L = 0) sice there is soe oil o the water. Thus, for =,,..., axiu reflectio occurs for wavelegths l ( )( ) = L , 55, = =

19 389 We ote that oly the 55 wavelegth falls withi the visible light rage. (b) As rearked above, axiu trasissio ito the water occurs for wavelegths give by F 4L L= + + HG I K J = which yields = 08, 736, 44 for the differet values of. We ote that oly the 44 wavelegth (blue) is i the visible rage, though we ight expect soe red cotributio sice the 736 is very close to the visible rage. 54. For costructive iterferece (which is obtaied for = 600 ) i this circustace, we require k k L = = where k = soe positive odd iteger ad is the idex of refractio of the thi fil. Rearragig ad pluggig i L = 7.7 ad the wavelegth value, this gives k k(600 ) k = = = = 0.55k. 4L 4(7.7 ).88 Sice we expect >, the k = is ruled out. However, k = 3 sees reasoable, sice it leads to =.65, which is close to the typical values foud i Table 34-. Takig this to be the correct idex of refractio for the thi fil, we ow cosider the destructive iterferece part of the questio. Now we have L = (iteger) dest /. Thus, dest = (900 )/(iteger). We ote that settig the iteger equal to yields a dest value outside the rage of the visible spectru. A siilar reark holds for settig the iteger equal to 3. Thus, we set it equal to ad obtai dest = We solve Eq with =.33 ad = 600 for =,, 3, : L = 3, 338, 564, 789, K Ad, we siilarly solve Eq with the sae ad = 450 : L = 0,69, 338, 508, 677, K The lowest uber these lists have i coo is L = 338.

20 390 CHAPTER The situatio is aalogous to that treated i Saple Proble 35-6, i the sese that the icidet light is i a low idex ediu, the thi fil of oil has soewhat higher =, ad the last layer (the glass plate) has the highest refractive idex. To see very little or o reflectio, accordig to the Saple Proble, the coditio l L= + where = 0,,, K ust hold. With = 500 ad =.30, the possible aswers for L are L = 96, 88, 48, 673, 865,... Ad, with = 700 ad the sae value of, the possible aswers for L are L = 35, 404, 673, 94,... The lowest uber these lists have i coo is L = I this setup, we have < ad < 3, ad the coditio for iiu trasissio (axiu reflectio) or destructive iterferece is L, 0,,,... L = + = + = The secod least thickess is (=) L 587 = + = 39. (.34) 58. I this setup, we have < ad > 3, ad the coditio for axiu trasissio (iiu reflectio) or costructive iterferece is L, 0,,,... L = + = + = The secod least thickess is (=) L 34 = + = 6. (.59) 59. I this setup, we have > ad < 3, ad the coditio for axiu trasissio (iiu reflectio) or costructive iterferece is

21 39 L, 0,,,... L = + = + = The secod least thickess is (=) L 48 = + = 48. (.46) 60. I this setup, we have > ad > 3, ad the coditio for iiu trasissio (axiu reflectio) or destructive iterferece is L, 0,,,... L = + = + = The third least thickess is (=) L 38 = + = 73. (.75) 6. I this setup, we have < ad < 3, ad the coditio for iiu trasissio (axiu reflectio) or destructive iterferece is L, 0,,,... L = + = + = The secod least thickess is (=) L 63 = + = 339. (.40) 6. I this setup, we have > ad > 3, ad the coditio for iiu trasissio (axiu reflectio) or destructive iterferece is L, 0,,,... L = + = + = The third least thickess is (=) L 6 = + = 478. (.60)

