Model use the ray model of light and the law of

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Model: Use the ray model of light and the law of reflection. Visualize: We only need one ray of light that leaves your toes and reflects in your eye. Solve: From the geometry of the diagram, the distance from your eye to the toes’ image is 2 2 2 (400 cm) (165 cm) 433 cm d = + = Assess: The light appears to come from your toes’ image.
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23.12. Model: Use the ray model of light and Snell’s law. Visualize: Solve: According to Snell’s law for the air-water and water-glass boundaries, air air water water sin sin n n θ θ = water water glass glass sin sin n n θ θ = From these two equations, we have air air glass glass sin sin n n θ θ = air glass air glass 1.0 sin sin sin60 1.50 n n θ θ = = ° 1 glass sin 60 sin 35 1.5 θ ° = = °
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23.13. Visualize: Use Snell’s law 1 1 2 2 sin sin . n n θ θ = We are given 2 CZ 25 . θ θ = = ° We look up in Table 23.1 1 oil 1.46 n n = = and 2 CZ 1.96. n n = = Solve: Solve the equation for 1 . θ 1 1 2 1 2 1 1.96 sin sin sin sin25 35 1.46 n n θ θ = = ° = ° Assess: Since the ray goes into a material with higher index of refraction we know it bends toward the normal, so we expect 1 2 ; θ θ > this is the case.
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23.14. Model: Use the ray model of light. The sun is a point source of light. Visualize: A ray that arrives at the diver 50 ° above horizontal refracted into the water at θ water = 40 ° . Solve: Using Snell’s law at the water-air boundary air air water water sin sin n n θ θ = water air water air 1.33 sin sin sin40 1.0 n n θ θ = = ° θ air = 58.7 ° Thus the height above the horizon is θ = 90 ° θ air = 31.3 ° 31 . ° Because the sun is far away from the fisherman (and the diver), the fisherman will see the sun at the same angle of 31 ° above the horizon.
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23.15. Model: Represent the laser beam with a single ray and use the ray model of light. Solve: Using Snell’s law at the air-water boundary, air air liquid liquid sin sin n n θ θ = air liquid air liquid sin sin37 1.0 1.37 sin sin26 n n θ θ ° = = = ° Assess: As expected, n liquid is larger than n air .
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23.16. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize: Solve: The critical angle of incidence is given by Equation 23.9: cladding 1 1 c core 1.48 sin sin 67.7 1.60 n n θ = = = ° Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is 90 ° 67.7 ° = 23.3 ° . Assess: We can have total internal reflection because n core > n cladding .
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23.17. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection.
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