to calculate the angular position θ of the first minimum. The minima are located by EXECUTE:

Transcription

1 DIFFRACTIO IDETIFY: Use y = xtnθ to clculte the ngulr position θ of the first minimum The minim re locted by m Eq(36): sin θ =, m =±, ±, First minimum mens m = nd sin θ = / nd = sin θ Use this eqution to clculte The centrl mximum is sketched in Figure 36 y = xtnθ y tnθ = = x = m θ = 675 rd Figure 36 = = = sin θ (75 )sin(675 rd) 56 nm EVALUATE: θ is smll so the pproximtion used to obtin Eq(363) is vlid nd this eqution could hve been used 36 IDETIFY: m The ngle is smll, so ym = x y = mm x x y = = = (6 m)(546 ) y 5 3 EVALUATE: The diffrction pttern is observed t distnce of 6 cm from the slit 363 IDETIFY: m The drk fringes re locted t ngles θ tht stisfy sin θ =, m =±, ±, The lrgest vlue of is 666 () Solve for m tht corresponds to = : m = = = 9 38 The lrgest vlue m 585 cn hve is 3 m =±, ±,, ± 3 gives 6 drk fringes 585 (b) For m =± 3, =± =± nd θ =± EVALUATE: When the slit width is decresed, there re fewer drk fringes When < there re no drk fringes nd the centrl mximum completely fills the screen 364 IDETIFY nd m / is very smll, so the pproximte expression ym = R is ccurte The distnce between the two drk fringes on either side of the centrl mximum is y R (633 )(35 m) y = = = 95 = 95 mm y = 59 mm 75 EVALUATE: When is decresed, the width y of the centrl mximum increses 365 IDETIFY: The minim re locted by = = cm x = 4 cm m 36-

2 36- Chpter 36 The ngle to the first minimum is θ = rcsin = rcsin 9 cm = 486 cm So the distnce from the centrl mximum to the first minimum is just y = x tn θ = (4 cm)tn(486 ) =± 454 cm EVALUATE: / is greter thn, so only the m = minimum is seen 366 IDETIFY: The ngle tht loctes the first diffrction minimum on one side of the centrl mximum is given by v = The time between crests is the period T f = nd = T f The time between crests is the period, so T = h () f v 8 km/h = = = h = = = 8 km T h f h 8 km (b) Afric-Antrctic: = nd θ = 45 km 8 km Austrli-Antrctic: = nd θ = 5 37 km EVALUATE: Diffrction effects re observed when the wvelength is bout the sme order of mgnitude s the dimensions of the opening through which the wve psses 367 IDETIFY: We cn model the hole in the concrete brrier s single slit tht will produce single-slit diffrction pttern of the wter wves on the shore For single-slit diffrction, the ngles t which destructive interference occurs re given by m = m/, where m =,, 3, () The frequency of the wter wves is f = 75 min = 5 s = 5 Hz, so their wvelength is = v/f = (5 cm/s)/(5 Hz) = cm At the first point for which destructive interference occurs, we hve tn θ = (63 m)/(3 m) θ = 84 sin θ = nd = /sin θ = ( cm)/(sin 84 ) = 638 cm (b) First find the ngles t which destructive interference occurs sin θ = / = ( cm)/(638 cm) θ = ± sin θ 3 = 3/ = 3( cm)/(638 cm) θ 3 = ±343 sin θ 4 = 4/ = 4( cm)/(638 cm) θ 4 = ±488 sin θ 5 = 5/ = 5( cm)/(638 cm) θ 5 = ±7 EVALUATE: These re lrge ngles, so we cnnot use the pproximtion tht θ m m/ 368 IDETIFY: m The minim re locted by = For prt (b) pply Eq(367) For the first minimum, m = The intensity t θ = is I m m 4 () = = sin9 = = = Thus = = 58 nm = 58 m (b) According to Eq(367), I sin [ π(sin θ) ] sin [ π(sin π / 4) ] = = = 8 I π(sin θ) π(sin π / 4) EVALUATE: I sin [( π / )(sin π / 4) ] If = /, for exmple, then t θ = 45, = = 8 As / decreses, I ( π /)(sin π /4) the screen becomes more uniformly illuminted 369 IDETIFY nd v= f gives The person hers no sound t ngles corresponding to diffrction minim The diffrction minim re locted by sin θ = m/, m=±, ±, Solve for θ = v/ f = (344 m/s)/(5 Hz) = 75 m; = m m=±, θ =± 6 ; m =±, θ =± 334 ; m =± 3, θ =± 556 ; no solution for lrger m EVALUATE: / = 8 so for the lrge wvelength sound wves diffrction by the doorwy is lrge effect Diffrction would not be observble for visible light becuse its wvelength is much smller nd / V

3 Diffrction IDETIFY: Compre E y to the expression E sin( ) y = Emx kx ωt nd determine k, nd from tht clculte m f = c/ The drk bnds re locted by = 8 c = 3 /s The first drk bnd corresponds to m = () E = E sin( kx ωt) π π π mx k = = = = 54 7 k 8 c 3 s 4 f = c f = = = 573 Hz (b) = 9 = = = m sin 86 (c) sin θ = m ( m=,, 3, ) 54 =± =± nd θ 9 m =± 74 EVALUATE: For m = 3, m is greter thn so only the first nd second drk bnds pper 36 IDETIFY nd sin θ = / loctes the first minimum y = xtnθ tn θ = y x= (365 cm) (4 cm) nd θ = 438 = θ = = μ sin (6 ) (sin 438 ) 9 m EVALUATE: θ = 74 rd nd = 67, so the pproximtion θ would not be ccurte 36 IDETIFY: m The ngle is smll, so ym = x pplies The width of the centrl mximum is y, so y = 3 mm x x (5 m)(5 ) 4 () y = = = = 47 y 3 5 x (5 m)(5 ) (b) = = = 47 = 4 cm y 3 x (5 m)(5 ) (c) = = = 47 y 3 EVALUATE: The rtio / stys constnt, so is smller when is smller 363 IDETIFY: Clculte the ngulr positions of the minim nd use y= xtnθ to clculte the distnce on the screen between them () The centrl bright fringe is shown in Figure 363 The first minimum is locted by = = = 35 θ = 89 rd Figure 363 y = xtn θ = (3 m)tn(89 rd) = 547 w= y = (547 ) = 9 = 9 mm (b) The first bright fringe on one side of the centrl mximum is shown in Figure 363b w= y y y = 547 (prt ()) = = 368 θ = 368 rd y = xtnθ = 85 Figure 363b w= y y = = 54 mm EVALUATE: The centrl bright fringe is twice s wide s the other bright fringes

