2λ, and so on. We may write this requirement succinctly as

Transcription

1 Chapter 35. The fact that wave W reflects two additioal times has o substative effect o the calculatios, sice two reflectios amout to a (/) = phase differece, which is 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 m)/4 =55 m usig the wavelegth value give i the problem. (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 m L L= = = = 30m We cosider waves W ad W with a iitial effective phase differece (i wavelegths) equal to, ad seek positios of the sliver that cause the wave to costructively iterfere (which correspods to a iteger-valued phase differece i 3 wavelegths). Thus, the extra distace L traveled by W must amout to,, ad so o. We may write this requiremet succictly as m + L= where m= 0,,,. 4 (a) Thus, the smallest value of L / that results i the fial waves beig exactly i phase is whe m = 0, which gives L / = / 4 = 0.5. (b) The secod smallest value of L / that results i the fial waves beig exactly i phase is whe m =, which gives L / = 3/ 4 = (c) The third smallest value of L / that results i the fial waves beig exactly i phase is whe m =, which gives L / = 5/4= (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 umber ad is the wavelegth i glass. Similarly, 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 umber ad is the wavelegth i plastic. The agular frequecies are the same sice the waves have the same wavelegth i air ad the frequecy of a wave does ot chage whe the wave eters aother medium. The phase differece is 347

2 348 CHAPTER 35 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. Similarly, = air /, where is the idex of refractio of the plastic. This meas that the phase differece is π φ φ = ( ) L. The value of L that makes this 5.65 rad is L = b g 9 φ φ l m air = p p b g c b air h g = (b) 5.65 rad is less tha π rad = 6.8 rad, the phase differece for completely costructive iterferece, ad greater tha π rad (= 3.4 rad), the phase differece for completely destructive iterferece. The iterferece is, therefore, itermediate, either completely costructive or completely destructive. It is, however, closer to completely costructive tha to completely destructive. 4. Note that Sell s law (the law of refractio) leads to θ = θ whe =. The graph idicates that θ = 30 (which is what the problem gives as the value of θ ) occurs at =.5. Thus, =.5, ad the speed with which light propagates i that medium is 6 m m s c v = = = m s. 5. Comparig the light speeds i sapphire ad diamod, we obtai Δ v= vs vd = c = ( ) = s d (a) The frequecy of yellow sodium light is m s m s. f m s m c = = = Hz. (b) Whe travelig through the glass, its wavelegth is 589m = = = 388 m..5

3 349 (c) The light speed whe travelig through the glass is ( )( ) v= f l = = 7. The idex of refractio is foud from Eq. 35-3: Hz m.97 0 m s. c = = v ms ms = (a) The time 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 Similarly for pulse : t L L L 633. L = + + =. c 59. c 65. c 50. c Thus, pulse travels through the plastic i less time. (b) The time differece (as a multiple of L/c) is Thus, the multiple is L 630. L 003. L Δt = t t = =. c c c 9. (a) We wish to set Eq. 35- equal to /, sice a half-wavelegth phase differece is equivalet to a π radias differece. Thus, L mi l 60 m = = b g b g = 550m = 55. μ m. (b) Sice a phase differece of 3 (wavelegths) is effectively the same as what we required i part (a), the L = 3l = 3Lmi = 355 = 465 b g b. μmg. μm. 0. (a) The exitig agle is 50º, the same as the icidet agle, due to what oe might call the trasitive ature of Sell s law: siθ = siθ = 3 siθ 3 =

4 350 CHAPTER 35 (b) Due to the fact that the speed (i a certai medium) is c/ (where is that medium s idex of refractio) ad that speed is distace divided by time (while it s costat), we fid t = L/c = (.45)(5 0 9 m)/( m/s) = s = 0.4 ps.. (a) Equatio 35- (i absolute value) yields L = l 6 c mh 9 b g m = 70.. (b) Similarly, L = l 6 c mh 9 b g m = 70.. (c) I this case, we obtai L = l 6 c35. 0 mh 9 b g m = 30.. (d) Sice their phase differeces were idetical, the brightess should be the same for (a) ad (b). Now, the phase differece i (c) differs from 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 same as that i (a) ad (b).. (a) We ote that ray travels a extra distace 4L more tha ray. To get the least possible L that will result i destructive iterferece, we set this extra distace equal to half of a wavelegth: 40.0 m 4L= L= = = 5.50 m. 8 8 (b) The ext case occurs whe that extra distace is set equal to 3. The result is 3 3(40.0 m) L = = = 57.5 m (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 similar 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 approximately equal to ). Thus, combiig these phase differeces with = μm, we have

5 35 L L L 3.50 μm 4.00 μm 3.50 μm μm μm ( ) + ( ) = (.60.40) + (.40) = (b) Sice the aswer i part (a) is closer to a iteger tha to a half-iteger, the iterferece is more early costructive tha destructive. 4. (a) For the maximum adjacet to the cetral oe, we set m = i Eq ad obtai ( )( l) ml θ = = = si si 0.00 rad. d m= 00l (b) Sice y = D ta θ (see Fig. 35-0(a)), we obtai y = (500 mm) ta (0.00 rad) = 5.0 mm. The separatio is Δy = y y 0 = y 0 = 5.0 mm. 5. The agular positios of the maxima of a two-slit iterferece patter are give by dsiθ = m, where d is the slit separatio, is the wavelegth, ad m is a iteger. If θ is small, si θ may be approximated by θ i radias. The, θ = m/d to good approximatio. The agular separatio of two adjacet maxima is Δθ = /d. Let ' be the wavelegth for which the agular separatio is greater by0.0%. The,.0/d = '/d. or ' =.0 =.0(589 m) = 648 m. 6. The distace betwee adjacet maxima is give by Δy = D/d (see Eqs ad 35-8). Dividig both sides by D, this becomes Δθ = /d with θ i radias. I the steps that follow, however, we will ed up with a expressio where degrees may be directly used. Thus, i the preset case, Δ θ 0.0 Δ θ = = = = = 0.5. d d Iterferece maxima occur at agles θ such that d si θ = m, where m is a iteger. Sice d =.0 m ad = 0.50 m, this meas that si θ = 0.5m. We wat all values of m (positive ad egative) for which 0.5m. 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 m = 4 ad a sigle value (+90 ) is associated with m = +4. There are sixtee differet agles i all ad, therefore, sixtee maxima. 8. (a) The phase differece (i wavelegths) is

