Chapter 35 Solutos 35.1 The Moo's radius is 1.74 10 6 m ad the Earth's radius is 6.37 10 6 m. The total distace traveled by the light is: d = (3.4 10 m 1.74 10 6 m 6.37 10 6 m) = 7.5 10 m This takes.51 s, so v = 7.5 10 m.51 s =.995 10 m/s = 99.5 Mm/s 35. x = ct ; c = x t = (1.50 10 km)(1000 m/km) (.0 mi)(60.0 s/mi) =.7 10 m/s = 7 Mm/s 35.3 The experimet is most covicig if the wheel turs fast eough to pass outgoig light through oe otch ad returig light through the ext: t = l c θ = ω t = ω l c so ω = cθ l = (.99 10 )[π /(70)] (11.45 10 3 = 114 rad/s ) The returig light would be blocked by a tooth at oe-half the agular speed, givig aother data poit. 35.4 (a) For the light beam to make it through both slots, the time for the light to travel the distace d must equal the time for the disk to rotate through the agle θ, if c is the speed of light, d c = θ ω, so c = dω θ (b) We are give that d =.50 m, c = dω θ =.50 m 100. π rad θ = = 91. 10 4 rad, 60. 0 10 3.49 10 4 rad s.91 10 4 rad ω = 5555 rev s = 3.00 10 ms= 300 Mm/s π rad 1.00 rev = 3.49 104 rad s 35.5 Usig Sell's law, si θ = si θ1 θ = 5.5 λ1 λ = = 44 m 1 1 000 by Harcourt, Ic. All rights reserved.
Chapter 35 Solutios 33 c 35.6 (a) f = = λ 300. 10 m/s 7 6.3 10 m = 4.74 10 14 Hz (b) λ glass λ air 63. m = = = 1.50 4 m (c) v glass air 300. 10 m/s c = = 1.50 = 00. 10 m/s = 00 Mm/s 35.7 si θ = si θ 1 1 si θ 1 = 1.333 si 45.0 si θ 1 = (1.33)(0.707) = 0.943 θ 1 = 70.5 19.5 above the horizo Figure for Goal Solutio Goal Solutio A uderwater scuba diver sees the Su at a apparet agle of 45.0 from the vertical. What is the actual directio of the Su? G : The sulight refracts as it eters the water from the air. Because the water has a higher idex of refractio, the light slows dow ad beds toward the vertical lie that is ormal to the iterface. Therefore, the elevatio agle of the Su above the water will be less tha 45 as show i the diagram to the right, eve though it appears to the diver that the su is 45 above the horizo. O : We ca use Sell s law of refractio to fid the precise agle of icidece. A : Sell s law is: siθ = siθ 1 1 which gives si θ 1 = 1.333 si 45.0 siθ 1 = (1.333)(0.707) = 0.943 The sulight is at θ 1 = 70.5 to the vertical, so the Su is 19.5 above the horizo. L : The calculated result agrees with our predictio. Whe applyig Sell s law, it is easy to mix up the idex values ad to cofuse agles-with-the-ormal ad agles-with-the-surface. Makig a sketch ad a predictio as we did here helps avoid careless mistakes. 000 by Harcourt, Ic. All rights reserved.
34 Chapter 35 Solutios *35. (a) si θ = si θ 1 1 1.00 si 30.0 = si 19.4 = 1.5 (c) c f = = λ 300. 10 m/s 7 6.3 10 m = 4.74 10 14 Hz i air ad i syrup. (d) v c = = 300. 10 m/s 1.5 = 19. 10 m/s = 19 Mm/s v (b) λ = = f 19. 10 14 m/s 474. 10 / s = 417 m 35.9 (a) Flit Glass: v c = = 300. 10 1.66 ms = 11. 10 m s = 11 Mm/s (b) Water: v c = = 300. 10 ms 1.333 = 5. 10 m s = 5 Mm/s (c) Cubic Zircoia: v c = = 300. 10.0 ms = 136. 10 m s = 136 Mm/s 35.10 si θ = si θ ; 1.333 si 37.0 = si 5.0 1 1 = 1.90 = c v ; v = c 1.90 = 1.5 10 m/s = 15 Mm/s 35.11 si θ = si θ ; θ = si 1 1 si θ 1 1 1 1 ( 100. )(si 30 ) θ = si 150. = 19.5 θ ad θ 3 are alterate iterior agles formed by the ray cuttig parallel ormals. So, θ 3 = θ = 19.5. 1.50 si θ 3 = (1.00) si θ 4 θ 4 = 30.0
Chapter 35 Solutios 35 λ 0 436 m 35.1 (a) Water λ = = = 1.333 37 m (b) λ 0 436 m Glass λ = = = 1.5 7 m *35.13 si θ1 = w si θ si θ 1 1 = si θ = si( 90. 0. 0 ) = 0. 66 1. 333 1. 333 1 θ 1 = si 0. 66 = 41. 5 d h = = ta θ 300. m ta 41.5 = 3.39 m 35.14 (a) From geometry, 1.5 m = d si 40.0 so d = 1.94 m (b) 50.0 above horizotal, or parallel to the icidet ray *35.15 The icidet light reaches the left-had mirror at distace (1.00 m) ta 5.00 = 0.075 m above its bottom edge. The reflected light first reaches the right-had mirror at height (0.075 m) = 0.175 m It bouces betwee the mirrors with this distace betwee poits of cotact with either. Sice 1.00 m 0.175 m = 5.7, the light reflects five times from the right-had mirror ad six times from the left. 000 by Harcourt, Ic. All rights reserved.