4 36-4 Chpter IDETIFY: sin( β / ) I = I β / π β = The ngle θ is smll, so tn θ y/ x () 4 π (45 ) sin π y π θ y (5 m ) y β = = = x (6 )(3 m) 3 β (5 m )( ) y = : = = 76 (b) (c) I I sin( β ) sin(76) 8 β I = = = 76 I 3 β (5 m )(3 ) y = 3 : = = 8 I I sin( β ) sin(8) β I = = = 8 I 3 β (5 m )(5 ) y = 5 : = = 38 I I sin( β ) sin(38) 59 β I = = = 38 I x (6 )(3 m) EVALUATE: The first minimum occurs t y = = = 4 mm The distnces in prts () 4 45 nd (b) re within the centrl mximum y = 5 mm is within the first secondry mximum 365 () IDETIFY: Use Eq(36) with m = to locte the ngulr position of the first minimum nd then use y = xtnθ to find its distnce from the center of the screen The diffrction pttern is sketched in Figure 365 = = 54 = θ = 5 rd Figure 365 y = xtn θ = (3 m)tn(5 rd) = 675 = 675 mm (b) IDETIFY nd Use Eqs(365) nd (366) to clculte the intensity t this point Midwy between the center of the centrl mximum nd the first minimum implies (675 mm) 3375 y = = m y 3375 tnθ = = = 5 ; θ = 5 rd x 3 m The phse ngle β t this point on the screen is π π β = sin θ = (4 )sin(5 rd) = π 54 sin β/ sin π / Then I = I = (6 W/m ) β/ π / I 6 6 (6 W/m ) 43 = = W/m π4 EVALUATE: The intensity t this point midwy between the center of the centrl mximum nd the first minimum is less thn hlf the mximum intensity Compre this result to the corresponding one for the two-slit pttern, Exercise 353

5 Diffrction IDETIFY: In the single-slit diffrction pttern, the intensity is mximum t the center nd zero t the drk spots At other points, it depends on the ngle t which one is observing the light Drk fringes occur when sin θ m = m/, where m =,, 3,, nd the intensity is given by sin β / I β /, where π β /= () At the mximum possible ngle, θ = 9, so m mx = (sin9 )/ = (5 mm)/(638 nm) = 395 Since m must be n integer nd sin θ must be, m mx = 39 The totl number of drk fringes is 39 on ech side of the centrl mximum for totl of 78 (b) The frthest drk fringe is for m = 39, giving 39 = (39)(638 nm)/(5 mm) θ 39 = ±88 (c) The next closer drk fringe occurs t 38 = (38)(638 nm)/(5 mm) θ 38 = 74 The ngle midwy these two extreme fringes is ( )/ = 7745, nd the intensity t this ngle is I = sin β / I β /, where πsin θ π(5 mm)sin(7745 ) β /= = = 5 rd, which 638 nm sin(5 rd) gives I = ( 85 W/m ) 5 rd = W/m EVALUATE: At the ngle in prt (c), the intensity is so low tht the light would be brely perceptible 367 IDETIFY nd Use Eq(366) to clculte nd use Eq(365) to clculte I θ = 35, β = 56 rd, = 5 () π β = so π π(5 )sin35 = = = 668 nm β 56 rd sin β / (sin( β / )) [sin(8 rd)] 936 (b) I = I = I = I = I β/ β (56 rd) EVALUATE: At the first minimum β = π rd nd t the point considered in the problem β = 78π rd, so the point is well outside the centrl mximum Since β is close to mπ with m = 8, this point is ner one of the minim The intensity here is much less thn I π 368 IDETIFY: Use β = to clculte β The totl intensity is given by drwing n rc of circle tht hs length E nd finding the length of the chord which connects the strting nd ending points of the curve π π () β = sin θ π = = From Figure 368, Ep π = E E p = E π 4I The intensity is I = I = = 45 I This grees with Eq(365) π π π π (b) β = π = = From Figure 368b, it is cler tht the totl mplitude is zero, s is the intensity This lso grees with Eq(365) π π 3 (c) β = 3 π = = From Figure 368c, Ep 3 π = E E p = E The intensity is 3π 4 I = I = I This grees with Eq(365) 3π 9π

6 36-6 Chpter 36 EVALUATE: In prt () the point is midwy between the center of the centrl mximum nd the first minimum In prt (b) the point is t the first mximum nd in (c) the point is pproximtely t the loction of the first secondry mximum The phsor digrms help illustrte the rpid decrese in intensity t successive mxim Figure IDETIFY: The spce between the skyscrpers behves like single slit nd diffrcts the rdio wves Cncelltion of the wves occurs when sin θ = m, m =,, 3,, nd the intensity of the wves is sin β / given by I β /, where π β /= () First find the wvelength of the wves: = c/f = (3 8 m/s)/(889 MHz) = 3375 m For no signl, sin θ = m m = : sin θ = ()(3375 m)/(5 m) θ = ±3 m = : sin θ = ()(3375 m)/(5 m) θ = ±67 m = 3: sin θ 3 = (3)(3375 m)/(5 m) θ 3 = ±44 m = 4: sin θ 4 = (4)(3375 m)/(5 m) θ 4 = ±64 sin β / (b) I β /, where πsin θ π(5 m)sin(5 ) β /= = = 7 rd 3375 m sin(7 rd) I = ( 35 W/m ) 7 rd = 8 W/m EVALUATE: The wvelength of the rdio wves is very long compred to tht of visible light, but it is still considerbly shorter thn the distnce between the buildings 36 IDETIFY: The net intensity is the product of the fctor due to single-slit diffrction nd the fctor due to double slit interference φ sin β / The double-slit fctor is IDS = I cos nd the single-slit fctor is ISS = β / () d = m = m/d = /d, = /d, 3 = 3/d, 4 = 4/d (b) At the interference bright fringes, cos πsin θ π( d/3) φ/ = nd β / = = π( d/3)( / d) At θ, sin θ = /d, so β / = = π /3 The intensity is therefore φ sin β/ sin π /3 I = I cos = I() β / π /3 = 684 I π( d/3)( / d) At θ, sin θ = /d, so β / = = π /3 Using the sme procedure s for θ, we hve I = sin π /3 I() π /3 = 7 I At θ 3, we get β /= π, which gives I 3 = since sin π = sin 4 π /3 At θ 4, sin θ 4 = 4/d, so β / = 4 π /3, which gives I4 = I = 47 I 4 π /3 (c) Since d = 3, every third interference mximum is missing (d) In Figure 36c in the textbook, every fourth interference mximum t the sides is missing becuse d = 4