6 35 CHAPTER 35 φ = d siθ/ = (4.4 µm)si(0 )/(0.500 µm) =.90. (b) Multiplyig this by π gives φ = 8. rad. (c) The result from part (a) is greater tha 5 (which would idicate the third miimum) ad is less tha 3 (which would correspod to the third side maximum). 9. The coditio for a maximum i the two-slit iterferece patter is d si θ = m, where d is the slit separatio, is the wavelegth, m is a iteger, ad θ is the agle made by the iterferig rays with the forward directio. If θ is small, si θ may be approximated by θ i radias. The, θ = m/d, ad the agular separatio of adjacet maxima, oe associated with the iteger m ad the other associated with the iteger m +, is give by Δθ = /d. The separatio o a scree a distace D away is give by Thus, Δy = D Δθ = D/d. 9 c500 0 mhb540. mg Δy = = m 3 m =.5 mm. 0. (a) We use Eq with m = 3: ml d F θ = H G I si K J = (b) θ = (0.6) (80 /π) =.4. si L NM 9 c mh O P 6 m Q P = 06. rad.. The maxima of a two-slit iterferece patter are at agles θ give by d si θ = m, where d is the slit separatio, is the wavelegth, ad m is a iteger. If θ is small, si θ may be replaced by θ i radias. The, dθ = m. The agular separatio of two maxima associated with differet wavelegths but the same value of m is Δθ = (m/d)( ), ad their separatio o a scree a distace D away is md Δy Dta Δθ DΔθ l l d L 30 = M b. mg O P =. c600 0 m mh 7 0 m m = = L NM N Q The small agle approximatio ta Δθ Δθ (i radias) is made. O QP b g

7 353. Imagie a y axis midway betwee the two sources i the figure. Thirty poits of destructive iterferece (to be cosidered i the xy plae of the figure) implies 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 must 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 miimum phase differece, we ca cout (say, i the first quadrat) the m 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 m = 7 (i the sese of Eq. 35-6) to the poit o the x axis itself (where the path differece for waves comig from the sources is simply equal to the separatio of the sources, d); this would correspod to θ = 90 i Eq Thus, d = ( 7 + d ) = 7.5 = 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 m, which is00 m 400m = 4 wavelegth. So the et phase differece betwee A ad B at the detector is zero. 4. (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 middle 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 limit 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 miimum 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 rightmost edge of the scree (which is assumed to go o forever), so φ is maximum at x =. (d) I cosiderig x =, we ca treat the rays from the sources as if they are essetially horizotal. I this way, we see that the differece betwee the path legths is simply the distace (d) betwee the sources. The problem specifies d = 6.00, or d/ = (e) Usig the Pythagorea theorem, we have D + ( x+ d) D + ( x d) φ = =.7

8 354 CHAPTER 35 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 itermediate but closer to a miimum tha to a maximum. 5. Let the distace i questio be x. The path differece (betwee rays origiatig from S ad S ad arrivig at poits o the x > 0 axis) is F I HG K J l, d + x x = m+ where we are requirig destructive iterferece (half-iteger wavelegth phase differeces) ad m = 0,,,. After some algebraic steps, we solve for the distace i terms of m: d bm + gl x =. m + l 4 To obtai the largest value of x, we set m = 0: x d ( 3.00) 4 4 b g 3 0 = = = 8.75 = 8.75(900 m) = m = 7.88μm. 6. (a) We use Eq to fid d: d siθ = m d = (4)(450 m)/si(90 ) = 800 m. For the third-order spectrum, the wavelegth that correspods to θ = 90 is = d si(90 )/3 = 600 m. Ay wavelegth greater tha this will ot be see. Thus, 600 m < θ 700 m are abset. (b) The slit separatio d eeds to be decreased. (c) I this case, the 400 m wavelegth i the m = 4 diffractio is to occur at 90. Thus d ew siθ = m d ew = (4)(400 m)/si(90 ) = 600 m. This represets a chage of Δd = d d ew = 00 m = 0.0 µm.

9 Cosider the two waves, oe from each slit, that produce the seveth bright frige i the absece of the mica. They are i phase at the slits ad travel differet distaces to the seveth bright frige, where they have a phase differece of πm = 4π. Now a piece of mica 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 px px px = l l l m where m is the wavelegth i the mica ad is the idex of refractio of the mica. The relatioship m = / is used to substitute for m. Sice the waves are ow i phase at the scree, pxb g = 4p l or 9 7l 7c550 0 mh 6 x = = = m The problem 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 some poit P o the x axis, which is, of course, a distace x from the top source ad (usig Pythagoras theorem) a distace d + x from the bottom source. The differece (i ormal legth uits) is therefore d + x x, or (expressed i radias) is π ( d + x x). We ote (lookig at the leftmost poit i the graph) that at x = 0, this latter quatity equals 6π, which meas d = 3. Usig this value for d, we ow must solve the coditio π b ( d ) x x g + = π. Straightforward algebra the leads to x = (35/4), ad usig = 400 m we fid x = 3500 m, or 3.5 μm. 9. The itesity is proportioal to the square of the resultat field amplitude. Let the electric field compoets of the two waves be writte as E = E0siωt E = E si( ωt+ φ), 0 where E 0 =.00, E 0 =.00, ad φ = 60. The resultat field is E E E = +. We use the phasor diagram to calculate the amplitude of E.

