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Unformatted text preview: montes (dam2772) – Homework8 – shih – (56310) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The distance of the lens in the human eye from the retina, on which the image is focused, is about 1 . 7 cm. To focus on a book 39 . 7 cm from the eye, what must the focal length of the eye be? Correct answer: 1 . 63019 cm. Explanation: The lens equation is 1 p + 1 q = 1 f . For the lens in the human eye, both p and q are positive. f = 1 1 p + 1 q = 1 1 (39 . 7 cm) + 1 (1 . 7 cm) = 1 . 63019 cm . 002 10.0 points At what angle above the horizon is the Sun if light from it is completely polarized upon re flecting from water? (Assume that the index of refraction of water is n = 1 . 4.) Correct answer: 35 . 5377 ◦ . Explanation: Complete polarization occurs at Brewster’s angle θ p given by the relation tan θ p = n , which yields θ p = arctan( n ) = arctan(1 . 4) = 54 . 4623 ◦ . So, the Sun is 90 ◦ 54 . 4623 ◦ = 35 . 5377 ◦ above the horizon. 003 (part 1 of 2) 10.0 points Hint: Construct a ray diagram. Given: A real object is located at “ p 1 = 11 7 f ” to the left of a convergent lens with a focal length f as shown in the figure below. 11 7 f f ( × f ) The image distance q 1 to the right of the lens is 1. q 1 = 5 2 f . 2. q 1 = 7 3 f . 3. q 1 = 11 4 f . correct 4. q 1 = 15 8 f . 5. q 1 = 13 7 f . 6. q 1 = 11 6 f . 7. q 1 = 13 6 f . 8. q 1 = 10 3 f . 9. q 1 = 13 4 f . 10. q 1 = 12 5 f . Explanation: montes (dam2772) – Homework8 – shih – (56310) 2 Basic Concepts: 1 p + 1 q = 1 f m = h ′ h = q p Converging Lens f > ∞ > p > f f < q < ∞ > m>∞ f > p >∞ < q < ∞ > m> 1 Convergent (concave) lenses f > 0 have real images q > 0 when the object ∞ > p > f . 11 7 f f 11 4 ( × f ) Basic Concepts: 1 p + 1 q = 1 f . Solution: 1 q 1 = 1 p 1 + 1 f = 7 11 f + 1 f = 11 7 11 f = 4 11 f q 1 = 11 4 f . 004 (part 2 of 2) 10.0 points A second convergent lens with the same focal length f is placed behind the first lens, as shown in the figure below (the first lens has the lighter image). 11 7 3 4 f f f f ( × f ) The second image location q ′ 2 , measured with respect to the second lens, is 1. q ′ 2 = 5 7 f . 2. q ′ 2 = 3 5 f . 3. q ′ 2 = 8 13 f . 4. q ′ 2 = 9 16 f . 5. q ′ 2 = 5 8 f . 6. q ′ 2 = 2 3 f . correct 7. q ′ 2 = 1 2 f . 8. q ′ 2 = 3 4 f . 9. q ′ 2 = 4 7 f . 10. q ′ 2 = 7 12 f . Explanation: 17 12 f 11 7 2 f 3 4 11 4 2 3 f f ( × f ) Given: Distance between lenses is d = 3 4 f , montes (dam2772) – Homework8 – shih – (56310) 3 and from Part 2, we have q 1 = 11 4 f . Solution: q 1 > d and for the second lens p 2 = q 1 + d = parenleftbigg 11 4 + 3 4 parenrightbigg f = 2 f ....
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This note was uploaded on 03/22/2012 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas.
 Spring '10
 TSOI
 Physics, Work

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