956_PartUniversity Physics Solution

956_PartUniversity Physics Solution - 33-14 Chapter 33...

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33-14 Chapter 33 EVALUATE: For total internal reflection the ray must be incident in the material of greater refractive index. glass water nn > , so that is the case here. Figure 33.50 33.51. IDENTIFY: Apply Snell&s law to the refraction of each ray as it emerges from the glass. The angle of incidence equals the angle 25.0 . A SET UP: The paths of the two rays are sketched in Figure 33.51. Figure 33.51 EXECUTE: sin sin aa bb θ = glas sin25.0 1.00sin b n °= glass sin sin25.0 b n sin 1.66sin25.0 0.7015 b = 44.55 b 90.0 45.45 b β = ° Then 90.0 90.0 25.0 45.45 19.55 . A δβ −° ° = ° The angle between the two rays is 2 39.1 . δ EVALUATE: The light is incident normally on the front face of the prism so the light is not bent as it enters the prism. 33.52. IDENTIFY: The ray shown in the figure that accompanies the problem is to be incident at the critical angle. SET UP: 90 b = ° . The incident angle for the ray in the figure is 60 ° . EXECUTE: sin sin = gives sin 1.62sin60 1.40. sin sin90 b b n n ⎛⎞ ° == = ⎜⎟ ° ⎝⎠ EVALUATE: Total internal reflection occurs only when the light is incident in the material of the greater refractive index. 33.53. IDENTIFY: No light enters the gas because total internal reflection must have occurred at the water-gas interface. SET UP: At the minimum value of S , the light strikes the water-gas interface at the critical angle. We apply Snell&s law, n a sin a = n b sin b , at that surface. EXECUTE: (a) In the water, = S R = (1.09 m)/(1.10 m) = 0.991 rad = 56.77±. This is the critical angle. So, using the refractive index for water from Table 33.1, we get n = (1.333) sin 56.77± = 1.12 (b) (i) The laser beam stays in the water all the time, so t = 2 R/v = 2 R / water c n = water Dn c = (2.20 m)(1.333)/(3.00 × 10 8 m/s) = 9.78 ns
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The Nature and Propagation of Light 33-15 (ii) The beam is in the water half the time and in the gas the other half of the time. t gas = gas Rn c = (1.10 m)(1.12)/(3.00 × 10 8 m/s) = 4.09 ns The total time is 4.09 ns + (9.78 ns)/2 = 8.98 ns EVALUATE: The gas must be under considerable pressure to have a refractive index as high as 1.12. 33.54. IDENTIFY: No light enters the water because total internal reflection must have occurred at the glass-water surface. SET UP: A little geometry tells us that θ is the angle of incidence at the glass-water face in the water. Also, = 59.2& must be the critical angle at that surface, so the angle of refraction is 90.0&. Snell±s law, n a sin a = n b sin b , applies at that glass-water surface, and the index of refraction is defined as n = . c v EXECUTE: Snell±s law at the glass-water surface gives n sin 59.2& = (1.333)(1.00), which gives n = 1.55. v = c n = (3.00 × 10 8 m/s)/1.55 = 1.93 × 10 8 m/s. EVALUATE: Notice that is not the angle of incidence at the reflector, but it is the angle of incidence at the glass- water surface. 33.55. (a) IDENTIFY: Apply Snell±s law to the refraction of the light as it enters the atmosphere. SET UP: The path of a ray from the sun is sketched in Figure 33.55. ab δ θθ = From the diagram sin b R R h = + arcsin b R Rh ⎛⎞ = ⎜⎟ + ⎝⎠ Figure 33.55 EXECUTE: Apply Snell±s law to the refraction that occurs at the top of the atmosphere: sin sin aa bb nn = ( a = vacuum of space, refractive, index 1.0; b = atmosphere, refractive index n ) sin sin so arcsin a Rn R == = ++ arcsin arcsin nR R
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956_PartUniversity Physics Solution - 33-14 Chapter 33...

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