This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 1 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 1 = n 2 sin 2 = n 3 sin 3 , Since 1 = 3 , sin 1 = sin 2 and n 1 = n 3 . Since sin 1 = sin 3 > sin 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 c = n r n i Given: n i = 1 . 333 n r = 1 . 00 y = 0 . 520 m Solution: c = sin 1 parenleftbigg n r n i parenrightbigg = sin 1 parenleftbigg 1 1 . 333 parenrightbigg = 48 . 6066 tan c = d 2 y = d 2 y d = 2 y (tan 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)....
View Full
Document
 Fall '08
 Turner
 Physics, Light

Click to edit the document details