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Transcript
Refraction Practice Problems 7. An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the vertical. What is actual direction of the Sun? 8. Light is incident normal to a 1.00-cm layer of water that lies on top of a flat ® Lucite plate with a thickness of 0.500 cm. How much more time is required for light to pass through this double layer than is required to traverse the same distance in air (nLucite = 1.59)? 9. A laser beam is incident at an angle of 30.0° to the vertical onto a solution of corn syrup in water. If the beam is refracted to 19.24° to the vertical, (a) what is the index of refraction of the syrup solution? Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find its (b) wavelength, (c) frequency, and (d) speed in the solution. Hint: The frequency stays constant when light travels from one material to another. 21. The light beam shown in Figure P22.21 makes an angle of 20.0° with the normal line NN’ in the linseed oil. Determine the angles θ and θ’. (The refractive index for linseed oil is 1.48.) Figure P22.21 22. A submarine is 300 m horizontally out from the shore and 100 m beneath the surface of the water. A laser beam is sent from the sub so that it strikes the surface of the water at a point 210 m from the shore. If the beam just strikes the top of a building standing directly at the water’s edge, find the height of the building. Solutions: 22.7 n1 sin 1 n2 sin 2 sin 1 1.333sin 45.0 sin 1 (1.333)(0.707) 0.943 1 70.5 22.8 19.5 above the horizontal The speed of light in water is vw ater c , and in Lucite® vLucite nw ater total time required to transverse the double layer is t1 c nLucite . Thus, the dwater dLucite dwaternwater dLucitenLucite vwater vLucite c The time to travel the same distance in air is t2 dwater dLucite , so the additional c time required for the double layer is t t1 t2 1.00 10 dw ater nw ater 1 dLucite nLucite 1 2 22.9 c m 1.333 1 0.500 10 (a) From Snell’s law, n2 (b) 2 (c) f (d) v2 0 n2 c 0 2 3.00 10 m s 8 m 1.59 1 n1 sin1 1.00 sin 30.0 1.52 sin 2 sin19.24 632.8 nm 417 nm 1.52 3.00 108 m s 4.74 1014 H z in air and in syrup 632.8 109 m c 3.00 108 m s 1.98 108 m s n2 1.52 2.09 1011 s 22.21 From Snell’s law, the angle of incidence at the air-oil interface is noil sin oil nair sin1 1.48 sin 20.0 sin 1 30.4 1.00 and the angle of refraction as the light enters the water is noilsinoil 48 sin20.0 1 1. sin 22.3 1.333 nwater sin1 22.22 The angle of incidence at the water surface is 1 tan 1 90.0 m 42.0 100 m Then, Snell’s law gives the angle of refraction as nwater sin1 333 sin 42.0 1 1. sin 63.1 nair 1.00 2 sin1 so the height of the building is h 210 m 210 m 107 m tan 2 tan63.1