10 356 CHAPTER 35 The phasor diagram is show o the right. The resultat amplitude E m is give by the trigoometric law of cosies: Thus, m ( ) E = E + E E E cos 80 φ. b g b g b gb g E m = cos 0 =. 65. Note: Summig over the horizotal compoets of the two fields gives E h = E 0 cos 0 + E 0 cos 60 =.00 + (.00) cos 60 =.00. Similarly, the sum over the vertical compoets is E v = E 0 si 0 + E 0 si 60 =.00si 0 + (.00)si 60 =.73. The resultat amplitude is E m = (.00) + (.73) =.65, which agrees with what we foud above. The phase agle relative to the phasor represetig E is.73 β = ta = Thus, the resultat field ca be writte as E = (.65)si( ωt ). 30. I addig these with the phasor method (as opposed to, say, trig idetities), we may set t = 0 ad add them 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 = yhkj = F H G I

11 357 b g b g. y = y+ y = yr si ωt+ β = 7. 4 si ωt+ 33. Quotig the aswer to two sigificat figures, we have y 7si ( ωt 3 ) I addig these with the phasor method (as opposed to, say, trig idetities), we may set t = 0 ad add them as vectors: y y h v b b g 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) 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/), we fid E + E = E cos( φ / )si( ωt+ φ/ ) 0 where E 0 =.00 µv/m, ω = rad/s, ad φ = 39.6 rad. This shows that the electric field amplitude of the resultat wave is E = E cos( φ / ) = (.00 μv/m) cos(9. rad) =.33 μv/m. 0 (b) Equatio 35- leads to at poit P, ad I = 4I cos ( φ / ) =.35 I 0 0 I = 4I cos (0) = 4 I ceter 0 0 at the ceter. Thus, I / I ceter =.35/ 4 = (c) The phase differece φ (i wavelegths) is gotte from φ i radias by dividig by π. Thus, φ = 39.6/π = 6.3 wavelegths. Thus, poit P is betwee the sixth side maximum (at which φ = 6 wavelegths) ad the seveth miimum (at which φ = 6 wavelegths).

12 358 CHAPTER 35 (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. With phasor techiques, this amouts to a vector additio problem R = A + B + C where (i magitude-agle otatio) A= b0 0 g, B= b5 45 g, ad C = b5 45 g, where the magitudes are uderstood to be i μv/m. We obtai the resultat (especially efficiet o a vector-capable calculator i polar mode): R = = 7. 0 b g b g b g b g which leads to b g b g ER = 7. μvmsi ωt where ω = rad/s. 34. (a) Referrig to Figure 35-0(a) makes clear that θ = ta (y/d) = ta (0.05/4) =.93. Thus, the phase differece at poit P is φ = dsiθ / = wavelegths, which meas it is betwee the cetral maximum (zero wavelegth differece) ad the first miimum ( wavelegth differece). Note that the above computatio could have bee simplified somewhat by avoidig the explicit use of the taget ad sie fuctios ad makig use of the small-agle approximatio (taθ siθ). (b) From Eq. 35-, we get (with φ = (0.397)(π) =.495 rad) at poit P ad I = 4I cos ( φ / ) = I 0 0 I = 4I cos (0) = 4I ceter 0 0 at the ceter. Thus, I / I ceter = / 4 = For complete destructive iterferece, we wat the waves reflected from the frot ad back of the coatig to differ i phase by a odd multiple of π rad. Each wave is icidet o a medium of higher idex of refractio from a medium of lower idex, so both suffer phase chages of π rad o reflectio. If L is the thickess of the coatig, the wave reflected from the back surface travels a distace L farther tha the wave reflected from 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 vacuum, ad the phase differece is L(π/). We solve

13 359 L F pi HG l K J = b + g m p for L. Here m is a iteger. The result is b m + L = gl. 4 To fid the least thickess for which destructive iterferece occurs, we take m = 0. The, m m. l L = = = ( ) 36. (a) O both sides of the soap is a medium with lower idex (air) ad we are examiig the reflected light, so the coditio for strog reflectio is Eq With legths i m, = L m + = 3360 for m = 0 0 for m = 67 for m = 480 for m = for m = for m = 5 from which we see the latter four values are i the give rage. (b) We ow tur to Eq ad obtai = L m = 680 for m = 840 for m = 560 for m = 3 40 for m = for m = 5 from which we see the latter three values are i the give rage. 37. Light reflected from the frot surface of the coatig suffers a phase chage of π rad while light reflected from the back surface does ot chage phase. If L is the thickess of the coatig, light reflected from the back surface travels a distace L farther tha light reflected from 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 vacuum, 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 multiple of π. We solve

14 360 L F πi π HG K J = mπ CHAPTER 35 for L. Here m is a iteger. The solutio is b m + L = g. 4 To fid the smallest coatig thickess, we take m = 0. The, L = = b 9 g m 8 = m. 38. (a) We are dealig with a thi film (material ) i a situatio where > > 3, lookig for strog reflectios; the appropriate coditio is the oe expressed by Eq Therefore, with legths i m ad L = 500 ad =.7, we have = L m = 700 for m = 850 for m = 567 for m = 3 45 for m = 4 from which we see the latter two values are i the give rage. The loger wavelegth (m=3) is = 567 m. (b) The shorter wavelegth (m = 4) is = 45 m. (c) We assume the temperature depedece of the refractio idex is egligible. From the proportioality evidet i the part (a) equatio, loger L meas loger. 39. For costructive iterferece, we use Eq : L = bm+ g. For the smallest value of L, let m = 0: 64m L0 = = = 7m = 0.7 μ m ( ) (b) For the secod smallest value, we set m = ad obtai ( + ) 3 L = = = 3L = m = 0.35μm. 0 ( μ )