7 Diffrction 36-7 EVALUATE: The result in this problem is different from tht in Figure 36c becuse in this cse d = 3, so every third interference mximum t the sides is missing Also the envelope of the intensity function decreses more rpidly here thn in Figure 36c becuse the first diffrction minimum is reched sooner, nd the decrese in intensity from one interference mximum to the next is fster for = d/3 thn for = d/4 36 () IDETIFY nd The interference fringes (mxim) re locted by dsin θ = m, with sin β / π m =, ±, ±, The intensity I in the diffrction pttern is given by I = I, with β = sin θ β / We wnt m =± 3 in the first eqution to give θ tht mkes I = in the second eqution π 3 d = m gives β = = π(3 / d) d I = sys sin β / = so β = π nd then π = π(3 / d) nd ( d/ ) = 3 β / (b) IDETIFY nd Fringes m =, ±, ± re within the centrl diffrction mximum nd the m =± 3 fringes coincide with the first diffrction minimum Find the vlue of m for the fringes tht coincide with the second diffrction minimum Second minimum implies β = 4 π π π m β = = = πm( / d) = π( m/3) d Then β = 4π sys 4π = π( m /3) nd m = 6 Therefore the m =± 4 nd m =+ 5 fringes re contined within the first diffrction mximum on one side of the centrl mximum; two fringes EVALUATE: The centrl mximum is twice s wide s the other mxim so it contins more fringes 36 IDETIFY nd Use Figure 364b in the textbook There is totlly destructive interference between slits whose phsors re in opposite directions By exmining the digrm, we see tht every fourth slit cncels ech other EVALUATE: The totl electric field is zero so the phsor digrm corresponds to point of zero intensity The first two mxim re t φ = nd φ = π, so this point is not midwy between two mxim 363 () IDETIFY nd If the slits re very nrrow then the centrl mximum of the diffrction pttern for ech slit completely fills the screen nd the intensity distribution is given solely by the two-slit interference The mxim re given by d = m so sin θ = m/ d Solve for θ 58 st order mximum: m =, so = = = 94 ; d 53 θ = 67 nd order mximum: m =, so = = 88 ; θ = 5 d (b) IDETIFY nd sin β / The intensity is given by Eq(36): I = I cos ( φ / ) Clculte φ nd β / β t ech θ from prt () πd πd m φ = = = πm, so cos ( φ/ ) = cos ( mπ) = d (Since the ngulr positions in prt () correspond to interference mxim) π π m 3 mm β = = = πm ( / d) = m π = m(3794 rd) d 53 mm sin(3794/ ) rd st order mximum: m =, so I I() = = 49I (3794/ ) rd sin3794 rd nd order mximum: m =, so I = I() = 56I 3794 rd EVALUATE: The first diffrction minimum is t n ngle θ given by sin θ = / so θ = 4 The first order fringe is within the centrl mximum nd the second order fringe is inside the first diffrction mximum on one side of the centrl mximum The intensity here t this second fringe is much less thn I 364 IDETIFY: A double-slit bright fringe is missing when it occurs t the sme ngle s double-slit drk fringe Single-slit diffrction drk fringes occur when sin θ = m, nd double-slit interference bright fringes occur when d sin θ = m

8 36-8 Chpter 36 () The ngles re the sme for cncelltion, so dividing the equtions gives d/ = m /m m /m = 7 m = 7m When m =, m = 7; when m =, m = 4, nd so forth, so every 7 th bright fringe is missing from the double-slit interference pttern EVALUATE: (b) The result is independent of the wvelength, so every 7 th fringe will be cncelled for ll wvelengths But the bright interference fringes occur when d sin θ = m, so the loction of the cncelled fringes does depend on the wvelength 365 IDETIFY nd The phsor digrms re similr to those in Fig364 An interference minimum occurs when the phsors dd to zero () The phsor digrm is given in Figure 365 Figure 365 There is destructive interference between the light through slits nd 3 nd between nd 4 (b) The phsor digrm is given in Figure 365b Figure 365b There is destructive interference between the light through slits nd nd between 3 nd 4 (c) The phsor digrm is given in Figure 365c Figure 365c There is destructive interference between light through slits nd 3 nd between nd 4 EVALUATE: Mxim occur when φ =, π, 4 π, etc Our digrms show tht there re three minim between the mxim t φ = nd φ = π This grees with the generl result tht for slits there re minim between ech pir of principl mxim 366 IDETIFY: A double-slit bright fringe is missing when it occurs t the sme ngle s double-slit drk fringe Single-slit diffrction drk fringes occur when sin θ = m, nd double-slit interference bright fringes occur when d sin θ = m () The ngle t which the first bright fringe occurs is given by tn θ = (53 mm)/(5 mm) θ = 357 d sin θ = nd d = /( ) = (638 nm)/sin(357 ) = 3 m = 3 mm (b) The 7 th double-slit interference bright fringe is just cncelled by the st diffrction drk fringe, so diff = / nd interf = 7/d The ngles re equl, so / = 7/d = d/7 = (3 mm)/7 = 48 mm EVALUATE: We cn generlize tht if d = n, where n is positive integer, then every n th double-slit bright fringe will be missing in the pttern md 367 IDETIFY: The diffrction minim re locted by = nd the two-slit interference mxim re locted mi by sin θ = The third bright bnd is missing becuse the first order single slit minimum occurs t the sme d ngle s the third order double slit mximum 3cm The pttern is sketched in Figure 367 tnθ =, so θ = 9 9 cm Single-slit drk spot: = nd Double-slit bright fringe: d = 3 nd 5 nm 5 4 = = = nm = 5 μ m (width) sin9 3 3(5 nm) 4 d = = = 45 nm = 45 μm (seprtion) sin9

