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Homework #6: Reflection and Refraction 1. The two mirrors in the figure below meet at a right angle. The beam of light in the vertical plane P strikes mirror 1 as shown. (a) Determine the distance the reflected light beam travels before striking mirror 2. (b) In what direction does the light beam travel after being reflected from mirror 2? 2. How many times will the incident beam shown in the figure above right be reflected by each of the parallel mirrors? 3. An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the vertical. What is actual direction of the Sun? 4. 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. 5. A ray of light is incident on the surface of a block of clear ice at an angle of 40.0° with the normal. Part of the light is reflected and part is refracted. Find the angle between the reflected and refracted light. 6. A flashlight on the bottom of a 4.00-m-deep swimming pool sends a ray upward and at an angle so that the ray strikes the surface of the water 2.00 m from the point directly above the flashlight. What angle (in air) does the emerging ray make with the water’s surface? 7. A beam of light both reflects and refracts at the surface between air and glass, as shown in the figure below. If the index of refraction of the glass is ng, find the angle of incidence, θ1, in the air that would result in the reflected ray and the refracted ray being perpendicular to each other. [Hint: Remember the identity sin(90° − θ) = cos θ.] 8. For 589-nm light, calculate the critical angle for the following materials surrounded by air: (a) diamond and (b) flint glass. 9. A beam of light is incident from air on the surface of a liquid. If the angle of incidence is 30.0° and the angle of refraction is 22.0°, find the critical angle for the liquid when surrounded by air. Solutions: 1. (a) From geometry, 1.25 m dsin40.0 , so d 1.94 m (b) 50.0 above horizontal , or parallel to the incident ray 2. The incident light reaches the left-hand mirror at distance 1.00 m tan5.00 0.087 5 m above its bottom edge. The reflected light first reaches the right-hand mirror at height 2 0.087 5 m 0.175 m It bounces between the mirrors with this distance between points of contact with either. Since 1.00 m 5.72 , the light reflects 0.175 m five tim es from the right-hand m irrorand six tim es from the left 3. n1 sin 1 n2 sin 2 sin 1 1.333sin 45.0 sin 1 (1.333)(0.707) 0.943 1 70.5 19.5 above the horizontal 4. (a) From Snell’s law, n2 (b) 2 (c) f (d) v2 0 n2 c 0 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 5. From Snell’s law, n1 sin1 00 sin 40.0 1 1. sin 29.4 1.309 n2 2 sin1 and from the law of reflection, 1 40.0 Hence, the angle between the reflected and refracted rays is 180 2 180 29.4 40.0 111 6. The angle of incidence is 2.00 m 26.6 4.00 m 1 tan1 Therefore, Snell’s law gives n1 sin1 n2 2 sin1 1.333 sin 26.6 sin 1 36.6 1.00 and the angle the refracted ray makes with the surface is 90.0 2 90.0 36.6 53.4 7. As shown at the right, 1 2 180 When 90 , this gives 2 90 1 Then, from Snell’s law sin1 ng sin 2 nair ng sin 90 1 ng cos1 Thus, when 90 , sin1 tan1 ng or 1 tan1 ng cos1 8. When surrounded by air n2 1.00 , the critical angle of a material is n2 1 1 sin n1 nm aterial c sin1 1 (a) For diamond, c sin 1 24.4 2.419 1 (b) For flint glass, c sin 1 37.0 1.66 9. Using Snell’s law, the index of refraction of the liquid is found to be nliquid nair sini 1.00 sin 30.0 1.33 sin r sin 22.0 n 1.00 Thus, c sin1 air sin1 48.5 nliquid 1.33