36 Chapter 35 Solutios *35.16 At etry, si θ = si θ or 1.00 si 30.0 = 1.50 si θ 1 1 θ = 19.5 The distace h the light travels i the medium is give by cos θ = (.00 cm) h ( 00. cm) or h = = 1. cm cos 19.5 The agle of deviatio upo etry is α = θ1 θ = 30. 0 19. 5 = 10. 5 The offset distace comes from si α = d h : d = (.1 cm) si 10.5 = 0.3 cm 00. cm *35.17 The distace, h, traveled by the light is h = = 1. cm cos 19.5 The speed of light i the material is Therefore, t h = = v 1. 10 m = 106. 10 00. 10 m/s v 10 c = = s = 106 ps 300. 10 m/s = 00. 10 1.50 m/s *35.1 Applyig Sell's law at the air-oil iterface, air si θ = si 0. 0 yields θ = 30.4 oil Applyig Sell's law at the oil-water iterface w si θ = oil si 0. 0 yields θ =. 3 *35.19 time differece = (time for light to travel 6.0 m i ice) (time to travel 6.0 m i air) t = 6.0 m v ice 6.0 m c but v = c t = (6.0 m) 1.309 1 c c (6.0 m) = (0.309) = 6.39 10 9 s = 6.39 s c
Chapter 35 Solutios 37 *35.0 Cosider glass with a idex of refractio of 1.5, which is 3 mm thick The speed of light i the glass is 3 10 m/s = 10 15. m/s The extra travel time is 3 10 3 m 10 m/s 3 10 3 m 3 10 m/s ~ 10 11 s For light of wavelegth 600 m i vacuum ad wavelegth 600 m = 400 m i glass, 1.5 the extra optical path, i wavelegths, is 3 10 4 10 3 7 m 3 10 m 6 10 3 7 m m ~ 103 wavelegths = *35.1 Refractio proceeds accordig to 100. si θ 166. si θ (1) 1 (a) For the ormal compoet of velocity to be costat, v 1 cos θ 1 = v cos θ = or c cos θ c 1. 66 cos θ () 1 We multiply Equatios (1) ad (), obtaiig: si θ cos θ = si θ cos θ 1 1 or si θ = si θ 1 The solutio θ1 = θ = 0 does ot satisfy Equatio () ad must be rejected. The physical solutio is θ = 10 θ or θ = 90. 0 θ. The Equatio (1) becomes: 1 1 which yields θ 1 = 5.9 si θ = 166. cos θ, or ta θ 1 = 166. 1 1 (b) Light eterig the glass slows dow ad makes a smaller agle with the ormal. Both effects reduce the velocity compoet parallel to the surface of the glass, so that compoet caot remai costat, or will remai costat oly i the trivial case θ 1 = θ = 0 35. See the sketch showig the path of the light ray. α ad γ are agles of icidece at mirrors 1 ad. For triagle abca, α + γ + β = 10 or β = 10 α + γ (1) Now for triagle bcdb, ( 90.0 α)+ ( 90.0 γ)+ θ = 10 or θ = α + γ () 000 by Harcourt, Ic. All rights reserved.