9 Diffrction 36-9 EVALUATE: ote tht d/ = 3 Figure IDETIFY: The mxim re locted by d = m The order corresponds to the vlues of m First-order: d = Fourth-order: d 4 = 4 d 4 4 =, sin θ4 = 4sin θ = 4sin894 nd θ4 = 384 d EVALUATE: We did not hve to solve for d 369 IDETIFY nd The bright bnds re t ngles θ given by dsin θ = m Solve for d nd then solve for θ for the specified order 4 () θ = 784 for m = 3 nd = 68 nm, so d = m/ = 86 cm The number of slits per cm is / d = 479 slits/cm (b) st order: m =, so sin θ = / d = (68 )/(86 ) nd θ = 9 nd order: m =, so = / d nd θ = 48 (c) For m= 4, = 4 / d is greter thn, so there is no 4th-order bright bnd EVALUATE: The ngulr position of the bright bnds for prticulr wvelength increses s the order increses 363 IDETIFY: The bright spots re locted by d = m Third-order mens m = 3 nd second-order mens m = m d constnt mrr mvv = r v m v v 4 nm v = sin θr = (sin 65 ) = 345 nd θ v = mr r 3 7 nm EVALUATE: The third-order line for prticulr occurs t lrger ngle thn the second-order line In given order, the line for violet light (4 nm) occurs t smller ngle thn the line for red light (7 nm) 363 IDETIFY nd Clculte d for the grting Use Eq(363) to clculte θ for the longest wvelength in the visible spectrum nd verify tht θ is smll Then use Eq(363) to relte the liner seprtion of lines on the screen to the difference in wvelength 5 () d = cm = 9 For = 7 nm, / d = 63 The first-order lines re locted t sin θ = / d; is smll enough for θ to be n excellent pproximtion (b) y = x / d, where x = 5 m The distnce on the screen between st order bright bnds for two different wvelengths is Δ y = x( Δ )/ d, so Δ = d Δ y x = = 5 ( ) / ( )(3 ) /(5 m) 33 nm EVALUATE: The smller d is (greter number of lines per cm) the smller the Δ tht cn be mesured 363 IDETIFY: The mxim re locted by d = m 35 slits mm d = =

10 36- Chpter 36 4 m = : θ4 = rcsin = rcsin 85 = d 86 7 θ7 = rcsin = rcsin 48 = Δ θ = = 63 d (4 ) m = 3: θ4 = rcsin = rcsin 48 = d (7 ) θ7 = rcsin = rcsin 473 = Δ θ = = 5 d 86 EVALUATE: Δθ is lrger in third order 3633 IDETIFY: The mxim re locted by d = m d = 6 m m[638 ] θ = rcsin = rcsin rcsin([396] m) = For m =, θ = 33 For d 6 m =, θ = 53 There re no other mxim EVALUATE: The reflective surfce produces the sme interference pttern s grting with slit seprtion d 3634 IDETIFY: The mxim re locted by d = m 5 slits cm d = = 5 5 d sin θ ( )sin35 () = = = 467 m m (467 ) (b) m = : θ = rcsin = rcsin = 78 d EVALUATE: Since the ngles re firly smll, the second-order devition is pproximtely twice the first-order devition 3635 IDETIFY: The mxim re locted by d = m 35 slits mm d = = m m(5 ) θ = rcsin = rcsin rcsin((8) m) = d 86 m= : θ = 5 ; m= : θ = 3 ; m= 3: θ = 33 EVALUATE: The ngles re not precisely proportionl to m, nd devite more from being proportionl s the ngles increse 3636 IDETIFY: The resolution is described by R = = m Mxim re locted by d = m Δ For 5 slits/mm, d = (5 slits mm) = (5, slits m) () = = = 8 slits m Δ ( ) m (b) θ = sin θ = sin (()(65645 )(5, m )) = 497 nd d θ = sin (()(6567 )(5, m )) = 46 Δ θ = 37 EVALUATE: dcos θ dθ = /, so for 8 slits the ngulr intervl Δθ between ech of these mxim nd the 656 first djcent minimum is Δ θ = = = 37 This is the sme s the ngulr d cos θ (8)( )cos 4 seprtion of the mxim for the two wvelengths nd 8 slits is just sufficient to resolve these two wvelengths in second order 3637 IDETIFY: The resolving power depends on the line density nd the width of the grting The resolving power is given by R = m = = /Δ () R = m = (5 lines/cm)(35 cm)() = 7,5 (b) The resolving power needed to resolve the sodium doublet is R = /Δ = (589 nm)/(58959 nm 589 nm) = 998

