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Unformatted text preview: lim (kl9356) Homework 22 Weathers (22202) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A coin is at the bottom of a beaker. The beaker is filled with 1 . 6 cm of water ( n 1 = 1 . 33) covered by 3 . 7 cm of liquid ( n 2 = 1 . 4) floating on the water. How deep does the coin appear to be from the upper surface of the liquid (near the top of the beaker)? Correct answer: 3 . 84586 cm. Explanation: For small angles sin tan = x . The appearance of the width of the coin x remains the same. Applying Snells Law, n 1 sin 1 = n 2 sin 2 n i x i = n f x f f = i n i since n f 1 for air and the apparent distance d = f . Thus d = 2 n 2 + 1 n 1 = 3 . 7 cm 1 . 4 + 1 . 6 cm 1 . 33 = (2 . 64286 cm) + (1 . 20301 cm) = 3 . 84586 cm . The coin appears to be closer to the surface by (1 . 6 cm) + (3 . 7 cm) (3 . 84586 cm) = 1 . 45414 cm . Alternate Solution: Light coming straight up from the coin falls on each in terface at 0 and continues straight up. Con sider light making a 1 angle (therefore we can use the small angle approximation) with the vertical in the water. It enters the liquid a distance (1 . 6 cm) tan 1 (1 . 6 cm) 1 u = 0 . 0279253 cm from the vertical ray. In the liquid its angle with the vertical is given by 1 . 33 sin 1 = 1 . 4 sin 2 2 = 0 . 950001 This same ray reaches air at distance (0 . 0279253 cm) + (3 . 7 cm) tan(0 . 950001 ) = 0 . 0892736 cm , from the vertical ray, and the angle of refrac tion is found from 1 . 4 sin(0 . 950001 ) = 1 sin 3 3 = 1 . 33 Your brain automatically finds the intersec tion of this ray with the vertical ray, at an apparent depth of . 0892736 cm tan(1 . 33 ) . 0892736 cm . 0232129 rad = 3 . 84586 cm . Rays for all other small angles with the ver tical appear to diverge from the image of the coin at this depth....
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 Spring '00
 Littler
 Magnetism, Work

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