36 - Kapoor(mk9499 – homework36 – Turner –(60230 1...

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Unformatted text preview: Kapoor (mk9499) – homework36 – Turner – (60230) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A thin lens has a crescent shape, which is defined by two spherical surfaces A and B . The radius of the surface A is | R A | = 3 a, and of surface B is | R B | = a. The lens is submerged in a medium. Assume: Index of refraction of the lens is 2.0 and the index of refraction of the medium is 1.0. A B a 3 a O C A C B The focal length f in the small angle ap- proximation is given by 1. f =- 1 3 a 2. f = 4 a 3. f = 3 a 4. f =- a 5. f =- 1 2 a 6. f = 5 2 a 7. f =- 1 4 a 8. f = 2 a 9. f = 3 2 a correct 10. f =- 3 4 a Explanation: R 1 is the x-coordinate of the center of the spherical surface A ; i.e. , R 1 =-| R A | =- 3 a . Similarly, R 2 =- a . The lens maker’s formula gives 1 f = ( n- 1) parenleftbigg 1 R 1- 1 R 2 parenrightbigg = (2- 1) parenleftbigg 1- 3 a- 1- a parenrightbigg = 2 3 a . Hence, f = 3 2 a . 002 (part 1 of 3) 10.0 points A cat is a distance d = 20 . 757 cm from a thin converging lens with focal length f = 11 . 1 cm. d lens How far from the lens is the image of the cat due only to this lens? Correct answer: 23 . 8586 cm. 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 Solution: Using the thin lens formula 1 s + 1 s ′ = 1 f , we can compute the position of the image which would be: x = parenleftbigg...
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36 - Kapoor(mk9499 – homework36 – Turner –(60230 1...

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