11 so this grting cn esily resolve the doublet (c) (i) R = /Δ Since R = 7,5 when m =, R = 7,5 = 35, for m = Therefore Δ = /R = (5878 nm)/35, = 68 nm min = +Δ = 5878 nm + 68 nm = nm (ii) mx = Δ = 5878 nm 68 nm = nm Diffrction 36- EVALUATE: (iii) Therefore the rnge of resolvble wvelengths is nm < < nm 3638 IDETIFY nd m Δ = 5878 nm slits = = = = 33 slits = = 75 m Δ ( nm 5878 nm) 78 cm cm cm EVALUATE: A smller number of slits would be needed to resolve these two lines in higher order 3639 IDETIFY nd The mxim occur t ngles θ given by Eq(366), dsin θ = m, where d is the spcing between djcent tomic plnes Solve for d second order sys m = m (85 ) d = = = 3 = 3 nm sin 5 EVALUATE: Our result is similr to d clculted in Exmple IDETIFY: The mxim re given by d = m, m =,, d d () m = nd = = (35 )sin5 = 8 = 8 nm This is n x ry m 8 (b) sin θ = m = m m(586) = m = : θ = 3 m = 3 : θ = 59 The eqution d [35 ] doesn't hve ny solutions for m > 3 EVALUATE: In this problem / d = IDETIFY: Ryleigh's criterion sys = D = 35 5 The best resolution is 3 rcseconds, which is bout (833 ) (55 ) () D = = = 46 m 5 sin(833 ) EVALUATE: (b) The Keck telescopes re ble to gther more light thn the Hle telescope, nd hence they cn detect finter objects However, their lrger size does not llow them to hve greter resolution tmospheric conditions limit the resolution 364 IDETIFY: Apply = D θ = (/ 6) (55 ) D = = = 3 = 3 mm sin(/6) EVALUATE: The lrger the dimeter the smller the ngle tht cn be resolved 3643 IDETIFY: Apply = D W θ =, where W = 8 km nd h = km θ is smll, so θ h 6 h D = = = (36 m) = 88 m 4 W 8 EVALUATE: D must be significntly lrger thn the wvelength, so much lrger dimeter is needed for microwves thn for visible wvelengths 3644 IDETIFY: Apply = D 8 θ is smll, so θ = rd 6 8 Dsin θ Dθ (8 )( ) = = = 656 m = 656 cm EVALUATE: corresponds to microwves

12 36- Chpter IDETIFY nd The ngulr size of the first drk ring is given by = / D (Eq367) Clculte θ, nd then the dimeter of the ring on the screen is (45 m) tn θ 6 = ; = θ = 4 rd 74 The rdius of the Airy disk (centrl bright spot) is r = (45 m) tnθ = 46 m The dimeter is r = 9 m= 9 cm EVALUATE: / D = 84 For this smll D the centrl diffrction mximum is brod 3646 IDETIFY: Ryleigh s criterion limits the ngulr resolution Ryleigh s criterion is sin θ θ = /D () Using Ryleigh s criterion θ = /D = ()(55 nm)/(35/4 mm) = 99 5 rd On the ber this ngle subtends distnce x θ = x/r nd x = Rθ = (5 m)(99 5 rd) = 9 4 m = 3 mm (b) At f/, D is 4/ times s lrge s t f/4 Since θ is proportionl to /D, nd x is proportionl to θ, x is /(4/) = /4 times s lrge s it ws t f/4 x = (9 mm)(/4) = 3 mm EVALUATE: A wide-ngle lens, such s one hving focl length of 8 mm, would hve much smller opening t f/ nd hence would hve n even less resolving bility 3647 IDETIFY nd Resolved by Ryleigh s criterion mens ngulr seprtion θ of the objects equls / D The ngulr seprtion θ of the objects is their liner seprtion divided by their distnce from the telescope 3 5, θ = where 593 is the distnce from erth to Jupiter Thus θ = (5 ) Then θ = nd D = = = 45 m D θ 46 EVALUATE: This is very lrge telescope mirror The greter the ngulr resolution the greter the dimeter the lens or mirror must be 3648 IDETIFY: Ryleigh s criterion sys θ res = D y D = 7 cm θ res =, where s is the distnce of the object from the lens nd y = 4 mm s y yd (4 )(7 ) = s = = = 49 m s D (55 ) EVALUATE: The focl length of the lens doesn t enter into the clcultion In prctice, it is difficult to chieve resolution tht is t the diffrction limit 3649 IDETIFY nd Let y be the seprtion between the two points being resolved nd let s be their distnce y from the telescope Then the limit of resolution corresponds to = D s () Let the two points being resolved be the opposite edges of the crter, so y is the dimeter of the 8 crter For the moon, s = 38 y = s D Hubble: D = 4 m nd = 4 nm gives the mximum resolution, so y = 77 m 6 Arecibo: D = 35 m nd = 75 m; y = yd (b) s = Let y 3 (the size of license plte) s = (3 m)(4 m) [()(4 )] = 5 km EVALUATE: D / is much lrger for the opticl telescope nd it hs much lrger resolution even though the dimeter of the rdio telescope is much lrger 365 IDETIFY: Apply = D θ is smll, so θ Smllest resolving ngle is for short-wvelength light (4 nm) 4 8, mi θ = () = 96 rd θ =, where R is the distnce to the str D 58 m R, mi 6, km R = = = 8 7 km θ 96 rd EVALUATE: This is less thn light yer, so there re no strs this close

13 Diffrction IDETIFY: Let y be the seprtion between the two points being resolved nd let s be their distnce from the telescope The limit of resolution corresponds to D = y s 6 s = 48 ly = 45 Assume visible light, with = 4 m 6 9 y = s D = (4 )(45 ( m) = EVALUATE: 8 The dimeter of Jupiter is 38, so the resolution is insufficient, by bout one order of mgnitude 365 IDETIFY: If the pprtus of Exercise 364 is plced in wter, then ll tht chnges is the wvelength = n For y<< x, the distnce between the two drk fringes on either side of the centrl mximum is D = y Let D= y be the seprtion of 59 found in Exercise 364 x x D 59 y = = = = = 444 = 444 mm n n 33 EVALUATE: The wter shortens the wvelength nd this decreses the width of the centrl mximum 3653 () IDETIFY nd sin β / The intensity in the diffrction pttern is given by Eq(365): I = I, β / where π β = sin θ Solve for θ tht gives I = I The ngles θ + nd θ re shown in Figure 3653 I = I so sin β / = β / Let x = β / ; the eqution for x is sin x = = 77 x Use tril nd error to find the vlue of x tht is solution to this eqution x (sin x) / x rd 84 5 rd 665 rd rd rd 777; thus x= 39 rd nd β = x= 78 rd (i) For (ii) For Figure 3653 Δ θ = θ θ = θ + + β = + π = 78 rd = 445 π rd =, + = 445 = ; θ+ = 78 ; Δ θ = θ+ = 56 = 5, + = 445 = 885; θ+ = 577 ; Δ θ = θ+ = 5 (iii) For =, + = 445 = 445; θ+ = 536 ; Δ θ = θ+ = 5 (b) IDETIFY nd = loctes the first minimum Solve for θ (i) For, sin θ ; θ 3 ; θ 6 = = = = (ii) For 5, sin θ ; θ 54 ; θ 3 = = 5 = =