22 39 CHAPTER I this setup, we have < ad > 3, ad the coditio for axiu trasissio (iiu reflectio) or costructive iterferece is Thus, we have 4L = + = = + L, 0,,,... 4L = 4(45 )(.59) = 639 ( = 0) = 4 L / 3 = 4(45 )(.59) / 3 = 880 ( = ). 4 L / 5 = 4(45 )(.59) / 5 = 58 ( = ) For the wavelegth to be i the visible rage, we choose =3 with = I this setup, we have < ad < 3, ad the coditio for axiu trasissio (iiu reflectio) or costructive iterferece is Thus, we obtai L = = = L, 0,,,... L = (380 )(.34) = 08 ( = ) =. L = (380 )(.34) = 509 ( = ) For the wavelegth to be i the visible rage, we choose = with = I this setup, we have > ad > 3, ad the coditio for iiu trasissio (axiu reflectio) or destructive iterferece is 4L L= + =, = 0,,,... + Therefore, 4L = 4(35 )(.75) = 75 ( = 0) = 4 L / 3 = 4(45 )(.59) / 3 = 758 ( = ). 4 L / 5 = 4(45 )(.59) / 5 = 455 ( = ) For the wavelegth to be i the visible rage, we choose = with = I this setup, we have > ad < 3, ad the coditio for axiu trasissio (iiu reflectio) or costructive iterferece is

23 393 Thus, we have 4L = + = = + L, 0,,,... 4L = 4(0 )(.46) = 6 ( = 0) =. 4 L / 3 = 4(0 )(.46) / 3 = 409 ( = ) For the wavelegth to be i the visible rage, we choose = with = I this setup, we have > ad > 3, ad the coditio for iiu trasissio (axiu reflectio) or destructive iterferece is 4L = + = = + L, 0,,,... Therefore, 4L = 4(85 )(.60) = 84 ( = 0) =. 4 L / 3 = 4(45 )(.59) / 3 = 608 ( = ) For the wavelegth to be i the visible rage, we choose = with = I this setup, we have < ad < 3, ad the coditio for axiu trasissio (iiu reflectio) or costructive iterferece is = L = = Thus, we have (with =) L, 0,,,... = L = (00 )(.40) = Assue the wedge-shaped fil is i air, so the wave reflected fro oe surface udergoes a phase chage of π rad while the wave reflected fro the other surface does ot. At a place where the fil thickess is L, the coditio for fully costructive iterferece is L = b + g where is the idex of refractio of the fil, is the wavelegth i vacuu, ad is a iteger. The eds of the fil are bright. Suppose the ed where the fil is arrow has thickess L ad the bright frige there correspods to =. Suppose the ed where the fil is thick has thickess L ad the bright frige there correspods to =. Sice there are te bright friges, = + 9. Subtract L = b + g fro L = b + 9+ g to obtai L = 9, where L = L L is the chage i the fil thickess over its legth. Thus,

24 394 CHAPTER 35 c L = = b g h = By the coditio = y where y is the thickess of the air-fil betwee the plates directly udereath the iddle of a dark bad), the edge of the plates (the edge where they are ot touchig) are y = 8/ = 400 apart (where we have assued that the iddle of the ith dark bad is at the edge). Icreasig that to y' = 3000 would correspod to ' = y'/ = 0 (couted as the eleveth dark bad, sice the first oe correspods to = 0). There are thus dark friges alog the top plate. 7. Cosider the iterferece of waves reflected fro the top ad botto surfaces of the air fil. The wave reflected fro the upper surface does ot chage phase o reflectio but the wave reflected fro the botto surface chages phase by π rad. At a place where the thickess of the air fil is L, the coditio for fully costructive iterferece is L= b+ g where (= 683 ) is the wavelegth ad is a iteger. This is satisfied for = 40: L = 9 b + g b40. 5gc h = = = At the thi ed of the air fil, there is a bright frige. It is associated with = 0. There are, therefore, 40 bright friges i all. 7. (a) The third setece of the proble iplies o = 9.5 i d o = o iitially. The, t = 5 s later, we have = 9.0 i d =. This eas d = d o d = ( o ) = 55. Thus, d divided by t gives 0.3 /s. (b) I this case, f = 6 so that d o d f = ( o f ) = 7 4 = 085 =.09 µ. 73. Usig the relatios of 35-7, we fid that the (vertical) chage betwee the ceter of oe dark bad ad the ext is y = = = = Thus, with the (horizotal) separatio of dark bads give by x =., we have

