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Unformatted text preview: Huang, Lynn – Homework 13 – Due: May 2 2008, 11:00 pm – Inst: Matt Mackie 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Note: HW is due at 11pm central time, which is midnite our time!! 001 (part 1 of 1) 10 points A dentist uses a concave mirror (focal length 3 . 2 cm) to examine some teeth. If the distance from the object to the mirror is 0 . 96 cm, what is the magnification of the tooth? Correct answer: 1 . 42857 . Explanation: 1 p + 1 q = 1 f = 2 R m = h h = q p Concave Mirror f > ∞ >p> f f <q < ∞ >m>∞ f >p>∞ <q < ∞ >m> 1 Let : f = 3 . 2 cm and o = 0 . 96 cm . From the mirror equation 1 o + 1 i = 1 f m = i o m = f f o = 3 . 2 cm 3 . 2 cm . 96 cm = 1 . 42857 . keywords: 002 (part 1 of 2) 10 points Shiny lawn spheres placed on pedestals are convex mirrors. One such sphere has a di ameter of 28 cm. A robin (whose height is 10 . 3 cm) sits in a tree 7 . 1 m from the sphere. As measured from the surface of the sphere, where is the image of the robin? Correct answer: 6 . 93166 cm. Explanation: This is a convex mirror, so the focal length is negative f = 1 2 r = 1 2 D 2 = D 4 = (28 cm) 4 = 7 cm . The lens equation is 1 f = 1 p + 1 q or 1 q = 1 f 1 p = p f pf . Therefore q = f p p f = ( 7 cm) (710 cm) (710 cm) ( 7 cm) = 6 . 93166 cm . 003 (part 2 of 2) 10 points What is the height of the robin’s image? Correct answer: 0 . 100558 cm. Explanation: The magnification is defined by M = h h = q p h = q h p = ( 6 . 93166 cm) (10 . 3 cm) (710 cm) = 0 . 100558 cm . keywords: Huang, Lynn – Homework 13 – Due: May 2 2008, 11:00 pm – Inst: Matt Mackie 2 004 (part 1 of 1) 0 points A shallow pool of a liquid 7 . 6 m deep is not the depth one expects when viewed from over head. The index of refraction of the liquid is 1 . 32. How deep does it appear to be? Correct answer: 575 . 758 cm. Explanation: Basic Concepts n 1 s 1 + n 2 s 2 = n 2 n 1 R Solution: Consider a point on the bottom of the pool as our object. In the refraction surface formula n 1 s 1 + n 2 s 2 = n 2 n 1 R . Because the water surface of a pool is a plane, R = ∞ , thus n 2 n 1 R = 0. The object s 1 is under the water, n 1 = 1 . 32 is the index of refraction of water, and n 2 = 1 is the index of refraction of air. Hence, s 2 = s 1 n 2 n 1 = (7 . 6 m)(1) (1 . 32) = 5 . 75758 m . The image is virtual and on the same side of the interface as the object at a distance d apparent =  5 . 75758 m  = 575 . 758 cm. Note: One may also determine the apparent depth in an alternative approach. First sketch a ray diagram for a finite incident angle where the refacted ray hits the bottom of the pool at P. Draw a vertical line OP, where O is at the surface of the water, OP is the real depth....
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This note was uploaded on 04/29/2010 for the course PHYS 2202 taught by Professor Chan during the Spring '05 term at Texas A&M.
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