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Unformatted text preview: Kapoor (mk9499) – oldmidterm 04 – Turner – (60230) 1 This printout should have 29 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A light ray passes through substances 1, 2, and 3, as shown. The indices of refraction for these three substances are n 1 , n 2 , and n 3 , respectively. Ray segments in 1 and 3 are parallel. x θ 1 θ 2 d 1 2 3 n 1 n 2 n 3 One can conclude that 1. n 1 must be equal to 1.00. 2. n 2 must be less than n 1 . 3. all three indices must be the same. 4. n 2 must be less than n 3 . 5. n 3 must be the same as n 1 . correct Explanation: n 1 sin θ 1 = n 2 sin θ 2 = n 3 sin θ 3 , Since θ 1 = θ 3 , sin θ 1 = sin θ 2 and n 1 = n 3 . Since sin θ 1 = sin θ 3 > sin θ 2 , n 2 > n 1 = n 3 . So, we can only conclude that n 3 must be the same as n 1 . 002 10.0 points A jewel thief decides to hide a stolen diamond by placing it at the bottom of a crystalclear fountain. He places a circular piece of wood on the surface of the water and anchors it directly above the diamond at the bottom of the fountain. d . 52 m If the fountain is 0.520 m deep, find the minimum diameter, d of the piece of wood that would prevent the diamond from being seen from outside the water. Correct answer: 1 . 17992 m. Explanation: Basic Concept: sin θ c = n r n i Given: n i = 1 . 333 n r = 1 . 00 Δ y = 0 . 520 m Solution: θ c = sin − 1 parenleftbigg n r n i parenrightbigg = sin − 1 parenleftbigg 1 1 . 333 parenrightbigg = 48 . 6066 ◦ tan θ c = d 2 Δ y = d 2Δ y d = 2Δ y (tan θ c ) = 2(0 . 52 m)(tan48 . 6066 ◦ ) = 1 . 17992 m 003 (part 1 of 2) 10.0 points Hint: Construct a ray diagram. Given: A real object is located at “ p 1 = 7 4 f ” to the left of a convergent lens with a focal length f as shown in the figure below. Kapoor (mk9499) – oldmidterm 04 – Turner – (60230) 2 7 4 f f ( × f ) The image distance q 1 to the right of the lens is 1. q 1 = 9 4 f . 2. q 1 = 12 5 f . 3. q 1 = 13 6 f . 4. q 1 = 15 7 f . 5. q 1 = 15 8 f . 6. q 1 = 10 3 f . 7. q 1 = 8 3 f . 8. q 1 = 11 6 f . 9. q 1 = 7 3 f . correct 10. q 1 = 13 7 f . Explanation: 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 . 7 4 f f 7 3 ( × f ) Basic Concepts: 1 p + 1 q = 1 f . Solution: 1 q 1 = − 1 p 1 + 1 f = − 4 7 f + 1 f = 7 − 4 7 f = 3 7 f q 1 = 7 3 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)....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Light

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