25 395 y θ ta θ = = rad. x Covertig this agle ito degrees, we arrive at θ = We apply Eq to both scearios: = 400 ad = air, ad = 4000 ad = vacuu =.00000: L = b400g ad L = b4000g. air Sice the L factor is the sae i both cases, we set the right had sides of these expressios equal to each other ad cacel the wavelegth. Fially, we obtai b g air = = We reark that this sae result ca be obtaied startig with Eq (which is developed i the textbook for a soewhat differet situatio) ad usig Eq to eliiate the L/ ter. 75. Cosider the iterferece patter fored by waves reflected fro the upper ad lower surfaces of the air wedge. The wave reflected fro the lower surface udergoes a π rad phase chage while the wave reflected fro the upper surface does ot. At a place where the thickess of the wedge is d, the coditio for a axiu i itesity is d = b+ g where is the wavelegth i air ad is a iteger. Therefore, d = ( + )/4. As the geoetry of Fig shows, d = R R r, where R is the radius of curvature of the les ad r is the radius of a Newto s rig. Thus, b + g = R R r. First, we rearrage the ters so the equatio becoes ( + ) l R r = R 4. Next, we square both sides, rearrage to solve for r, the take the square root. We get r = b g b g + R +. 6 If R is uch larger tha a wavelegth, the first ter doiates the secod ad

26 396 CHAPTER 35 r = b + g R. 76. (a) We fid fro the last forula obtaied i Proble 35-75: r = = R c b gc which (roudig dow) yields = 33. Sice the first bright frige correspods to = 0, = 33 correspods to the thirty-fourth bright frige. (b) We ow replace by = / w. Thus, h 9 h 3 (.33)( 0 0 ) 9 ( 5.0 )( ) r r w = = = = 45. Rl Rl This correspods to the forty-sixth bright frige (see reark at the ed of our solutio i part (a)). b 77. We solve for usig the forula r = + g R obtaied i Proble ad fid = r /R /. Now, whe is chaged to + 0, r becoes r', so + 0 = r' /R /. Takig the differece betwee the two equatios above, we eliiate ad fid b g b g R = r r = c 06. c c c h = 00c. 78. The tie to chage fro oe iiu to the ext is t = s. This ivolves a chage i thickess L = / (see Eq ), ad thus a chage of volue V = πr² L = πr² dv dt = πr² t = π(0.080)² (550 x 0-9 ) (.40) () usig SI uits. Thus, the rate of chage of volue is /s. 79. A shift of oe frige correspods to a chage i the optical path legth of oe wavelegth. Whe the irror oves a distace d the path legth chages by d sice the light traverses the irror ar twice. Let N be the uber of friges shifted. The, d = N ad

27 397 3 c h d = = N 79 7 = = Accordig to Eq , the uber of friges shifted ( N) due to the isertio of the fil of thickess L is N = (L / ) ( ). Therefore, N L = = = b g b gb. g 5. µ. 40 b. 8. Let φ be the phase differece of the waves i the two ars whe the tube has air i it, ad let φ be the phase differece whe the tube is evacuated. These are differet because the wavelegth i air is differet fro the wavelegth i vacuu. If is the wavelegth i vacuu, the the wavelegth i air is /, where is the idex of refractio of air. This eas Lπ π 4πb g L φ φ = L g NM O QP = where L is the legth of the tube. The factor arises because the light traverses the tube twice, oce o the way to a irror ad oce after reflectio fro the irror. Each shift by oe frige correspods to a chage i phase of π rad, so if the iterferece patter shifts by N friges as the tube is evacuated, ad b g 4π L = Nπ c c N = + = + L h h = We deote the two wavelegths as ad ', respectively. We apply Eq to both wavelegths ad take the differece: We ow require N' N = ad solve for L: L L N N = = L F. HG I K J L = = = = µ (a) Applyig the law of refractio, we obtai si θ / si θ = si θ / si 30 = v s /v d. Cosequetly,

