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# EXAM 4 - Platt David Quiz 4 Due Dec 6 2005 10:00 pm Inst...

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Platt, David – Quiz 4 – Due: Dec 6 2005, 10:00 pm – Inst: Ken Shih 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points In our everyday environment, diffraction is much more evident for sound waves than for light waves. Why is this so? 1. We see light more often than hear sound in our everyday environment. 2. None of these. 3. Light waves travel much faster than sound waves. 4. Light waves have much shorter wave- length than sound waves. correct Explanation: Diffraction around ordinary-sized objects is most pronounced for waves with a wavelength as big or bigger than the objects. The wave- length of sound waves is relatively long, and for light extremely short. Hence the diffrac- tion of sound is more evident in our everyday environment. 002 (part 1 of 1) 10 points The thin composite lens has | R 1 | = R and | R 2 | = R 2 , as shown in the figure. R 1 R 2 n 2 n 1 The lens has focal length 1. f = n 1 4 ( n 2 - n 1 ) R . 2. f = n 1 4 ( n 1 - n 2 ) R . 3. f = n 1 2 ( n 1 - n 2 ) R . 4. f = n 2 n 1 - n 2 R . 5. f = n 2 2 ( n 2 - n 1 ) R . 6. f = n 1 2 ( n 2 - n 1 ) R . 7. f = n 2 n 2 - n 1 R . 8. f = n 1 n 2 - n 1 R . 9. f = n 1 n 1 - n 2 R . correct 10. f = n 2 2 ( n 1 - n 2 ) R . Explanation: Use the lens makers equation 1 f = n 2 n 1 - 1 1 R 1 - 1 R 2 = n 2 - n 1 n 1 1 R - 2 R = n 1 - n 2 n 1 1 R f = n 1 n 1 - n 2 R . 003 (part 1 of 1) 10 points Assume 515 nm light in the eye and a pupil diameter of 2 . 5 mm. What are the approximate dimensions of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 219 km above the Earth? Correct answer: 55 . 0391 m. Explanation: If the dimensions of the smallest object on Earth that astronauts can resolve by eye is x , then the Rayleigh’s criterion yields x D = 1 . 22 λ d , where D is the distance from the astronaut to the Earth, λ is the wavelength of the light, and d is the diameter of the astronaut’s pupil. Therefore x = 1 . 22 λD d

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Platt, David – Quiz 4 – Due: Dec 6 2005, 10:00 pm – Inst: Ken Shih 2 x = (1 . 22)(515 nm )(219 km ) 2 . 5 mm = 55 . 0391 m . 004 (part 1 of 2) 10 points A beam of light with a wavelength 660 nm in air travels in water. The index of refraction of water is 1 . 33. What is the wavelength in water? Correct answer: 496 . 241 nm. Explanation: Let : n w = 1 . 33 , and λ 0 = 660 nm . The wavelength of light in water is λ w = λ 0 n w = 660 nm 1 . 33 = 496 . 241 nm . 005 (part 2 of 2) 10 points Does a swimmer underwater observe the same color or a different color for this light? 1. It is not possible to know what a swimmer underwater will observe. 2. A swimmer underwater observes a differ- ent color because the wavelength changes. 3. A swimmer underwater observes a the same color because the wavelength does not change. 4. A swimmer underwater observes a differ- ent color because the frequency changes.
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EXAM 4 - Platt David Quiz 4 Due Dec 6 2005 10:00 pm Inst...

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