3 Chapter 35 Solutios Substitutig Equatio () ito Equatio (1) gives β = 10 θ Note: From Equatio (), γ = θ α. Thus, the ray will follow a path like that show oly if α < θ. For α > θ, γ is egative ad multiple reflectios from each mirror will occur before the icidet ad reflected rays itersect. 35.3 Let (x) be the idex of refractio at distace x below the top of the atmosphere ad x= h = be its value at the plaet surface. The, x = 1.000 + 1.000 h x (a) The total time required to traverse the atmosphere is h dx t = = 0 v 0 h 3 c x dx = 1 h 1.000 + 1.000 c h x 0 dx = h c + 1.000 ch 0. 0 10 m 1. 005 + 1. 000 t = 3.00 10 m s = 66. µs h = h c + 1.000 (b) The travel time i the absece of a atmosphere would be h / c. Thus, the time i the presece of a atmosphere is + 1.000 = 1.005 times larger or 0.50% loger. 35.4 Let (x) be the idex of refractio at distace x below the top of the atmosphere ad x= h = be its value at the plaet surface. The, x = 1.000 + 1.000 h x (a) The total time required to traverse the atmosphere is h dx t = = 0 v 0 h c x dx = 1 h 1.000 + 1.000 c h x 0 dx = h c + 1.000 ch h = h c + 1.000 (b) The travel time i the absece of a atmosphere would be h / c. Thus, the time i the presece of a atmosphere is + 1. 000 times larger
Chapter 35 Solutios 39 35.5 From Fig. 35.0 v = 1.470 at 400 m ad r = 1.45 at 700 m The ( 1. 00)si θ = 1. 470 si θv ad ( 1. 00)si θ 1. 45 si θ δ r δ v = θ r θ v = si 1 si θ 1.45 si θ si 1 1.470 δ = si 1 si 30.0 1.45 si 30.0 si 1 1.470 = 0.171 = r 35.6 1 si θ 1 = si θ so θ = si 1 1 si θ 1 θ = si 1 (1.00)(si 30.0 ) = 19.5 1. 50 θ 3 = ([(90.0 19.5 ) + 60.0 ] 10 )+ 90.0 = 40.5 3 si θ 3 = 4 si θ 4 so θ 4 = si 1 3 si θ 3 = si 1 (1.50)(si 40.5 ) = 77.1 1.00 4 35.7 Takig Φ to be the apex agle ad δ mi to be the agle of miimum deviatio, from Equatio 35.9, the idex of refractio of the prism material is si Φ+δ mi = si Φ Solvig for δ mi, δ mi = si 1 si Φ Φ= si 1 [(.0)si( 5.0 )] 50.0 = 6. 35. (700 m) = 1.45 (a) (1.00) si 75.0 = 1.45 si θ ; θ = 41.5 (b) Let θ 3 + β = 90.0, θ + α = 90.0 ; the α + β + 60.0 =10 So 60.0 θ θ 3 = 0 60.0 41.5 =θ 3 = 1.5 (c) 1.45 si 1.5 = 1.00 si θ 4 θ 4 = 7.6 (d) γ = (θ 1 θ ) + [β (90.0 θ 4 )] γ = 75.0 41.5 +(90.0 1.5 ) (90.0 7.6 ) = 4.6 000 by Harcourt, Ic. All rights reserved.
330 Chapter 35 Solutios 35.9 For the icomig ray, si θ1 si θ = Usig the figure to the right, (θ ) violet = si 1 si 50.0 1.66 = 7.4 (θ ) red = si 1 si 50.0 1.6 =. For the outgoig ray, θ 3 = 60.0 θ ad siθ 4 = siθ 3 (θ 4 ) violet = si 1 [1.66 si 3.5 ] = 63.17 (θ 4 ) red = si 1 [1.6 si 31.7 ] = 5.56 The dispersio is the differece θ 4 =(θ 4 ) violet (θ 4 ) red = 63.17 5.56 = 4.61 si Φ+δ mi 35.30 = si Φ Φ+δ mi For small Φ, δ mi Φ so is also a small agle. approximatio ( siθ θ whe θ << 1 rad), we have: The, usig the small agle ( Φ+δ mi) = Φ+δ mi Φ Φ or δ mi ( 1)Φ where Φ is i radias. = 35.31 At the first refractio, 100. si θ1 si θ The critical agle at the secod surface is give by si θ 3 = 1.00, or θ 3 = si 1 1.00 1. 50 = 41.. But, θ = 60.0 θ 3. Thus, to avoid total iteral reflectio at the secod surface (i.e., have θ 3 < 41. ), it is ecessary that θ > 1.. Sice si θ 1 = si θ, this requiremet becomes si θ 1 > ( 1. 50)si( 1. )= 0.46, or θ 1 > 7.9
Chapter 35 Solutios 331 35.3 At the first refractio, ( 1.00)si θ 1 = si θ. The critical agle at the secod surface is give by si θ 3 = 1.00, or θ 3 = si 1 ( 1.00 ) But ( 90.0 θ )+ ( 90.0 θ 3 )+Φ=10, which gives θ =Φ θ 3. ad avoid total iteral reflectio at the secod surface, it is Thus, to have θ 3 < si 1 1.00 ecessary that θ >Φ si 1 ( 1.00 ). Sice si θ = si θ, this requiremet becomes 1 si θ 1 > si Φ si 1 1.00 or θ 1 > si 1 si Φ si 1 1.00 Through the applicatio of trigoometric idetities, θ 1 > si 1 1 si Φ cos Φ 35.33 = si δ + φ si φ / ta 1 φ si 5 ( )= 1.544 cos5 so ad φ = 1.1 1.544si( 1 φ )= si( 5 + 1 φ )= cos( 1 φ )si 5 +si( 1 φ )cos5 *35.34 Note for use i every part: Φ+ ( 90.0 θ )+ ( 90.0 θ 3 )= 10 so θ 3 =Φ θ At the first surface is α = θ 1 θ At exit, the deviatio is β = θ 4 θ 3 The total deviatio is therefore δ = α + β = θ 1 + θ 4 θ θ 3 = θ 1 + θ 4 Φ (a) At etry: 1 si θ 1 = si θ or θ = si 1 si 4.6 1. 50 = 30.0 Thus, θ 3 = 60. 0 30. 0 = 30. 0 At exit: 1.50 si 30.0 =1.00 si θ 4 or θ 4 = si 1 [ 1.50 si( 30.0 )]= 4.6 so the path through the prism is symmetric whe θ 1 = 4. 6. (b) δ = 4. 6 + 4. 6 60. 0 = 37. si 45. 6 (c) At etry: siθ = θ =. 4 θ3 = 60. 0. 4 = 31. 6 150. At exit: si θ = 150. si 316. θ 517. 45. 6 51. 7 60. 0 37.3 = δ = + = 4 4 si 51. 6 (d) At etry: si θ = θ = 31. 5 θ3 = 60. 0 31. 5 =. 5 150. At exit: si θ = 150. si ( 5. ) θ = 457. δ = 51. 6 + 45. 7 60. 0 = 37.3 4 4 000 by Harcourt, Ic. All rights reserved.