15 The icidet light is i a low idex medium, the thi film of acetoe has somewhat higher =, ad the last layer (the glass plate) has the highest refractive idex. To see very little or o reflectio, the coditio L= ( m+ ) l where m= 0,,, must hold. This is the same as Eq , which was developed for the opposite situatio (costructive iterferece) regardig a thi film surrouded o both sides by air (a very differet cotext from the oe i this problem). By aalogy, we expect Eq to apply i this problem to reflectio maxima. A more careful aalysis such as that give i Sectio 35-7 bears this out. Thus, usig Eq with =.5 ad = 700 m yields L = 0, 80m, 560m, 840m,0m, for the first several m values. Ad the equatio show above (equivalet to Eq ) gives, with = 600 m, L = 0m,360m,600m,840m,080m, for the first several m values. The lowest umber these lists have i commo is L = 840 m. 4. I this setup, we have < ad > 3, ad the coditio for destructive iterferece is L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 34 m = + = 6 m. (.59) 4. I this setup, we have > ad > 3, ad the coditio for costructive iterferece is 4L L= m+ =, m= 0,,,... m+ Thus, we get 4L = 4(85 m)(.60) = 84 m ( m = 0) =. 4 L / 3 = 4(85 m)(.60) / 3 = 608 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 608 m.

16 36 CHAPTER Whe a thi film of thickess L ad idex of refractio is placed betwee materials ad 3 such that > ad 3 > where ad 3 are the idexes of refractio of the materials, the geeral coditio for destructive iterferece for a thi film is L = = = m L m, m 0,,,... where is the wavelegth of light as measured i air. Thus, we have, for m = = L = (00 m)(.40) = 560 m. 44. I this setup, we have < ad < 3, ad the coditio for costructive iterferece is L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 587 m = + = 39 m. (.34) 45. I this setup, we have > ad > 3, ad the coditio for costructive iterferece is L m, 0,,,... L m = + = + m = The third least thickess is (m = ) L 6 m = + = 478 m. (.60) 46. I this setup, we have < ad > 3, ad the coditio for destructive iterferece is Therefore, 4L = + = = m+ L m, m 0,,,...

17 363 4L = 4(45 m)(.59) = 639 m ( m = 0) = 4 L / 3 = 4(45 m)(.59) / 3 = 880 m ( m = ). 4 L / 5 = 4(45 m)(.59) / 5 = 58 m ( m = ) For the wavelegth to be i the visible rage, we choose m = 3 with = 58 m. 47. I this setup, we have < ad < 3, ad the coditio for destructive iterferece is L L= m =, m= 0,,,... m Thus, we have L = (380 m)(.34) = 08 m ( m = ) =. L = (380 m)(.34) = 509 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 509 m. 48. I this setup, we have < ad < 3, ad the coditio for costructive iterferece is L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 63 m = + = 339 m. (.40) 49. I this setup, we have > ad > 3, ad the coditio for costructive iterferece is L m, 0,,,... L m = + = + m = The third least thickess is (m = ) L 38 m = + = 73 m. (.75) 50. I this setup, we have > ad < 3, ad the coditio for destructive iterferece is

18 364 CHAPTER 35 L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 48 m = + = 48 m. (.46) 5. I this setup, we have > ad < 3, ad the coditio for destructive iterferece is Thus, 4L L= m+ = m= m+, 0,,,... 4L = 4(0 m)(.46) = 6 m ( m = 0) =. 4 L / 3 = 4(0 m)(.46) / 3 = 409 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 409 m. 5. I this setup, we have > ad > 3, ad the coditio for costructive iterferece is 4L L= m+ =, m= 0,,,... m+ Thus, we have 4L = 4(35 m)(.75) = 75 m ( m = 0) = 4 L / 3 = 4(35 m)(.75) / 3 = 758 m ( m = ). 4 L / 5 = 4(35 m)(.75) / 5 = 455 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 455 m. 53. We solve Eq with =.33 ad = 600 m for m =,, 3, : L = 3 m, 338m, 564m, 789m, Ad, we similarly solve Eq with the same ad = 450 m: L = 0,69m, 338m, 508m, 677 m, The lowest umber these lists have i commo is L = 338 m.

19 The situatio is aalogous to that treated i Sample Problem Thi-film iterferece of a coatig o a glass les, i the sese that the icidet light is i a low idex medium, the thi film of oil has somewhat higher =, ad the last layer (the glass plate) has the highest refractive idex. To see very little or o reflectio, accordig to the Sample Problem, the coditio l L= m+ where m= 0,,, must hold. With = 500 m ad =.30, the possible aswers for L are L = 96 m, 88 m, 48m, 673m, 865 m,... Ad, with = 700 m ad the same value of, the possible aswers for L are L = 35m, 404m, 673m, 94m,... The lowest umber these lists have i commo is L = 673 m. 55. The idex of refractio of oil is greater tha that of the air, but smaller tha that of the water. Let the idices of refractio of the air, oil, ad water be,, ad 3, respectively. Sice < ad < 3, there is a phase chage of π rad from both surfaces. Sice the secod wave travels a additioal distace of L, the phase differece is π φ = ( L) where = / is the wavelegth i the oil. The coditio for costructive iterferece is π ( L) = m π, or L= m, m= 0,,,... (a) For m =,,..., maximum reflectio occurs for wavelegths ( )( ) = L m 04m, 55m, 368m,... m = m = We ote that oly the 55 m wavelegth falls withi the visible light rage. (b) Maximum trasmissio ito the water occurs for wavelegths for which reflectio is a miimum. The coditio for such destructive iterferece is give by