28 398 CHAPTER 35 ( ) v si s si 30 s si si. vd 4.0 s θ = = = (b) The agle of icidece is gradually reduced due to refractio, such as show i the calculatio above (fro 30 to ). Evetually after ay refractios, θ will be virtually zero. This is why ost waves coe i oral to a shore. 84. Whe the depth of the liquid (L liq ) is zero, the phase differece φ is 60 wavelegths; this ust equal the differece betwee the uber of wavelegths i legth L = 40 µ (sice the liquid iitially fills the hole) of the plastic (for ray r ) ad the uber i that sae legth of the air (for ray r ). That is, L L =. plastic air 60 (a) Sice = ad air = (to good approxiatio), we fid plastic =.6. (b) The slope of the graph ca be used to deterie liq, but we show a approach ore closely based o the above equatio: L plastic L liq = 0 which akes use of the leftost poit of the graph. This readily yields liq = (a) The path legth differece betwee Rays ad is 7d d = 5d. For this to correspod to a half-wavelegth requires 5d = /, so that d = (b) The above requireet becoes 5d = / i the presece of the solutio, with =.38. Therefore, d = (a) The iiu path legth differece occurs whe both rays are early vertical. This would correspod to a poit as far up i the picture as possible. Treatig the scree as if it exteded forever, the the poit is at y =. (b) Whe both rays are early vertical, there is o path legth differece betwee the. Thus at y =, the phase differece is φ = 0. (c) At y = 0 (where the scree crosses the x axis) both rays are horizotal, with the ray fro S beig loger tha the oe fro S by distace d. (d) Sice the proble specifies d = 6.00, the the phase differece here is φ = 6.00 wavelegths ad is at its axiu value.

29 399 (e) With D = 0, use of the Pythagorea theore leads to φ = L L = d² + (d + D)² d² + D² = 5.80 which eas the rays reachig the poit y = d have a phase differece of roughly 5.8 wavelegths. (f) The result of the previous part is iterediate closer to 6 (costructive iterferece) tha to 5 (destructive iterferece). 87. The wave that goes directly to the receiver travels a distace L ad the reflected wave travels a distace L. Sice the idex of refractio of water is greater tha that of air this last wave suffers a phase chage o reflectio of half a wavelegth. To obtai costructive iterferece at the receiver, the differece L L ust be a odd ultiple of a half wavelegth. Cosider the diagra below. The right triagle o the left, fored by the vertical lie fro the water to the trasitter T, the ray icidet o the water, ad the water lie, gives D a = a/ ta θ. The right triagle o the right, fored by the vertical lie fro the water to the receiver R, the reflected ray, ad the water lie leads to D = x/taθ. Sice D a + D b = D, b ta θ = a+ x. D We use the idetity si θ = ta θ / ( + ta θ) to show that si ( ) / ( ) This eas L θ = a+ x D + a+ x. a b g a a D + a+ x = = siθ a+ x ad Therefore, L ( ) x x D + a+ x b = =. siθ a+ x ba+ xgd + ba+ xg L = La + Lb = = D + ba+ xg. a+ x

30 400 CHAPTER 35 Usig the bioial theore, with D large ad a + x sall, we approxiate this L D+ a+ x / D. The distace traveled by the direct wave is L = D + ba xg. Usig the bioial theore, we approxiate this expressio: expressio: ( ) ( ) L D+ a x / D. Thus, L L D a + ax + x + D D a ax + x D ax =. D Settig this equal to b + g, where is zero or a positive iteger, we fid x = + D a b gb g. 88. (a) Sice P is equidistat fro S ad S we coclude the sources are ot i phase with each other. Their phase differece is φ source = 0.60 π rad, which ay be expressed i ters of wavelegths (thikig of the π correspodece i discussig a full cycle) as φ source = (0.60 π / π) = 0.3 (with S leadig as the proble states). Now S is closer to P tha S is. Source S is 80 ( 80/400 = 0. ) fro P while source S is 360 ( 360/400 = 3.4 ) fro P. Here we fid a differece of φ path = 3. (with S leadig sice it is closer). Thus, the et differece is or.90 wavelegths. φ et = φ path φ source =.90, (b) A whole uber (like 3 wavelegths) would ea fully costructive, so our result is of the followig ature: iterediate, but close to fully costructive. 89. We ifer fro Saple Proble 35-, that (with agle i radias) θ = d for adjacet friges. With the wavelegth chage (' = / by Eq. 35-8), this equatio becoes θ = d. Dividig oe equatio by the other, the requireet of radias ca ow be relaxed ad we obtai θ = = θ.