33 Chapter 35 Solutios 35.35 si θ = 1. From Table 35.1, (a) θ = si (b) θ = si (c) θ = si. 419 = 1 1 166. = 1 1 1. 309 = 1 1 4.4 37.0 49. 35.36 si θc = ; θc = si 1 1 1 (a) Diamod: θ c = si (b) Flit glass: θ c = si 1 1. 333. 419 = 1 1. 333 166. = 33.4 53.4 (c) Ice: Sice > 1, there is o critical agle. 35.37 si θ c = 1 (Equatio 35.10) = 1 si. = (1.0003)(0.999) = 1.000 0 *35.3 si θ c air 100. = = = 0. 735 θ c = 47.3 136. pipe Geometry shows that the agle of refractio at the ed is θ r = 90.0 θ c = 90.0 47.3 = 4.7 The, Sell's law at the ed, 1.00 si θ = 1.36 si 4.7 gives θ = 67. 35.39 For total iteral reflectio, 1 si θ 1 = si 90. 0 (1.50) si θ 1 = (1.33)(1.00) or θ 1 = 6.4
Chapter 35 Solutios 333 35.40 To avoid iteral reflectio ad come out through the vertical face, light iside the cube must have θ 1 3 < si ( 1/ ) 1 So θ > 90. 0 si ( 1/ ) But θ < 90. 0 ad si θ < 1 1 1 1 I the critical case, si ( 1/ ) = 90. 0 si ( 1/ ) 1/ = si 45.0 = 1.41 35.41 From Sell's law, si θ = si θ 1 1 At the extreme agle of viewig, θ = 90.0 (1.59)(si θ 1 ) = (1.00) si 90.0 So θ 1 = 39.0 Therefore, the depth of the air bubble is r d ta θ 1 < d < r p ta θ 1 or 1.0 cm < d < 1.17 cm *35.4 (a) si θ si θ 1 = v v 1 ad θ = 90. 0 at the critical agle si 90. 0 150 ms = si 343 ms θ c so θ c = si 1 0. 15 = 10.7 (b) Soud ca be totally reflected if it is travelig i the medium where it travels slower: air (c) Soud i air fallig o the wall from most directios is 100% reflected, so the wall is a good mirror. 000 by Harcourt, Ic. All rights reserved.
334 Chapter 35 Solutios *35.43 For plastic with idex of refractio 1. 4 surrouded by air, the critical agle for total iteral reflectio is give by θ c = si 1 1 1 si 1 1. 4 = 44. I the gasolie gauge, skylight from above travels dow the plastic. The rays close to the vertical are totally reflected from both the sides of the slab ad from facets at the lower ed of the plastic, where it is ot immersed i gasolie. This light returs up iside the plastic ad makes it look bright. Where the plastic is immersed i gasolie, with idex of refractio about 1.50, total iteral reflectio should ot happe. The light passes out of the lower ed of the plastic with little reflected, makig this part of the gauge look dark. To frustrate total iteral reflectio i the gasolie, the idex of refractio of the plastic should be <.1, sice θ c = si 1 1.50 (.1) = 45.0. *35.44 Assume the lifeguard s path makes agle θ 1 with the orthsouth ormal to the shorelie, ad agle θ with this ormal i the water. By Fermat s priciple, his path should follow the law of refractio: si θ 1 = v 1 = 7.00 m s si θ v 1. 40 m s = 5.00 or θ = si 1 si θ 1 5 The lifeguard o lad travels eastward a distace x = ( 16. 0 m) ta 1. m x =(. m) ta θ further east. Thus, 6 0 m = ( 16 0 m ) θ + ( 0 0 m) travels 6 0 0 0 or 6.0 m = ( 16.0 m)ta θ 1 + ( 0.0 m)ta si 1 si θ 1 5 We home i o the solutio as follows: θ 1 (deg) 50.0 60.0 54.0 54. 54.1 right-had side. m 31. m 5.3 m 5.99 m 6.003 m θ. The i the water, he.. ta. ta θ 1 The lifeguard should start ruig at 54. east of orth. *35.45 Let the air ad glass be medium 1 ad, respectively. By Sell's law, si θ = 1 si θ 1 or 1.56 si θ = si θ 1 But the coditios of the problem are such that θ 1 = θ. 1.56 si θ = si θ We ow use the double-agle trig idetity suggested. 1.56 si θ = si θ cos θ or cos θ = Thus, θ = 3.7 ad θ 1 = θ = 77.5 1. 56 = 0.70
