# Visualize solve the critical angle of incidence is

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Visualize: Solve: The critical angle of incidence is given by Equation 23.9: 1 1 oil c glass 1.46 sin sin 76.7 1.50 n n θ = = = ° Assess: The critical angle exists because n oil < n glass .

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23.18. Model: Represent the can as a point source and use the ray model of light. Visualize: Paraxial rays from the can refract into the water and enter into the fish’s eye. Solve: The object distance from the edge of the aquarium is s . From the water side, the can appears to be at an image distance s = 30 cm. Using Equation 23.13, 2 water 1 air 1.33 1.0 n n s s s s n n ′ = = = 30 cm 23 cm 1.33 s = =
23.19. Model: Represent the beetle as a point source and use the ray model of light. Visualize: Paraxial rays from the beetle refract into the air and then enter into the observer’s eye. The rays in the air when extended into the plastic appear to be coming from the beetle at a shallower location, a distance s from the plastic-air boundary. Solve: The actual object distance is s and the image distance is s = 2.0 cm. Using Equation 23.13, 2 air 1 plastic n n s s s n n ′ = = 1.0 2.0 cm 1.59 s = s = 3.2 cm Assess: The beetle is much deeper in the plastic than it appears to be.

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23.20. Model: Represent the diver’s head and toes as point sources. Use the ray model of light. Visualize: Paraxial rays from the head and the toes of the diver refract into the air and then enter into your eyes. When these refracted rays are extended into the water, the head and the toes appear elevated toward you. Solve: Using Equation 23.13, 2 air T T T 1 water n n s s s n n ′ = = air H H water n s s n ′ = Subtracting the two equations, her apparent height is ( ) ( ) air H T H T water 1.0 150 cm 113 cm 1.33 n s s s s n = = =
23.21. Model: Represent the aquarium’s wall as a point source, and use the ray model of light. Visualize: Paraxial rays from the outer edge (O) are refracted into the water and then enter into the fish’s eye. When extended into the wall, these rays will appear to be coming from O rather from O. The point on the inside edge (I) of the wall will not change its apparent location. Solve: We are given that s O s I = 4.00 mm and O I 3.50 mm. s s = Using Equation 23.13, water O O wall n s s n ′ = water I I wall n s s n ′ = ( ) water O I O I wall n s s s s n = ( ) wall 1.33 3.50 mm 4.00 mm n = ( ) wall 4.00 mm 1.33 1.52 3.50 mm n = =

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23.22. Model: Use the ray model of light. Visualize: Solve: Using Snell’s law, air red red sin30 sin n n θ ° = 1 red sin30 sin 19.2 1.52 θ ° = = ° air violet violet sin30 sin n n θ ° = 1 violet sin30 sin 18.8 1.55 θ ° = = ° Thus the angular spread is red violet 19.2 18.8 0.4 θ θ θ Δ = = °− ° = °
23.23. Model: Use the ray model of light and the phenomenon of dispersion.

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