20 366 CHAPTER 35 F I l L= m+ HG K J l = 4L m + which yields = 08 m, 736 m, 44 m for the differet values of m. We ote that oly the 44-m wavelegth (blue) is i the visible rage, though we might expect some red cotributio sice the 736 m is very close to the visible rage. Note: A light ray reflected by a material chages phase by π rad (or 80 ) if the refractive idex of the material is greater tha that of the medium i which the light is travelig. Otherwise, there is o phase chage. Note that refractio at a iterface does ot cause a phase shift. 56. For costructive iterferece (which is obtaied for = 600 m) i this circumstace, we require k k L = = where k = some positive odd iteger ad is the idex of refractio of the thi film. Rearragig ad pluggig i L = 7.7 m ad the wavelegth value, this gives k k(600 m) k = = = = 0.55k. 4L 4(7.7 m).88 Sice we expect >, the k = is ruled out. However, k = 3 seems 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 film, we ow cosider the destructive iterferece part of the questio. Now we have L = (iteger) dest /. Thus, dest = (900 m)/(iteger). We ote that settig the iteger equal to yields a dest value outside the rage of the visible spectrum. A similar remark holds for settig the iteger equal to 3. Thus, we set it equal to ad obtai dest = 450 m. 57. I this setup, we have > ad > 3, ad the coditio for miimum trasmissio (maximum reflectio) or destructive iterferece is 4L L m m = + =, = 0,,,... m+ Therefore,

21 367 4L = 4(85 m)(.60) = 84 m ( m = 0) = 4 L / 3 = 4(45 m)(.59) / 3 = 608 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 608 m. 58. I this setup, we have > ad > 3, ad the coditio for miimum trasmissio (maximum reflectio) or destructive iterferece is L m, 0,,,... L m = + = + m = The third least thickess is (m = ) L 38 m = + = 73 m. (.75) 59. I this setup, we have < ad > 3, ad the coditio for maximum trasmissio (miimum reflectio) or costructive iterferece is Thus, we have 4L L= m+ = m= m+, 0,,,... 4L = 4(45 m)(.59) = 639 m ( m = 0) = 4 L / 3 = 4(45 m)(.59) / 3 = 880 m ( m = ). 4 L / 5 = 4(45 m)(.59) / 5 = 58 m ( m = ) For the wavelegth to be i the visible rage, we choose m = 3 with = 58 m. 60. I this setup, we have < ad < 3, ad the coditio for maximum trasmissio (miimum reflectio) or costructive iterferece is Thus, we obtai L = = = m L m, m 0,,,... L = (380 m)(.34) = 08 m ( m = ) =. L = (380 m)(.34) = 509 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 509 m.

22 368 CHAPTER I this setup, we have > ad > 3, ad the coditio for miimum trasmissio (maximum reflectio) or destructive iterferece is Therefore, 4L L= m+ = m= m+, 0,,,... 4L = 4(35 m)(.75) = 75 m ( m = 0) = 4 L / 3 = 4(45 m)(.59) / 3 = 758 m ( m = ) 4 L / 5 = 4(45 m)(.59) / 5 = 455 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 455 m. 6. I this setup, we have < ad > 3, ad the coditio for maximum trasmissio (miimum reflectio) or costructive iterferece is L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 34 m = + = 6 m. (.59) 63. I this setup, we have > ad < 3, ad the coditio for maximum trasmissio (miimum reflectio) or costructive iterferece is L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 48 m = + = 48 m. (.46) 64. I this setup, we have > ad < 3, ad the coditio for maximum trasmissio (miimum reflectio) or costructive iterferece is Thus, we have 4L L= m+ = m= m+, 0,,,...

23 369 4L = 4(0 m)(.46) = 6 m ( m = 0) = 4 L / 3 = 4(0 m)(.46) / 3 = 409 m ( m = ) For the wavelegth to be i the visible rage, we choose m = with = 409 m. 65. I this setup, we have < ad < 3, ad the coditio for miimum trasmissio (maximum reflectio) or destructive iterferece is L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 63 m = + = 339 m. (.40) 66. I this setup, we have < ad < 3, ad the coditio for maximum trasmissio (miimum reflectio) or costructive iterferece is = L = m = Thus, we have (with m =) L m, m 0,,,... = L = (00 m)(.40) = 560 m. 67. I this setup, we have < ad < 3, ad the coditio for miimum trasmissio (maximum reflectio) or destructive iterferece is L m, 0,,,... L m = + = + m = The secod least thickess is (m = ) L 587 m = + = 39 m. (.34) 68. I this setup, we have > ad > 3, ad the coditio for miimum trasmissio (maximum reflectio) or destructive iterferece is L m, 0,,,... L m = + = + m =

