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Ch24-h4-solutions

# Ch24-h4-solutions - pan(zp695 Ch24-h4 chiu(56565 This...

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pan (zp695) – Ch24-h4 – chiu – (56565) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The index of refraction of a particular liquid is 1 . 33. At what speed would a wave crest in a beam of light travel through this medium? Remember that c = 3 × 10 8 m / s. Correct answer: 2 . 25564 × 10 8 m / s. Explanation: We know that the light will travel slower than c , so we have v = c n = 3 × 10 8 m / s 1 . 33 = 2 . 25564 × 10 8 m / s . 002 10.0 points If a diver who is underwater shines a flash- light upward toward the surface, at an angle of 41 degrees from the normal, at what angle does the light emerge from the water? In- dices of refraction: water, 1 . 33; air, 1 . 00029. Remember that c = 3 × 10 8 m / s. Correct answer: 60 . 7277 degrees. Explanation: Since water has a higher index of refraction than air, a beam of light will bend away from the normal as it passes from the water to air. Define medium 1 to be the water and medium 2 to be the air. n 1 sin θ 1 = n 2 sin θ 2 sin θ 2 = n 1 n 2 sin θ 1 = 1 . 33 1 . 00029 sin 41 = 0 . 872306 θ 2 = arcsin(0 . 872306) = 60 . 7277 . 003 10.0 points If a beam of light from a medium with a higher index of refraction emerges into a medium with a lower index of refraction, what hap- pens? 1. The emerging beam bends toward the normal (smaller angle). 2. The emerging beam bends away from the normal (larger angle). correct 3. The emerging beam does not bend at all. Explanation: The correct choice is that the emerging beam bends away from the normal. Let medium 1 be the higher index medium. Then we have n 1 sin θ 1 = n 2 sin θ 2 sin θ 2 = n 1 n 2 sin θ 1 If n 1 > n 2 , then sin θ 2 > sin θ 1 , and θ 2 > θ 1 . 004 (part 1 of 2) 10.0 points To get total internal reflection at the interface of water (refractive index 1 . 33) and a plastic whose refractive index is 1 . 45: Which material must the light start in? 1. Water 2. It doesn’t matter 3. Plastic correct Explanation: The light must start in the plastic. The emerging beam must bend away from the nor- mal, thus the beam must start in the material of higher index and be transmitted to the ma- terial of lower index. 005 (part 2 of 2) 10.0 points What is the critical angle? Correct answer: 66 . 5261 degrees. Explanation:

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pan (zp695) – Ch24-h4 – chiu – (56565) 2 The critical angle is the angle of the incom- ing ray that refracts at an angle of 90 relative to the vertical. Apply Snell’s law to calculate the angle of the ray in the plastic. In the fol- lowing section, 1 refers to the plastic and 2 refers to the water. n 1 sin θ 1 = n 2 sin θ 2 sin θ 1 = n 2 n 1 sin θ 2 = 1 . 33 1 . 45 sin 90 = 0 . 917241 θ 1 = arcsin(0 . 917241) = 66 . 5261 . 006 (part 1 of 3) 10.0 points A spotlight mounted beside a swimming pool sends a beam of light at an angle into the pool, as shown in the figure below.
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