homework 8

homework 8 - montes(dam2772 – Homework8 – shih...

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Unformatted text preview: montes (dam2772) – Homework8 – shih – (56310) 1 This print-out should have 20 questions. Multiple-choice 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.

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homework 8 - montes(dam2772 – Homework8 – shih...

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