14 36-4 Chpter 36 (iii) For =, sin θ = ; θ = 574 ; θ = 5 EVALUATE: Either definition of the width shows tht the centrl mximum gets nrrower s the slit gets wider 3654 IDETIFY: The two holes behve like double slits nd cuse the sound wves to interfere fter they pss through the holes The motion of the spekers cuses Doppler shift in the wvelength of the sound The wvelength of the sound tht strikes the wll is = v s T s, nd destructive interference first occurs where sin θ = / () First find the wvelength of the sound tht strikes the openings in the wll = v s T s = v/f s v s /f s = (v v s )/f s = (344 m/s 8 m/s)/(5 Hz) = m Destructive interference first occurs where d sin θ = /, which gives d = /( ) = ( m)/( sin 7 ) = 48 m (b) = v/f = (344 m/s)/(5 Hz) = 75 m = /d = (75 m)/[(48 m)] θ = ±67 EVALUATE: The moving source produces sound of shorter wvelength thn the sttionry source, so the ngles t which destructive interference occurs re smller for the moving source thn for the sttionry source 3655 IDETIFY nd sin θ = / loctes the first drk bnd In the liquid the wvelength chnges nd this chnges the ngulr position of the first diffrction minimum ir liquid sin θir = ; liquid = liquid liquid = ir = 4836 ir = / n (Eq335), so ir n = ir / liquid = / 4836 = 7 EVALUATE: Light trvels fster in ir nd n must be > The smller in the liquid reduces θ tht locted the first drk bnd 3656 IDETIFY: d =, so the bright fringes re locted by = Red: sin R = 7 nm Violet: sin V = 4 nm R 7 = V 4 sin( θv + 5 ) 7 θr θv = 5 θr = θv + 5 = Using trig identify from Appendix B, V 4 V cos5 + cosθv sin5 = 74 cos5 + cotθ sin5 = 7 4 V V tnθv = 33 θv = 83 nd θr = θv + 5 = = 333 Then sin θ R = 7 nm gives R sin = = = 784 lines m = 784 lines cm The spectrum begins t 8 3 nd ends t nm 7 EVALUATE: As is incresed, the ngulr rnge of the visible spectrum increses 3657 () IDETIFY nd The ngulr position of the first minimum is given by = m (Eq36), with m = The distnce of the minimum from the center of the pttern is given by y = xtn θ 54 = = = 5 ; θ = 5 rd 36 y = xtn θ = ( m) tn(5 rd) = 8 = 8 mm (ote tht θ is smll enough for θ tn θ, nd Eq(363) pplies) (b) IDETIFY nd Find the phse ngle β where I = I / Then use Eq(366) to solve for θ nd y = xtnθ to find the distnce From prt () of Problem 3653, I = I when β = 78 rd π β β = (Eq(366)), so sin θ = π βx (78 rd)(54 )( m) 4 y= xtnθ x = = 796 = 796 mm π π(36 ) EVALUATE: The point where I = I / is not midwy between the center of the centrl mximum nd the first minimum; see Exercise 365

15 sinγ 3658 IDETIFY: I = I γ respect to γ is zero Diffrction 36-5 The mximum intensity occurs when the derivtive of the intensity function with d sinγ d = cosγ = dγ dγ γ γ di d sinγ sinγ cosγ sinγ cosγ sinγ = I = = dγ dγ γ γ γ γ γ γ γ cosγ = sin γ γ = tnγ (b) The grph in Figure 3658 is plot of f( γ ) = γ tn γ When f ( γ ) equls zero, there is n intensity mximum Getting estimtes from the grph, nd then using tril nd error to nrrow in on the vlue, we find tht the three smllest γ -vlues re γ = 449 rd 773 rd, nd 9 rd EVALUATE: γ = is the centrl mximum The three vlues of γ we found re the loctions of the first three secondry mxim The first four minim re t γ = 34 rd, 68 rd, 94 rd, nd 6 rd The mxim re between djcent minim, but not precisely midwy between them Figure IDETIFY nd Relte the phse difference between djcent slits to the sum of the phsors for ll slits The πd πdθ phse difference between djcent slits is φ θ θ φ Thus θ = π d A principl mximum occurs when φ = φmx = m π, where m is n integer, since then ll the phsors dd The first minim on either side of the m th ± principl mximum occur when φ = φ min= m π ± ( π / ) nd the phsor digrm for slits forms closed loop nd the resultnt phsor is zero The ngulr position of principl mximum is ± ± θ = φmx The ngulr position of the djcent minimum is θ min= φ min π d π d + π π θ min= φmx + = θ + = θ + πd πd d π θ min= φmx = θ π d d + The ngulr width of the principl mximum is θ min θ min=, s ws to be shown d EVALUATE: The ngulr width of the principl mximum decreses like / s increses 366 IDETIFY: The chnge in wvelength of the H α line is due to Doppler shift in the wvelength due to the motion of the glxy From Eqution 63, the Doppler effect formul for light is fr = c v fs c+ v First find the wvelength of the light using the grting informtion = d sin θ = [/(575,8 lines/m)] sin 34 = 69 7 m = 69 nm c v Using Eqution 63, we hve f R = f S In this cse, f R is the frequency of the 69-nm light tht the c+ v cosmologist mesures, nd f S is the frequency of the 6563-nm light of the H α line obtined in the lbortory

