# B we will now repeat the above calculation with x 25

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(b) We will now repeat the above calculation with x = 25 cm. From the geometry of the diagram at side A, 21.80 φ = ° and air 68.20 . θ = ° Using Snell’s law at the air-water boundary, water A 44.28 θ = ° and water B 45.72 . θ = ° Because water B c , θ θ < the ray will be refracted into the air. The angle of refraction is calculated as follows: air air B water water B sin sin n n θ θ = 1 air B 1.33sin45.72 sin 72.2 1 θ ° = = ° (c) Using the critical angle for the water-air boundary found in part (a), θ water A = 90 ° 48.75 ° = 41.25 ° . According to Snell’s law, air air water water A sin sin n n θ θ = 1 air 1.33sin41.25 sin 61.27 1.0 θ ° = = ° 90 61.27 28.73 φ = °− ° = ° The minimum value of x for which the laser beam passes through side B and emerges into the air is calculated as follows: 10 cm tan x φ = 10 cm 18.2 cm 18 cm tan28.73 x = = °

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23.53. Model: Use the ray model of light. Assume the bonfire is a point source right at the corner of the lake. Visualize: Solve: (a) Light rays from the fire enter the lake right at the edge. Even though the rays in air are incident on the surface at a range of angles from 0° up to 90°, the larger index of refraction of water causes the rays to travel downward in the water with angles θ c , the critical angle. Some of these rays can reach a fish that is deep in the lake, but a shallow fish out from shore is in the “exclusion zone” that is not reached by any rays from the fire. Thus a fish needs to be deep to see the light from the fire. (b) The shallowest fish that can see the fire is one that receives light rays refracting into the water at the critical angle θ c . These are rays that were incident on the water’s surface at 90°. The critical angle for a water-air boundary is 1 c 1.00 sin 48.75 1.33 θ = = ° The fish is 20 m from shore, so its depth is 20 m 17.5 m 18 m tan(48.75 ) d = = ° That is, a fish 20 m from shore must be at least 18 m deep to see the fire.
23.54. Model: Use the ray model of light. Assume that the target is a point source of light. Visualize: Solve: From the geometry of the figure with θ air = 60 ° , 1 air tan 2.0 m x θ = ( )( ) 1 2.0 m tan60 3.464 m x = ° = Let us find the horizontal distance x 2 by applying Snell’s law to the air-water boundary. We have water water air air sin sin n n θ θ = 1 water sin60 sin 40.63 1.33 θ ° = = ° Using the geometry of the diagram, 2 water tan 1.0 m x θ = ( ) 2 1.0 m tan40.63 0.858 m x = ° = To determine θ target , we note that target 1 2 3.0 m 3.0 m tan 0.6941 3.464 m 0.858 m x x θ = = = + + θ target = 35 °

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23.55. Model: Use the ray model of light and the phenomena of refraction and dispersion. Visualize: The refractive index of violet light is greater than the refractive index of red light. The violet wavelength thus gets refracted more than the red wavelength.
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