Chapter 35 Solutios 335 *35.46 (a) θ1 = θ1 = 30.0 si θ = si θ 1 1 (1.00) si 30.0 = 1.55 siθ θ = 1. (b) θ 1 = θ 1 = 30.0 θ = si 1 1 si θ 1 = si 1 155. si 300. 1 = 50. (c) ad (d) The other etries are computed similarly, ad are show i the table below. (c) air ito glass, agles i degrees (d) glass ito air, agles i degrees icidece reflectio refractio icidece reflectio refractio 0 0 0 0 0 0 10.0 10.0 6.43 10.0 10.0 15.6 0.0 0.0 1.7 0.0 0.0 3.0 30.0 30.0 1. 30.0 30.0 50. 40.0 40.0 4.5 40.0 40.0 5.1 50.0 50.0 9.6 50.0 50.0 oe* 60.0 60.0 34.0 60.0 60.0 oe* 70.0 70.0 37.3 70.0 70.0 oe* 0.0 0.0 39.4 0.0 0.0 oe* 90.0 90.0 40. 90.0 90.0 oe* *total iteral reflectio 35.47 For water, si θ c = 1 4/3 = 3 4 Thus θ c = si 1 (0.750) = 4.6 ad d = [(1.00 m)ta θ c ] d = (.00 m)ta 4.6 =.7 m Figure for Goal Solutio 000 by Harcourt, Ic. All rights reserved.
336 Chapter 35 Solutios Goal Solutio A small uderwater pool light is 1.00 m below the surface. The light emergig from the water forms a circle o the water's surface. What is the diameter of this circle? G : Oly the light that is directed upwards ad hits the water s surface at less tha the critical agle will be trasmitted to the air so that someoe outside ca see it. The light that hits the surface farther from the ceter at a agle greater tha θ c will be totally reflected withi the water, uable to be see from the outside. From the diagram above, the diameter of this circle of light appears to be about m. O : We ca apply Sell s law to fid the critical agle, ad the diameter ca the be foud from the geometry. A : The critical agle is foud whe the refracted ray just grazes the surface (θ = 90 ). The idex of refractio of water is = 1.33, ad 1 = 1.00 for air, so siθ c = si 90 gives θ c = si 1 1 1.333 = si 1 ( 0.750)= 4.6 1 The radius the satisfies taθ c = r (1.00 m) So the diameter is d = r = 1.00 ( m)ta 4.6 =.7 m L : Oly the light rays withi a 97. coe above the lamp escape the water ad ca be see by a outside observer (Note: this agle does ot deped o the depth of the light source). The path of a light ray is always reversible, so if a perso were located beeath the water, they could see the whole hemisphere above the water surface withi this coe; this is a good experimet to try the ext time you go swimmig! *35.4 Call θ 1 the agle of icidece ad of reflectio o the left face ad θ those agles o the right face. Let α represet the complemet of θ 1 ad β be the complemet of θ. Now α = γ ad β = δ because they are pairs of alterate iterior agles. We have A = γ + δ = α + β ad B = α + A + β = α + β + A = A
Chapter 35 Solutios 337 *35.49 (a) We see the Su swigig aroud a circle i the exteded plae of our parallel of latitude. Its agular speed is ω = θ π rad = t 6 400 s = 7.7 10 5 rad s The directio of sulight crossig the cell from the widow chages at this rate, movig o the opposite wall at speed v = rω = (.37 m) ( 7.7 10 5 rad s)= 1.7 10 4 ms= 0.17 mm s (b) The mirror folds ito the cell the motio that would occur i a room twice as wide: v = rω = ( 0.174 mm s)= 0.345 mm s (c) ad (d) As the Su moves southward ad upward at 50.0, we may regard the corer of the widow as fixed, ad both patches of light move orthward ad dowward at 50.0. *35.50 By Sell's law, 1 si θ 1 = si θ With v = c, This is also true for soud. Here, c si θ 1 = c si θ or v 1 v si 1. 0 si θ = 340 m/s 1510 m/s si θ1 si θ = v1 v θ = arcsi (4.44 si 1.0 ) = 67.4 *35.51 (a) = c v =.99 10 ms 61.15 km 1.00 h hr 3600 s 1.00 103 m 1.00 km = 1.76 10 7 ( 7 ) 1 = (b) 1 si θ 1 = si θ so 1. 76 10 si θ 1. 00 si 90. 0 θ 1 = 35. 10 6 degree This problem is misleadig. The speed of eergy trasport is slow, but the speed of the wavefrot advace is ormally fast. The codesate's idex of refractio is ot far from uity. 000 by Harcourt, Ic. All rights reserved.