31 40 Therefore, with =.33 ad θ = 0.30, we fid θ ' = (a) The graph shows part of a periodic patter of half-cycle legth = 0.4. Thus if we set =.0 + =.8 the the axiu at =.0 should repeat itself there. (b) Cotiuig the reasoig of part (a), addig aother half-cycle legth we get.8+ =. for the aswer. (c) Sice = 0.4 represets a half-cycle, the / represets a quarter-cycle. To accuulate a total chage of.0.0 =.0 (see proble stateet), the we eed + / = 5/4 th of a cycle, which correspods to.5 wavelegths. 9. We ote that φ = 60 = π 3 rad. The phasors rotate with costat agular velocity φ π /3 rad ω = = = 6 t.5 0 s rad/s. Sice we are workig with light waves travelig i a ediu (presuably air) where the wave speed is approxiately c, the kc = ω (where k = π/), which leads to = πc ω = (a) The differece i wavelegths, with ad without the =.4 aterial, is foud usig Eq. 35-9: L N = ( ) =.43. The result is equal to a phase shift of (.43)(360 ) = 4.4, or (b) ore eaigfully -- a shift of = (a) A path legth differece of / produces the first dark bad, of 3/ produces the secod dark bad, ad so o. Therefore, the fourth dark bad correspods to a path legth differece of 7/ = 750 =.75 µ. (b) I the sall agle approxiatio (which we assue holds here), the friges are equally spaced, so that if y deotes the distace fro oe axiu to the ext, the the distace fro the iddle of the patter to the fourth dark bad ust be 6.8 = 3.5 y. Therefore, we obtai y = 6.8/3.5 = We ote that ray travels a extra distace 4L ore tha ray. For costructive iterferece (which is obtaied for = 60 ) we require

32 40 CHAPTER 35 4L = where = soe positive iteger. For destructive iterferece (which is obtaied for = 496 ) we require 4L = k where k = soe positive odd iteger. Equatig these two equatios (sice their left-had sides are equal) ad rearragig, we obtai k = = =.5. We ote that this coditio is satisfied for k = 5 ad =. It is satisfied for soe larger values, too, but recallig that we wat the least possible value for L we choose the solutio set (k, ) = (5, ). Pluggig back ito either of the equatios above, we obtai the distace L: 4L = L = = I the case of a distat scree the agle θ is close to zero so si θ θ. Thus fro Eq. 35-4, F θ θ = H G I si K J = =, d d d or d / θ = /0.08 rad = = 33 µ. 96. We use Eq for costructive iterferece: L = ( + /), or b gb g L = = =, where = 0,,,. The oly value of which, whe substituted ito the equatio above, would yield a wavelegth which falls withi the visible light rage is =. Therefore, 30 = = (a) Lookig at the figure (where a portio of a periodic patter is show) we see that half of the periodic patter is of legth L = 750 (judgig fro the axiu at x = 0 to the iiu at x = 750 ); this suggests that the wavelegth (the full legth of the periodic patter) is = L = 500. A axiu should be reached agai at x = 500 (ad at x = 3000, x = 4500, ).

33 403 (b) Fro our discussio i part (b), we expect a iiu to be reached at each value x = (500 ), where =,, 3. For istace, for = we would fid the iiu at x = 50. (c) With = 500 (foud i part (a)), we ca express x = 00 as x = 00/500 = 0.80 wavelegth. 98. We use the forula obtaied i Saple Proble 35-6: L i L = = = = ( ) i For the first axiu = 0 ad for the teth oe = 9. The separatio is We solve for the wavelegth: d y = = 9D y =(D/d) = 9D/d. 3 3 c hc h c h 7 = = The idex of refractio of fused quartz at = 550 is about.459, obtaied fro Fig Thus, fro Eq. 35-3, we fid s c v = = = s. 0 s. 0. We adapt Eq. 35- to the o-reflective coatig o a glass les: I = I ax cos (φ/), where φ = (π/)( L) + π. (a) At = 450 I I ax ( )( ) φ πl π π π = = + = + = 450 cos cos cos %. (b) At = 650 I I ax ( )( ) π π = + = 650 cos %.