24 370 CHAPTER 35 The third least thickess is (m = ) L 6 m = + = 478 m. (.60) 69. Assume the wedge-shaped film is i air, so the wave reflected from oe surface udergoes a phase chage of π rad while the wave reflected from the other surface does ot. At a place where the film thickess is L, the coditio for fully costructive iterferece is L = bm + gl, where is the idex of refractio of the film, is the wavelegth i vacuum, ad m is a iteger. The eds of the film are bright. Suppose the ed where the film is arrow has thickess L ad the bright frige there correspods to m = m. Suppose the ed where the film is thick has thickess L ad the bright frige there correspods to m = m. Sice there are te bright friges, m = m + 9. Subtract L = bm + gl from L = bm+ 9+ gl to obtai ΔL = 9, where ΔL = L L is the chage i the film thickess over its legth. Thus, 9l ΔL = = 50. c b g 9 m h = (a) The third setece of the problem implies m o = 9.5 i d o = m o iitially. The, Δt = 5 s later, we have m = 9.0 i d = m. This meas 6 m. Δd = d o d = ( m o m ) = 55 m. Thus, Δd divided by Δt gives 0.3 m/s. (b) I this case, m f = 6 so that d o d f = (m o m f ) = 7 4 = 085 m =.09 µm. 7. Usig the relatios of Sectio 35-7, we fid that the (vertical) chage betwee the ceter of oe dark bad ad the ext is Δ y = = = = 500 m 50 m mm. Thus, with the (horizotal) separatio of dark bads give by Δx =. mm, we have Δy θ ta θ = = rad. Δx

25 37 Covertig this agle ito degrees, we arrive at θ = We apply Eq to both scearios: m = 400 ad = air, ad m = 4000 ad = vacuum =.00000: l l L = b400g ad L = b4000g air Sice the L factor is the same 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 remark that this same result ca be obtaied startig with Eq (which is developed i the textbook for a somewhat differet situatio) ad usig Eq to elimiate the L/ term. 73. Cosider the iterferece of waves reflected from the top ad bottom surfaces of the air film. The wave reflected from the upper surface does ot chage phase o reflectio but the wave reflected from the bottom surface chages phase by π rad. At a place where the thickess of the air film is L, the coditio for fully costructive iterferece is L= bm+ gl where (= 683 m) is the wavelegth ad m is a iteger. This is satisfied for m = 40: L = 9 bm + gl b40. 5gc683 0 mh = = m = 0.048mm. At the thi ed of the air film, there is a bright frige. It is associated with m = 0. There are, therefore, 40 bright friges i all. 74. By the coditio m = y where y is the thickess of the air film betwee the plates directly udereath the middle of a dark bad), the edges of the plates (the edges where they are ot touchig) are y = 8/ = 400 m apart (where we have assumed that the middle of the ith dark bad is at the edge). Icreasig that to y' = 3000 m would correspod to m' = y'/ = 0 (couted as the eleveth dark bad, sice the first oe correspods to m = 0). There are thus dark friges alog the top plate. 75. Cosider the iterferece patter formed by waves reflected from the upper ad lower surfaces of the air wedge. The wave reflected from the lower surface udergoes a π rad phase chage while the wave reflected from the upper surface does ot. At a place where the thickess of the wedge is d, the coditio for a maximum i itesity is d = m+ gl, where is the wavelegth i air ad m is a iteger. Therefore, b d = (m + )/4.

26 37 CHAPTER 35 As the geometry 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 m + gl4 = R R r. First, we rearrage the terms so the equatio becomes ( m + ) R r = R 4 Next, we square both sides, rearrage to solve for r, the take the square root. We get l. r = b g b g m+ Rl m+ l. 6 If R is much larger tha a wavelegth, the first term domiates the secod ad ( m+ ) R r =, m= 0,,, Note: Similarly, oe may show that the radii of the dark friges are give by r = mr. 76. (a) We fid m from the last formula obtaied i Problem 35-75: m r = = Rl 0 0 m 9 b5. 0 mgc589 0 mh 3 c which (roudig dow) yields m = 33. Sice the first bright frige correspods to m = 0, m = 33 correspods to the thirty-fourth bright frige. (b) We ow replace by = / w. Thus, h m 3 (.33)( 0 0 m) 9 ( 5.0 m)( m) r r w = = = = 45. Rl Rl This correspods to the forty-sixth bright frige (see the remark at the ed of our solutio i part (a)). 77. We solve for m usig the formula r = bm+ g Rl obtaied i Problem ad fid m = r /R /. Now, whe m is chaged to m + 0, r becomes r', so

27 373 m + 0 = r' /R /. Takig the differece betwee the two equatios above, we elimiate m ad fid R = r r = 0l b cm g b06. cmg 7 0c546 0 cmh = 00cm. 78. The time to chage from oe miimum to the ext is Δt = s. This ivolves a chage i thickess ΔL = / (see Eq ), ad thus a chage of volume π r² ΔV = πr²δl = dv π r² = dt Δ t π(0.080)² (550 x 0-9 ) = (.40) () usig SI uits. Thus, the rate of chage of volume is.67 0 m 3 /s. 79. A shift of oe frige correspods to a chage i the optical path legth of oe wavelegth. Whe the mirror moves a distace d, the path legth chages by d sice the light traverses the mirror arm twice. Let N be the umber of friges shifted. The, d = N ad 3 d c mh 7 = = = m= 588m. N Accordig to Eq , the umber of friges shifted (ΔN) due to the isertio of the film of thickess L is ΔN = (L / ) ( ). Therefore, ΔN L = b g = b gb g g 589 m b = 5. μ m. 8. Let φ be the phase differece of the waves i the two arms 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 from the wavelegth i vacuum. If is the wavelegth i vacuum, the the wavelegth i air is /, where is the idex of refractio of air. This meas π π 4π ( ) L φ φ = L = where L is the legth of the tube. The factor arises because the light traverses the tube twice, oce o the way to a mirror ad oce after reflectio from the mirror. 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,

28 374 ad b g 4π L = Nπ c c h h 9 N m = + = + L m = CHAPTER We apply Eq to both wavelegths ad take the differece: L L N N = = L. We ow require N N = ad solve for L: L = = = = m m μ m 9 m. 83. (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 = 50.0 m. (b) The above requiremet becomes 5d = / i the presece of the solutio, with =.38. Therefore, d = 36. m. 84. (a) The miimum 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 them. 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 from S beig loger tha the oe from S by distace d. (d) Sice the problem specifies d = 6.00, the the phase differece here is φ = 6.00 wavelegths ad is at its maximum value. (e) With D = 0, use of the Pythagorea theorem leads to φ = L L = d² + (d + D)² d² + D² = 5.80 which meas the rays reachig the poit y = d have a phase differece of roughly 5.8 wavelegths.