33 Chapter 35 Solutios *35.5 Violet light: Red Light: ( 1. 00) si 5. 0 = 1. 69 si θ θ = 14. 490 y v = ( 5. 00 cm ) ta = ( 5. 00 cm) ta 14. 490 θ = 1.6 cm ( 1. 00) si 5. 0 = 1. 64 si θ θ = 14. 915 y R = = 5. 00 cm ta 14. 915 1. 331 cm The emerget beams are both at 5.0 from the ormal. Thus, w = y cos 5. 0 where y = 1.331 cm 1.6 cm = 0.0396 cm w = ( 0. 396 ) cos 5. 0 = mm 0.359 mm 35.53 Horizotal light rays from the settig Su pass above the hiker. The light rays are twice refracted ad oce reflected, as i Figure (b) below, by just the certai special raidrops at 40.0 to 4.0 from the hiker's shadow, ad reach the hiker as the raibow. The hiker sees a greater percetage of the violet ier edge, so we cosider the red outer edge. The radius R of the circle of droplets is Figure (a) R = (.00 km)(si 4.0 ) = 5.35 km The the agle φ, betwee the vertical ad the radius where the bow touches the groud, is give by 00. km.00 km cos φ = = = 0. 374 R 53. 5 km or φ = 6.1 The agle filled by the visible bow is 360 ( 6.1 ) = 4, so the visible bow is 4 360 = 6.% of a circle Figure (b)
Chapter 35 Solutios 339 4 35.54 From Sell s law, ( 100. ) si θ1 = si θ 3 so x = Rsi θ = r si θ r R = si θ si θ 1 = 3 4 1 apparet depth actual depth = z d = r cos θ 1 R cos θ = 3 4 d z cos θ 1 r x R Fish 1 si θ θ θ 1 eye Fish at depth d Image at depth z But si θ = 3 4 si θ 1 = 9 ( 16 1 cos θ 1 ) So z d = 3 4 cos θ 1 1 9 16 + 9 = 3 4 16 cos θ 1 cos θ 1 7 + 9 cos θ 1 16 or z = 3d cos θ 1 7 + 9 cos θ 1 = θ givig θ 31 35.55 As the beam eters the slab, 100. si 500. 14. si =.. The beam the strikes the top of the slab at x 1 = 155. mm ta( 31. ) from the left ed. Thereafter, the beam strikes a face each time it has traveled a distace of x 1 alog the legth of the slab. Sice the slab is 40 mm log, the beam has a additioal 40 mm x 1 to travel after the first reflectio. The umber of additioal reflectios is 3.10 mm ta 31. 40 mm x 1 40 mm 1.55 mm ta 31. = = 1.5 x 1 or 1 reflectios sice the aswer must be a iteger. The total umber of reflectios made i the slab is the. *35.56 (a) (b) If medium 1 is glass ad medium is air, There is o differece S 1 = 1 S 1 + 1 S 1 = 1 S 1 + 1 1. 5 1.00 = 1. 5 + 1.00 1.00 1. 5 = 1.00 + 1. 5 = 0.046 = 0.046; (c) S 1 = 1.76 107 1.00 S 1 1.76 10 7 + 1.00 = 1.76 107 + 1.00.00 1.76 10 7 + 1.00 S 1 = S 1 00. + 00. 100. 100. = 1.00.7 10 7 or 100% 7 176 10 100 7.. 176. 10 + 100. This suggests he appearace would be very shiy, reflectig practically all icidet light. See, however, the ote cocludig the solutio to problem 35.51. 000 by Harcourt, Ic. All rights reserved.