16 36-6 Chpter 36 ( ) fr / fs Solving for v gives v = c Since f = c, f = c/, which gives f R /f S = S / R Substituting this into the + ( fr/ fs) eqution for v, we get S 6563 nm R v = c = 69 nm 8 ( 3 /s ) = 5 7 m/s, 6563 nm S + + R 69 nm which is 5% the speed of light EVALUATE: Since v is positive, the glxy is moving wy from us We cn lso see this becuse the wvelength hs incresed due to the motion 366 IDETIFY nd Drw the specified phsor digrms There is totlly destructive interference between two slits when their phsors re in opposite directions () For eight slits, the phsor digrms must hve eight vectors The digrms for ech specified vlue of φ re sketched in Figure 366 In ech cse the phsors ll sum to zero (b) The dditionl phsor digrms for φ = 3 π / nd 3 π / 4 re sketched in Figure 366b 3π 5π 7π 3 π For φ =, φ =,nd φ =, totlly destructive interference occurs between slits four prt For φ =, totlly destructive interference occurs with every second slit EVALUATE: At minimum the phsors for ll slits sum to zero Figure IDETIFY: Mxim re given by d = m d is the seprtion between crystl plnes m 5 nm () θ = rcsin = rcsin m = rcsin(6 m) d (8nm) For m= : θ = 8, m= : θ = 63, m= 3: θ = 47, nd m= 4 : θ = 64 o lrger m vlues yield nswers (b) If the seprtion d = m, then θ = rcsin rcsin(334 m) = So for m= : θ = 83, m= : θ = 388, nd m= 3: θ = 7 o lrger m vlues yield nswers EVALUATE: In prt (b), where d is smller, the mxim for ech m re t lrger θ 3663 IDETIFY nd In ech cse consider the relevnt phsor digrm () For the mxim to occur for slits, the sum of ll the phse differences between the slits must dd to zero (the phsor digrm closes on itself) This requires tht, dding up ll the reltive phse shifts, π m φ = π m, for some integer m Therefore φ =, for m not n integer multiple of, which would give mximum π m (b) The sum of phse shifts φ = brings you full circle bck to the mximum, so only the previous phses yield minim between ech pir of principl mxim EVALUATE: The minim between ech pir of principl mxim cuse the mxim to become shrper s increses 3664 IDETIFY: Set d = in the expressions for φ nd β nd use the results in Eq(36) Figure 3664 shows pir of slits whose width nd seprtion re equl

17 Diffrction 36-7 Figure 3664 shows tht the two slits re equivlent to single slit of width πd π φ = sin θ, so β sin θ φ = = So then the intensity is sin ( β/) (sin( β/)cos( β/)) sin β sin ( β /) π ( ) I = Icos ( β /) I = = I = I, where β = sin θ, ( β/) β β ( β /) which is Eq (355) with double the slit width EVALUATE: In Chpter 35 we considered the limit where << d > d is not possible Figure IDETIFY nd The condition for n intensity mximum is dsin θ = m, m=, ±, ±, Third order mens m = 3 The longest observble wvelength is the one tht gives θ = 9 nd hence θ = 5 65 lines/cm so 65 lines/m nd d = m = d sin θ (538 )() = = = 53 = 53 nm m 3 EVALUATE: The longest wvelength tht cn be obtined decreses s the order increses 3666 IDETIFY nd As the rys first rech the slits there is lredy phse difference between djcent slits of πd This, dded to the usul phse difference introduced fter pssing through the slits, yields the condition for n intensity mximum For mximum the totl phse difference must equl π m πd πd + = πm d( + sin θ ) = m (b) 6 slits mm d = = For θ =, m = : θ = rcsin() = 65 m = : θ = rcsin = rcsin 9 = d m = : θ = rcsin = rcsin 9 = d 67 For θ =, m = : θ = rcsin( sin ) = 65 m = : θ = rcsin sin 7 = m = : θ = rcsin sin = EVALUATE: When θ >, the mxim re shifted downwrd on the screen, towrd more negtive ngles 3667 IDETIFY: m The mxim re given by d = m We need = in order for ll the visible d wvelengths re to be seen For 65 slits mm d = =

18 36-8 Chpter 36 3 = 4 : m= : = 6; m= : = 5; m= 3: = 78 d d d 3 = 7 : m= : = 46; m= : = 9; m= 3: = 37 So, the third order does not contin the violet d d d end of the spectrum, nd therefore only the first nd second order diffrction ptterns contin ll colors of the spectrum EVALUATE: θ for ech mximum is lrger for longer wvelengths 3668 IDETIFY: Apply = D Δx 8 θ is smll, so, where Δx is the size of the detil nd R = 7 ly ly = 94 km = cf / R 5 8 Δ x R () cr ()(3 km s)(7 ly) sin θ = Δ x = = = = 6 ly 3 9 D R D Df (77 km)(665 Hz) 3 (94 km ly)( 6 ly) = 94 km EVALUATE: = 8 cm / D is very smll, so Δx is very smll Still, R is very lrge nd R Δxis mny orders of mgnitude lrger thn the dimeter of the sun 3669 IDETIFY nd Add the phses between djcent sources st () d sin θ = m Plce mximum t or θ = 9 d = If d <, this puts the first mximum beyond Thus, if d < there is only single principl mximum (b) At principl mximum when δ =, the phse difference due to the pth difference between djcent slits d is Φ pth = π This just scles πrdins by the frction the wvelength is of the pth difference between djcent sources If we dd reltive phse δ between sources, we still must mintin totl phse difference of zero to keep our principl mximum πd δ πd Φ pth ± δ = =± δ or θ = sin 8 m (c) m d = = (count the number of spces between 5 points) Let θ = 45 Also recll f = c, so 4 δ 9 π ( m)(88 Hz)sin 45 mx 8 =± =± 6 rdins (3 s) EVALUATE: δ must vry over wider rnge in order to sweep the bem through greter ngle 367 IDETIFY: The wvelength of the light is smller under wter thn it is in ir, which will ffect the resolving power of the lens, by Ryleigh s criterion The wvelength under wter is = /n, nd for smll ngles Ryleigh s criterion is θ = /D () In ir the wvelength is = c/f = (3 8 m/s)/(6 4 Hz) = 5 7 m In wter the wvelength is = /n = (5 7 m)/33 = m With the lens open ll the wy, we hve D = f/8 = (35 mm)/8 = (35 m)/8 In the wter, we hve sin θ θ = /D = ()(376 7 m)/[(35 m)/8] = rd Clling w the width of the resolvble detil, we hve θ = w/r w = Rθ = (75 mm)(367 5 rd) = mm (b) θ = /D = ()(5 7 m)/[(35 m)/8] = rd w = Rθ = (75 mm)(488 5 rd) = 34 mm EVALUATE: Due to the reduced wvelength underwter, the resolution of the lens is better under wter thn in ir 367 IDETIFY nd Resolved by Ryleigh s criterion mens the ngulr seprtion θ of the objects is given by θ = / D θ = y/ s, where y = 75 m is the distnce between the two objects nd s is their distnce from the stronut (her ltitude) y = s D yd (75 m)(4 ) 5 s = = = 49 = 49 km (5 ) EVALUATE: In prctice, this diffrction limit of resolution is not chieved Defects of vision nd distortion by the erth s tmosphere limit the resolution more thn diffrction does