29 375 (f) The result of the previous part is itermediate closer to 6 (costructive iterferece) tha to 5 (destructive iterferece). 85. The agular positios of the maxima of a two-slit iterferece patter are give by Δ L = dsiθ = m, where ΔL is the path-legth differece, d is the slit separatio, is the wavelegth, ad m is a iteger. If θ is small, si θ may be approximated by θ i radias. The, θ = m/d to good approximatio. The agular separatio of two adjacet maxima is Δθ = /d. Whe the arragemet is immersed i water, the wavelegth chages to ' = /, ad the equatio above becomes Δ θ =. d Dividig the equatio by Δθ = /d, we obtai Δθ l = = Δθ l. Therefore, with =.33 ad Δθ = 0.30, we fid Δθ ' = 0.3. Note that the agular separatio decreases with icreasig idex of refractio; the greater the value of, the smaller the value of Δθ. 86. (a) The graph shows part of a periodic patter of half-cycle legth Δ = 0.4. Thus if we set =.0 + Δ =.8 the the maximum at =.0 should repeat itself there. (b) Cotiuig the reasoig of part (a), addig aother half-cycle legth we get.8+δ =. for the aswer. (c) Sice Δ = 0.4 represets a half-cycle, the Δ/ represets a quarter-cycle. To accumulate a total chage of.0.0 =.0 (see problem statemet), the we eed Δ + Δ/ = 5/4 th of a cycle, which correspods to.5 wavelegths. 87. Whe the iterferece betwee two waves is completely destructive, their phase differece is give by φ = (m+ ) π, m= 0,,,... The equivalet coditio is that their path-legth differece is a odd multiple of /, where is the wavelegth of the light. (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 ΔL = 750 m (judgig from the maximum at x = 0 to the miimum at x = 750 m); this suggests that the wavelegth (the full legth of the periodic patter) is = ΔL = 500 m. A maximum should be reached agai at x = 500 m (ad at x = 3000 m, x = 4500 m, ).

30 376 CHAPTER 35 (b) From our discussio i part (b), we expect a miimum to be reached at each value x = 750 m + (500 m), where =,, 3,. For istace, for = we would fid the miimum at x = 50 m. (c) With = 500 m (foud i part (a)), we ca express x = 00 m as x = 00/500 = 0.80 wavelegth. 88. (a) The differece i wavelegths, with ad without the =.4 material, is foud usig Eq. 35-9: L Δ N = ( ) =.43. The result is equal to a phase shift of (.43)(360 ) = 4.4, or (b) more meaigfully, a shift of = 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 must be a odd multiple of a half wavelegth. Cosider the diagram o the right. The right triagle o the left, formed by the vertical lie from the water to the trasmitter T, the ray icidet o the water, ad the water lie, gives D a = a/ ta θ. The right triagle o the right, formed by the vertical lie from the water to the receiver R, the reflected ray, ad the water lie leads to Db = x/ taθ. Sice D a + D b = D, ta θ = a+ x. D We use the idetity si θ = ta θ / ( + ta θ) to show that This meas si ( ) / ( ) L θ = a+ x D + a+ x. a a a D + a+ x = = siθ a+ x b g ad

31 377 Therefore, L ( ) x x D + a+ x b = =. siθ a+ x L = L + L = a b b g b g b g. a+ x D + a+ x a+ x = D + a+ x Usig the biomial theorem, with D large ad a + x small, we approximate this expressio: L ( ) D+ a+ x / D. The distace traveled by the direct wave is L = D + ba xg. Usig the biomial theorem, we approximate this expressio: L ( ) D+ a x / D. Thus, L L D a + ax + x + D b D a ax + x D ax =. D Settig this equal to m + g, where m is zero or a positive iteger, we fid x = m+ D a b gb g. 90. (a) Sice P is equidistat from S ad S we coclude the sources are ot i phase with each other. Their phase differece is Δφ source = 0.60 π rad, which may be expressed i terms of wavelegths (thikig of the π correspodece i discussig a full cycle) as Δφ source = (0.60 π / π) = 0.3 (with S leadig as the problem states). Now S is closer to P tha S is. Source S is 80 m ( 80/400 = 0. ) from P while source S is 360 m ( 360/400 = 3.4 ) from P. Here we fid a differece of Δφ path = 3. (with S leadig sice it is closer). Thus, the et differece is or.90 wavelegths. Δφ et = Δφ path Δφ source =.90, (b) A whole umber (like 3 wavelegths) would mea fully costructive, so our result is of the followig ature: itermediate, but close to fully costructive. 9. (a) Applyig the law of refractio, we obtai si θ / si θ = si θ / si 30 = v s /v d. Cosequetly, ( ) v si m s si 30 s si si. vd 4.0 m s θ = = =