340 Chapter 35 Solutios *35.57 (a) With 1 = 1 ad =, the reflected fractioal itesity is S 1 = 1 S 1 + 1. The remaiig itesity must be trasmitted: S = 1 1 S 1 + 1 = ( + 1) 1 + 1 = + + 1 + 1 ( + 1) = 4 + 1 (b) At etry, S = 1 1 S 1 + 1 = 0. = 4.419.419 + 1 At exit, Overall, S 3 S = 0. S 3 = S 3 S = 0. S 1 S S 1 = 0.65 or 6.5% 4 *35.5 Defie T = as the trasmissio coefficiet for oe ( + 1) ecouter with a iterface. For diamod ad air, it is 0., as i problem 57. As show i the figure, the total amout trasmitted is + T ( 1 T) 4 + T ( 1 T) 6 +... T + T 1 T +... + T 1 T We have 1 T = 1 0. = 0.17 so the total trasmissio is [ + ( 0.17 ) 4 + ( 0.17 ) 6 +...] ( 0.) 1 + 0.17 + ( 0.17) 4 + ( 0.17) 6 +... F = ( 0.17) + ( 0.17) 4 + ( 0.17) 6 +..., ad F = 1 + ( 0.17) + ( 0.17) 4 + ( 0.17) 6 +... = F. To sum this series, defie F = 1 + 0.17 Note that 0.17 1 + 0.17 The, 1 = F ( 0.17) F or F = 1 1 0.17 The overall trasmissio is the ( 0.) (. ) 1 0 17 = 0. 706 or 70.6%
Chapter 35 Solutios 341 35.59 si 4. 0 = si 90. 0 so = 1 4 0 = 149. si. si 1. 0 si θ 1 = si 1. 0 ad si θ 1 = si 4. 0 θ 1 = 7.5 Figure for Goal Solutio Goal Solutio The light beam show i Figure P35.59 strikes surface at the critical agle. Determie the agle of icidece θ 1. G : From the diagram it appears that the agle of icidece is about 40. O : We ca fid θ 1 by applyig Sell s law at the first iterface where the light is refracted. At surface, kowig that the 4.0 agle of reflectio is the critical agle, we ca work backwards to fid θ 1. A : Defie 1 to be the idex of refractio of the surroudig medium ad to be that for the prism material. We ca use the critical agle of 4.0 to fid the ratio 1 : si 4.0 = 1 si 90.0 So, 1 = = 1. 49 1 si 4.0 Call the agle of refractio θ at the surface 1. The ray iside the prism forms a triagle with surfaces 1 ad, so the sum of the iterior agles of this triagle must be 10. Thus, ( 90.0 θ )+ 60.0 + ( 90.0 4.0 )= 10 Therefore, θ = 1.0 Applyig Sell s law at surface 1, 1 siθ 1 = si 1.0 siθ 1 = ( 1 )si θ = ( 1. 49)si 1.0 θ 1 = 7.5 L : The result is a bit less tha the 40.0 we expected, but this is probably because the figure is ot draw to scale. This problem was a bit tricky because it required four key cocepts (refractio, reflectio, critical agle, ad geometry) i order to fid the solutio. Oe practical extesio of this problem is to cosider what would happe to the exitig light if the agle of icidece were varied slightly. Would all the light still be reflected off surface, or would some light be refracted ad pass through this secod surface? 000 by Harcourt, Ic. All rights reserved.
34 Chapter 35 Solutios 35.60 Light passig the top of the pole makes a agle of icidece φ 1 = 90.0 θ. It falls o the water surface at distace s 1 = (L d) ta θ from the pole, ad has a agle of refractio φ from (1.00)si φ 1 = si φ. The s = dta φ ad the whole shadow legth is s 1 + s = L d ta θ + dta si φ 1 si 1 s 1 + s = L d cos θ + dta si 1 ta θ =.00 m ta 40.0 + (.00 m)ta cos 40.0 si 1 1. 33 = 3.79 m 35.61 (a) For polystyree surrouded by air, iteral reflectio requires θ 3 = si 1 1.00 1. 49 = 4. The from the geometry, θ = 90.0 θ 3 = 47. From Sell's law, si θ 1 = ( 1. 49)si 47. =1.10 This has o solutio. Therefore, total iteral reflectio always happes. (b) For polystyree surrouded by water, θ 3 = si 1 1. 33 1. 49 ad θ = 6. From Sell's law, θ 1 = 30.3 (c) No iteral refractio is possible sice the beam is iitially travelig i a medium of lower idex of refractio. *35.6 δ = θ1 θ = 10. 0 ad 1 si θ 1 = si θ with 1 = 1, = 4 3 Thus, θ 1 = si 1 ( si θ ) = si 1 [ si(θ 1 10.0 ) ]
Chapter 35 Solutios 343 (You ca use a calculator to home i o a approximate solutio to this equatio, testig differet values of θ 1 util you fid that θ 1 = 36.5. Alteratively, you ca solve for θ 1 exactly, as show below.) We are give that si θ 1 = 4 3 si(θ 1 10.0 ) This is the sie of a differece, so 3 4 si θ 1 = si θ 1 cos 10.0 cos θ 1 si 10.0 3 Rearragig, si 10. 0 cos θ1 = cos 10. 0 si θ 1 4 si 10. 0 = ta θ1 ad θ 1 = ta 1 0. 740 = 36.5 cos 10. 0 0. 750 400. cm 35.63 ta θ 1 = h ad 00. cm ta θ = h ta θ1.00 ta θ = 4.00 ta θ = si θ1 si θ = 400. ( 1 si θ ) ( 1 si θ ) 1 (1) Sell's law i this case is: si θ = si θ 1 1 si θ = 1. 333 si θ 1 Squarig both sides, si θ 1 = 1.777 si θ () Substitutig () ito (1), Defiig x = si θ, 1.777 si θ 1 1.777 si θ = 4.00 si θ 1 si θ 0.444 ( 1 1.777x) = 1 ( 1 x) Solvig for x, 0.444 0.444x = 1 1.777x ad x = 0.417 From x we ca solve for θ : θ = si 1 0.417 = 40. (.00 cm) (.00 cm) Thus, the height is h = = ta θ ta(. ) = 40.37 cm 000 by Harcourt, Ic. All rights reserved.