34 404 CHAPTER (a) We use y = D/d (see Saple Proble 35-). Because of the placeet of the irror i the proble D = (0.0 ) = 40.0, which we express i illieters i the calculatio below: D d = = y 4 6 c hc h = (b) I this case the iterferece patter will be shifted. At the locatio of the origial cetral axiu, the phase differece is ow.5 wavelegths, so there is ow a iiu istead of a axiu. 03. Let the positio of the irror easured fro the poit at which d = d be x. We assue the bea-splittig echais is such that the two waves iterfere costructively for x = 0 (with soe bea-splitters, this would ot be the case). We ca adapt Eq to this situatio by icorporatig a factor of (sice the iterferoeter utilizes directly reflected light i cotrast to the double-slit experiet) ad eliiatig the si θ factor. Thus, the phase differece betwee the two light paths is φ = (πx/) = 4πx/. The fro Eq. 35- (writig 4I 0 as I ) we fid φ πx I Icos Icos. = F H G I K J = F H G I K J 04. (a) Sice > 3, this case has o π-phase shift, ad the coditio for costructive iterferece is = L. We solve for L: b g b g b 55 L = = = 55. g 69. For the iiu value of L, let = to obtai L i = 69. (b) The light of wavelegth (other tha 55 ) that would also be preferetially trasitted satisfies ' = L, or b gb g = L = = 55. Here = 34,,,Κ (ote that ' = correspods to the = 55 light, so it should ot be icluded here). Sice the iiu value of ' is, oe ca easily verify that o ' will give a value of which falls ito the visible light rage. So o other parts of the visible spectru will be preferetially trasitted. They are, i fact, reflected. (c) For a sharp reductio of trasissio let

35 405 L 55 = =, + + where ' = 0,,, 3,. I the visible light rage ' = ad = 350. This correspods to the blue-violet light. 05. The phase differece i radias is b g The proble iplies = 5, so the thickess is 06. We use Eq : t = = t = b g 3 = = 80. µ. L = 6 +, L = The differece betwee these, usig the fact that = air =.0, is 480 L6 L 6 = ( 0) = 400 =.4µ..0 ( ) 07. (a) Every tie oe ore destructive (costructive) frige appears the icrease i thickess of the air gap is /. Now that there are 6 ore destructive friges i additio to the oe at poit A, the thickess at B is t B = 6(/) = 3(600 ) =.80 µ. (b) We ust ow replace by ' = / w. Sice t B is uchaged t B = N('/) = N(/ w ), or t 3 B w b gw N = = = 6w = 633 b. g= 8. Coutig the oe at poit A, a total of ie dark friges will be observed. 08. Let the = 0 bright frige o the scree be a distace y fro the cetral axiu. The fro Fig. 35-0(a) r r = y+ d + D y d + D = 0 b g b g

36 406 CHAPTER 35 fro which we ay solve for y. To the order of (d/d) we fid y = y + 0 yy c + d 4h, D where y 0 = 0D/d. Thus, we fid the percet error as follows: c 4h 0 d 589. µ 0. 0 D 8 D 000µ 8 40 y0 y0 + d yd which yields 0.03%. = F H G I K J + F H GI K J = F H G I K J + F H G I K J 09. A light ray travelig directly alog the cetral axis reaches the ed i tie L L tdirect = = v c For the ray takig the critical zig-zag path, oly its velocity copoet alog the core axis directio cotributes to reachig the other ed of the fiber. That copoet is v cos θ ', so the tie of travel for this ray is. t zig zag L = = L v cosθ c si / ( θ ) usig results fro the previous solutio. Pluggig i siθ = we obtai L L tzig zag = =. c / c b g ad siplifyig, The differece is L L L t = tzig zag tdirect = = c c c. With =.58, =.53 ad L = 300, we obtai L (.58)(300 ).58 8 t = = s 5.6 s 8 = = c /s The iiu speed of the electro is.