32 378 CHAPTER 35 (b) The agle of icidece is gradually reduced due to refractio, such as show i the calculatio above (from 30 to ). Evetually after several refractios, θ will be virtually zero. This is why most waves come i ormal to a shore. 9. Whe the depth of the liquid (L liq ) is zero, the phase differece φ is 60 wavelegths; this must equal the differece betwee the umber of wavelegths i legth L = 40 µm (sice the liquid iitially fills the hole) of the plastic (for ray r ) ad the umber i that same legth of the air (for ray r ). That is, L L =. plastic air 60 (a) Sice = m ad air = (to good approximatio), we fid plastic =.6. (b) The slope of the graph ca be used to determie liq, but we show a approach more closely based o the above equatio: L plastic L liq = 0 which makes use of the leftmost poit of the graph. This readily yields liq = The coditio for a miimum i the two-slit iterferece patter is d si θ = (m + ½), where d is the slit separatio, is the wavelegth, m is a iteger, ad θ is the agle made by the iterferig rays with the forward directio. If θ is small, si θ may be approximated by θ i radias. The, θ = (m + ½)/d, ad the distace from the miimum to the cetral frige is D y = Dtaθ Dsiθ Dθ = m+, d where D is the distace from the slits to the scree. For the first miimum m = 0 ad for the teth oe, m = 9. The separatio is D D 9D Δ y = 9 + =. d d d We solve for the wavelegth: 3 3 dδy c05. 0 mhc8 0 mh 7 l = = = m = 600 m. 9D m c h

33 379 Note: The distace betwee two adjacet dark friges, oe associated with the iteger m ad the other associated with the iteger m +, is Δy = Dθ = D/d. 94. A light ray travelig directly alog the cetral axis reaches the ed i time L L tdirect = = v c For the ray takig the critical zig-zag path, oly its velocity compoet alog the core axis directio cotributes to reachig the other ed of the fiber. That compoet is v cos θ ', so the time of travel for this ray is. t zig zag L = = L v cosθ c si / ( θ ) usig results from the previous solutio. Pluggig i siθ = we obtai L L tzig zag = =. c / c b g ad simplifyig, The differece is L L L Δ t = tzig zag tdirect = = c c c. With =.58, =.53, ad L = 300 m, we obtai t L (.58)(300 m).58 8 Δ = = s 5.6 s 8 = = c m/s Whe the iterferece betwee two waves is completely destructive, their phase differece is give by φ = (m+ ) π, m= 0,,,... The equivalet coditio is that their path-legth differece is a odd multiple of /, where is the wavelegth of the light. (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 m =.75 μm..

34 380 CHAPTER 35 (b) I the small agle approximatio (which we assume holds here), the friges are equally spaced, so that if Δy deotes the distace from oe maximum to the ext, the the distace from the middle of the patter to the fourth dark bad must be 6.8 mm = 3.5 Δy. Therefore, we obtai Δy = 6.8/3.5 = 4.8 mm. Note: The distace from the mth maximum to the cetral frige is D ybright = Dtaθ Dsiθ Dθ = m. d Similarly, the distace from the mth miimum to the cetral frige is y dark D = m+. d 96. We use the formula obtaied i Sample Problem Thi-film iterferece of a coatig o a glass les: Lmi Lmi = = = 0.00 = ( ) 97. Let the positio of the mirror measured from the poit at which d = d be x. We assume the beam-splittig mechaism is such that the two waves iterfere costructively for x = 0 (with some beam-splitters, this would ot be the case). We ca adapt Eq to this situatio by icorporatig a factor of (sice the iterferometer utilizes directly reflected light i cotrast to the double-slit experimet) ad elimiatig the si θ factor. Thus, the phase differece betwee the two light paths is Δφ = (πx/) = 4πx/. The from Eq. 35- (writig 4I 0 as I m ) we fid The itesity Δφ πx I Imcos Imcos. = F H G I K J = F H G I K J I / I m as a fuctio of x / is plotted below. From the figure, we see that the itesity is at a maximum whe

35 38 Similarly, the coditio for miima is m x=, m= 0,,,... x= ( m+ ), m= 0,,, We ote that ray travels a extra distace 4L more tha ray. For costructive iterferece (which is obtaied for = 60 m) we require 4L = m where m = some positive iteger. For destructive iterferece (which is obtaied for = 496 m) we require 4L = k where k = some positive odd iteger. Equatig these two equatios (sice their left-had sides are equal) ad rearragig, we obtai k = m = m =.5 m. We ote that this coditio is satisfied for k = 5 ad m =. It is satisfied for some larger values, too, but recallig that we wat the least possible value for L, we choose the solutio set (k, m) = (5, ). Pluggig back ito either of the equatios above, we obtai the distace L: 4L = L = = 30.0 m. 99. (a) Straightforward applicatio of Eq = c/ vad v = Δx/Δt yields the result: pistol with a time equal to Δt = Δx/c = s = 4.0 ps. (b) For pistol, the travel time is equal to s. (c) For pistol 3, the travel time is equal to s. (d) For pistol 4, the travel time is equal to s. (e) We see that the blast from pistol 4 arrives first. 00. We use Eq for costructive iterferece: L = (m + /), or b gb L m 30 m l = = =, m+ m+ m+ g

36 38 CHAPTER 35 where m = 0,,,. The oly value of m which, whe substituted ito the equatio above, would yield a wavelegth that falls withi the visible light rage is m =. Therefore, 30m l = = 49 m I the case of a distat scree the agle θ is close to zero so si θ θ. Thus from Eq. 35-4, m si Δθ Δ θ =Δ = Δ m =, d d d or d /Δθ = m/0.08 rad = m = 33 μm. 0. We ote that Δφ = 60 = π 3 rad. The phasors rotate with costat agular velocity Δφ π /3 rad ω = = = 6 Δ t.5 0 s rad/s. Sice we are workig with light waves travelig i a medium (presumably air) where the wave speed is approximately c, the kc = ω (where k = π/), which leads to = πc ω = 450 m.