344 Chapter 35 Solutios 35.64 Observe i the sketch that the agle of icidece at poit P is γ, ad usig triagle OPQ: si γ = L / R. Also, cos γ = 1 si γ = R L Applyig Sell s law at poit P, ( 1.00)si γ = si φ Thus, si φ = si γ = L R ad cos φ = 1 si φ = R L From triagle OPS, φ + ( α + 90.0 )+ ( 90.0 γ)= 10 or the agle of icidece at poit S is α = γ φ. The, applyig Sell s law at poit S gives ( 1.00)si θ = si α = si( γ φ), or R R L si θ = [ si γ cos φ cos γ si φ]= R R L R R L R L R siθ = L R R L R L ad θ = si 1 L R R L R L 35.65 To derive the law of reflectio, locate poit O so that the time of travel from poit A to poit B will be miimum. The total light path is L= asec θ + bsec θ 1 The time of travel is t = 1 v (a sec θ 1 + b sec θ ) If poit O is displaced by dx, the dt = 1 v (a sec θ 1 ta θ 1 dθ 1 + b sec θ ta θ dθ ) = 0 (1) (sice for miimum time dt = 0). Also, c + d = a ta θ 1 + b ta θ = costat so, a sec θ 1 dθ 1 + b sec θ dθ = 0 () Divide equatios (1) ad () to fid θ 1 = θ
Chapter 35 Solutios 345 35.66 As show i the sketch, the agle of icidece at poit A is: θ = si 1 d = si 1 1.00 m = 30.0 R.00 m If the emergig ray is to be parallel to the icidet ray, the path must be symmetric about the ceter lie CB of the cylider. I the isosceles triagle ABC, γ = α ad β = 10 θ. Therefore, α + β + γ = 10 becomes α + 10 θ = 10 or α = θ = 15.0 The, applyig Sell s law at poit A, si α 100. si θ = si θ si 30. 0 or = = si α si 15. 0 = 1.93 35.67 (a) At the boudary of the air ad glass, the critical agle is give by si θ c = 1 Cosider the critical ray PB B : ta θ c = d 4 t or si θ c cos θ c = d 4t Squarig the last equatio gives: si θ c cos θ c = si θ c 1 si = d θ c 4t Sice si θ c = 1, this becomes 1 1 = d 4t or = 1 + ( 4t d) (b) Solvig for d, d = 4t 1 Thus, if = 1.5 ad t = 0.600 cm, 1 4 0.600 cm d = 1. 5 =.10 cm (c) Sice violet light has a larger idex of refractio, it will lead to a smaller critical agle ad the ier edge of the white halo will be tiged with violet light. 000 by Harcourt, Ic. All rights reserved.
346 Chapter 35 Solutios 35.6 From the sketch, observe that the agle of icidece at poit A is the same as the prism agle θ at poit O. Give that θ = 60.0, applicatio of Sell s law at poit A gives 1. 50 si β = 1. 00 si 60. 0 or β = 35.3 From triagle AOB, we calculate the agle of icidece (ad reflectio) at poit B. θ + ( 90. 0 β)+ ( 90. 0 γ)= 10 so γ = θ β = 60. 0 35. 3 = 4. 7 Now, usig triagle BCQ : ( 90. 0 γ)+ ( 90. 0 δ)+ ( 90. 0 θ)= 10 Thus the agle of icidece at poit C is δ = ( 90. 0 θ) γ = 30. 0 4. 7 = 5. 30 Fially, Sell s law applied at poit C gives 100. si φ = 150. si 530. or φ = si 1 ( 1. 50 si 5. 30 )= 7.96 35.69 (a) Give that θ1 = 45. 0 ad θ = 76. 0, Sell s law at the first surface gives si α = ( 100. ) si 450. (1) Observe that the agle of icidece at the secod surface is β = 90. 0 α. Thus, Sell s law at the secod surface yields = si β = si 90. 0 α 1. 00 si 76. 0, or cos α = si 76. 0 () Dividig Equatio (1) by Equatio (), si 45. 0 ta α =. si 76. 0 = 0 79 or α = 36.1 si 45. 0 si 45. 0 The, from Equatio (1), = = si α si 36. 1 = 1.0 (b) From the sketch, observe that the distace the light travels i the plastic is d = L si α. Also, the speed of light i the plastic is v = c, so the time required to travel through the plastic is ( 0.500 m) t = d v = L c si α = 1. 0 ( 3.00 10 ms)si 36.1 = 3.40 10 9 s = 3.40 s
Chapter 35 Solutios 347 35.70 si θ 1 si θ si θ 1 /si θ 0.174 0.131 1.3304 0.34 0.61 1.319 0.500 0.379 1.3177 0.643 0.40 1.335 0.766 0.576 1.39 0.66 0.647 1.3390 0.940 0.711 1.30 0.95 0.740 1.3315 The straightess of the graph lie demostrates Sell's proportioality. The slope of the lie is = 1. 376 ± 0. 01 ad = 1.3 ± 0.% 000 by Harcourt, Ic. All rights reserved.