19 Diffrction IDETIFY: Apply sin D θ = Δx θ is smll, so, where Δx is the size of the detils nd R is the distnce to the erth R 5 ly = DΔ x (6 )(5 ) 7 () R = = = 3 = 3ly 5 ()( ) 5 5 R ()( )(4 ly)(94 ly) 8 (b) Δ x = = = 484 km This is bout, times the D m dimeter of the erth! ot enough resolution to see n erth-like plnet! Δx is bout 3 times the distnce from the erth to the sun 5 5 ()( )(59 ly)(94 ly) 6 (c) Δ x = = 3 = 3 km 6 6 Δx 3 km = = 89 ; Δx is smll compred to the size of the plnet 5 D 38 km plnet EVALUATE: The very lrge dimeter of Plnet Imger llows it to resolve plnet-sized detil t gret distnces 3673 IDETIFY nd Follow the steps specified in the problem () From the segment dy, the frction of the mplitude of E tht gets through is dy dy E de = E sin( kx ωt) (b) The pth difference between ech little piece is Edy y sin θ kx= k( D y sin θ) de = sin( k( D y sin θ) ωt) This cn be rewritten s Edy de = (sin( kd ωt)cos( ky sin θ) + sin( ky sin θ)cos( kd ωt)) (c) So the totl mplitude is given by the integrl over the slit of the bove E E = de = dy (sin( kd ωt) cos( ky sin θ) sin( ky sin θ) + cos( kd ωt)) But the second term integrtes to zero, so we hve: E sin( ky sin θ ) E = sin( kd ωt) dy (cos( ky sin )) E sin ( kd t) θ = ω ksin θ sin( k(sin θ) ) sin( π(sin θ) ) E = Esin( kd ωt) = E sin( kd ωt) k(sin θ) π(sin θ) sin [] At θ =, = E = E sin( kd ωt) [ ] sin( k(sin θ)/ ) sin( β ) I E I = I = I (d) Since, where we hve used k(sin θ)/ β I = E sin ( kx ωt ) EVALUATE: The sme result for I( θ ) is obtined s ws obtined using phsors 3674 IDETIFY nd Follow the steps specified in the problem () Ech source cn be thought of s trveling wve evluted t x= R with mximum mplitude of E However, ech successive source will pick up n extr phse from its respective pthlength to point d sin θ P φ = π which is just π, the mximum phse, scled by whtever frction the pth difference, d sin θ, is of the wvelength, By dding up the contributions from ech source (including the ccumulting phse difference) this gives the expression provided i( kr ωt+ nφ) (b) e = cos( kr ωt+ nφ ) + isin( kr ωt+ nφ ) The rel prt is just cos ( kr ωt + nφ ) So, i( kr ωt+ nφ) Re Ee = Ecos( kr ωt + nφ ) (ote: Re mens the rel prt of ) But this is just n= n= E cos( kr ωt) + E cos( kr ωt + φ) + E cos( kr ωt+ φ) + + E cos( kr ωt+ ( ) φ)

20 36- Chpter 36 i( kr ωt+ nφ ) iωt + ikr inφ i( kr ωt) inφ inφ iφ n n x (c) E e = E e e e = Ee e e = (e ) But recll x = n= n= n= n= n= n= x iφ iφ iφ/ iφ / iφ / iφ/ iφ/ iφ iφ n e e e ( e e ) i( ) φ / ( e e ) Let x = e so (e ) = (nice trick!) But = = e iφ iφ iφ/ iφ/ iφ/ iφ/ iφ/ n= e e e ( e e ) ( e e ) Putting everything together: iφ/ iφ/ i( kr ωt+ nφ) i( kr ωt+ ( ) φ/) ( e e ) Ee = E e iφ/ iφ/ n= ( e e ) cos φ/ + isin φ/ cos φ/ + isin φ/ = E [ cos( kr ωt+ ( ) φ/) + isin( kr ωt + ( ) φ/) ] cos φ/ + isin φ/ cos φ/ + isin φ/ sin( φ/) Tking only the rel prt gives E cos( kr ωt+ ( ) φ/) = E sin φ/ sin ( φ / ) (d) I = E = I (The cos term goes to v sin ( φ / ) in the time verge nd is included in the definition of I ) E I sin ( / ) I(sin / cos / ) EVALUATE: (e) = I = φ φ φ φ I 4I cos sin φ/ = sin φ/ = Looking t Eq(359), E I I E but for us I = 4 sin ( φ / ) 3675 IDETIFY nd From Problem 3674, I = I Use this result to obtin ech result specified sin φ / in the problem sin ( φ/ ) cos( φ/ ) () lim I Use l'hopitl's ˆ rule: lim = lim = So lim I = I φ φ sin φ/ φ cos( φ/ ) φ π (b) The loction of the first minimum is when the numertor first goes to zero t φmin = π or φmin = The width of the centrl mximum goes like φ min, so it is proportionl to φ (c) Whenever = nπ where n is n integer, the numertor goes to zero, giving minimum in intensity Tht is, nπ I is minimum wherever φ = This is true ssuming tht the denomintor doesn t go to zero s well, which φ occurs when = mπ, where m is n integer When both go to zero, using the result from prt(), there is mximum Tht is, if n is n integer, there will be mximum (d) From prt (c), if n is n integer we get mximum Thus, there will be minim (Plces where n is not n integer for fixed nd integer n ) For exmple, n = will be mximum, but n=,, will be minim with nother mximum t n= φ π 3π (e) Between mxim is hlf-integer multiple of π ie,, etc) nd if is odd then sin ( φ / ) sin φ /, so I I EVALUATE: These results show tht the principl mxim become shrper s the number of slits is incresed