PROBLEM 27 & 28 May 02 2010

PROBLEM 27 & 28 May 02 2010 - Solving problems 27...

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Solving problems 27 & 28 May 1, 2010 Problem 27 (like 26.23): At what radius does an external observer see the image of a small Jurassic insect (much smaller than the dragon fly in the picture) located in the center of an amber sphere 30 mm in radius? Take the index of refraction of amber to be 1.543. a) 0 mm ; b) 30 mm ; c) 15.4 mm; d) 1.543 mm; e) 3.08 mm. Answer: The object and virtual image coincide at the center of the sphere so the image appears at a radius of 0 mm. What is the length of the image if the insect inside is 1 mm long? a) 1 mm ; b) 30 mm ; c) 15.4 mm; d) 1.543 mm; e) 3.08 mm. Answer: A ray from the top of the object at height h has an angle with respect to the radius R h / in the small angle approximation as suggested in the drawing. A ray from the top of the image has an inside angle R h / ' ' where n ' by Snell’s law. Thus h’=nh and the image height is 1.543 mm. The image is
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This note was uploaded on 09/23/2010 for the course PH 02c taught by Professor Mile during the Spring '04 term at Riverside Community College.

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