# Solve because the angle 2 θ is very small 3 20 10 m

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Solve: Because the angle 2 θ is very small, 3 2.0 10 m tan2 2 5.0 m θ θ × = ( ) 1 180 rad degrees 0.011 5000 5000 θ π = = = °

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23.45. 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: For angles θ water that are less than the critical angle, light will be refracted into the air. Solve: Snell’s law at the water-air boundary is air air water water sin sin . n n θ θ = Because the maximum angle of θ air is 90 , ° we have ( ) water 1.0 sin90 1.33sin θ ° = 1 water 1 sin 48.75 1.33 θ = = ° Applying Snell’s law again to the glass-water boundary, glass glass water water sin sin n n θ θ = ( ) 1 1 water glass water glass 1.33 sin48.75 sin sin sin 42 1.50 n n θ θ ° = = = ° Thus 42 ° is the maximum angle of incidence onto the glass for which the ray emerges into the air.
23.46. Model: Use the ray model of light. Visualize: Solve: When the plastic is in place, the microscope focuses on the virtual image of the dot. From the figure, we note that s = 1.0 cm and s = 1.0 cm 0.4 cm = 0.6 cm. The rays are paraxial, and the object and image distances are measured relative to the plastic-air boundary. Using Equation 23.13, air plastic n s s n ′ = ( ) plastic 1.0 0.6 cm 1.0 cm n = plastic 1.0 cm 1.67 0.6 cm n = =

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23.47. Model: Use the ray model of light and the law of refraction. Solve: Snell’s law at the air-glass boundary is air air glass glass sin sin . n n θ θ = We require 1 glass air 2 , θ θ = so ( ) air glass glass glass sin 2 sin n n θ θ = ( ) air glass glass glass glass 2sin cos sin n n θ θ θ = glass 1 1 glass air air 1.50 cos cos 41.4 82.8 2 2 1.0 n n θ θ = = = ° ⇒ = ° ×
23.48. Model: Use the ray model of light and the law of refraction. Visualize: Solve: (a) The ray of light strikes the meter stick at empty P , which is a distance L from the zero mark of the meter stick. So, tan60 50 cm L ° = ( ) 50 cm tan60 86.6 cm L = ° = (b) The ray of light refracts at P half and strikes the meter stick a distance 1 2 x x + from the zero of the meter stick. We can find x 1 from the triangle P full P half O : 1 tan60 25 cm x ° = ( ) 1 25 cm tan 60 43.30 cm x = ° = We also have ( ) 2 half 25 cm tan . x φ = Using Snell’s law, air water half sin60 sin n n φ ° = 1 half sin60 sin 40.63 1.33 φ ° = = ° ( ) 2 25 cm tan40.63 21.45 cm x = ° = 1 2 43.30 cm 21.45 cm 64.8 cm x x + = + = (c) The ray of light experiences refraction at P full and the angle of refraction is the same as in part (b). We get ( ) 3 full 3 tan 50 cm tan40.63 42.9 cm 50 cm x x φ = = ° =

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23.49. Model: Use the ray model of light. Light undergoes total internal reflection if it is incident on a boundary at an angle greater than the critical angle. Visualize: Solve: (a) To reach your eye, a light ray must refract through the top surface of the water and into the air. You can see in the figure that rays coming from the bottom of the tank are incident on the top surface at fairly small
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