37 407 v i = c/ = ( /s)/.54 = /s.. (a) The path legth differece is 0.5 µ = 500, which is represets 500/400 =.5 wavelegths that is, a eaigful differece of 0.5 wavelegths. I agular easure, this correspods to a phase differece of (0.5)π = π/ radias.6 rad. (b) Whe a differece of idex of refractio is ivolved, the approach used i Eq is quite useful. I this approach, we cout the wavelegths betwee S ad the origi L L N = + where = (roudig off the idex of air), L = 5.0 µ, ' =.5 ad L' =.5 µ. This yields N = 8.5 wavelegths. The uber of wavelegths betwee S ad the origi is (with L = 6.0 µ) give by L N = = Thus, N N = 3.5 wavelegths, which gives us a eaigful differece of 0.5 wavelegth ad which coverts to a phase of π/4 radia 0.79 rad.. Usig Eq with the sall-agle approxiatio (illustrated i Saple Proble 35-), we arrive at b+ g D y = d for the positio of the ( + ) th dark bad (a siple way to get this is by averagig the expressios i Eq ad Eq. 35-8). Thus, with =, y = 0.0 ad d = 800, we fid D = (a) We are dealig with a syetric situatio (with the fil idex =.5 beig less tha that of the aterials boudig it), ad with reflected light, so Eqs ad -37 apply with their stated applicability. Both ca be writte i the for L = { half-iteger for bright iteger for dark Thus, we fid L/ = 3, so that we fid the iddle of a dark bad at the left edge of the figure. Sice there is othig beyod this "iddle" the a ore appropriate phrasig is that there is half of a dark bad ext to the left edge, beig darkest precisely at the edge. (b) The right edge, where they touch, satisfies the dark reflectio coditio for L = 0 (where = 0), so there is (essetially half of) a dark bad at the right ed. (c) Coutig half-bads ad whole bads alike, we fid four dark bads: ( = 0,,, 3).

38 408 CHAPTER (a) I this case, the fil has a saller idex aterial o oe side (air) ad a larger idex aterial o the other (glass), ad we are dealig (i part (a)) with strogly trasitted light, so the coditio is give by Eq (which would give dark reflectio i this sceario) for =.5 ad = 0. F HG I K J = L = + 0 (b) Now, we are dealig with strogly reflected light, so the coditio is give by Eq (which would give o trasissio i this sceario) L = = 0 for =.5 ad = (the = 0 optio is excluded i the proble stateet). 5. (a) Straightforward applicatio of Eq ad v = x/ t yields the result: fil. (b) The traversal tie is equal to s. (c) Use of Eq leads to the uber of wavelegths: N = L + L + L 3 3= (a) Followig Saple Proble 35-, we have L N N = = 87 b g. which represets a eaigful differece of 0.87 wavelegth. (b) The result i part (a) is closer to wavelegth (costructive iterferece) tha it is to / wavelegth (destructive iterferece) so the latter choice applies. (c) This would isert a ± / wavelegth ito the previous result resultig i a eaigful differece (betwee the two rays) equal to = 0.37 wavelegth. (d) The result i part (c) is closer to the destructive iterferece coditio. Thus, there is iterediate illuiatio but closer to darkess. 7. (a) With = 0.5 µ, Eq leads to

39 409 bgb g µ θ = si = µ (b) Decreasig the frequecy eas icreasig the wavelegth which iplies y icreases, ad the third side bright frige oves away fro the ceter of the patter. Qualitatively, this is easily see with Eq Oe should exercise cautio i appealig to Eq here, due to the fact the sall agle approxiatio is ot justified i this proble. (c) The ew wavelegth is 0.5/0.9 = µ, which produces a ew agle of bgb g µ θ = si = µ Usig y = D ta θ for the old ad ew agles, ad subtractig, we fid b g. y = D ta ta = Light reflected fro the upper oil surface (i cotact with air) chages phase by π rad. Light reflected fro the lower surface (i cotact with glass) chages phase by π rad if the idex of refractio of the oil is less tha that of the glass ad does ot chage phase if the idex of refractio of the oil is greater tha that of the glass. First, suppose the idex of refractio of the oil is greater tha the idex of refractio of the glass. The coditio for fully destructive iterferece is o d =, where d is the thickess of the oil fil, o is the idex of refractio of the oil, is the wavelegth i vacuu, ad is a iteger. For the shorter wavelegth, o d = ad for the loger, o d =. Sice is less tha, is greater tha, ad sice fully destructive iterferece does ot occur for ay wavelegths betwee, = +. Solvig ( + ) = for, we obtai = = = Sice ust be a iteger, the oil caot have a idex of refractio that is greater tha that of the glass. Now suppose the idex of refractio of the oil is less tha that of the glass. The coditio for fully destructive iterferece is the o d = ( + ). For the shorter wavelegth, o d = ( + ), ad for the loger, o d = ( + ). Agai, = + so ( + 3) = ( + ). This eas the value of is