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Unformatted text preview: Kapoor (mk9499) – oldhomework 35 – Turner – (60230) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 9 . 5 cm focal length converging lens is used to magnify small newspaper print 2 mm high. Calculate the height of the image for the maximum angular magnification for a normal eye. Assume the eye focuses at the near point of 25 cm . Correct answer: 7 . 26316 mm. Explanation: Basic Concepts: M = h ′ h Angular magnification m = θ θ Lens equation 1 p + 1 q = 1 f . Solution: The size of an image on the retina depends upon the angular magnification. The angular magnification is the ratio of the angle subtended by an object with a lens in use “ θ ” to that subtended by the object placed at the near point with no lens “ θ ”. The angular magnification is a maximum when the image is at the near point of the eye; i.e. , when q = − 25 cm for a normal eye. The object distance can be found from the thin lens formula p = − q f f − q = − ( − 25 cm) (9 . 5 cm) (9 . 5 cm) − ( − 25 cm) = 6 . 88406 cm . If we make small angle approximations, θ ≃ − h ( − 25 cm) , θ ≃ h p , then the angular magnification becomes m = θ θ = h p h q = 1 − q f . The new size of the image is then h ′ = m h = bracketleftbigg 1 − q f bracketrightbigg h = bracketleftbigg 1 − ( − 25 cm) (9 . 5 cm) bracketrightbigg (2 mm) = 7 . 26316 mm . 002 (part 1 of 3) 10.0 points A cat is a distance d = 18 . 984 cm from a thin converging lens with focal length f = 11 . 3 cm. d lens How far from the lens is the image of the cat due only to this lens? 1. f parenleftbigg 1 − f d parenrightbigg 2. 1 f − 1 d 3. parenleftbigg 1 d − 1 f parenrightbigg − 1 4. parenleftbigg 1 f + 1 d parenrightbigg − 1 5. parenleftbigg 1 f − 1 d parenrightbigg − 1 correct 6. f d parenleftbigg 1 f − 1 d parenrightbigg Explanation: Kapoor (mk9499) – oldhomework 35 – Turner – (60230) 2 Solutions: Using the thin lens formula: 1 s + 1 s ′ = 1 f , we can compute the position of the image which would be x = parenleftbigg 1 f − 1 d parenrightbigg − 1 = parenleftbigg 1 11 . 3 cm − 1 18 . 984 cm parenrightbigg − 1 = 27 . 9176 cm 003 (part 2 of 3) 10.0 points Now a glass ball of radius R = 10 cm and in dex of refraction n = 1 . 57 is inserted between the lens and our observer, with one edge a dis tance ℓ = 18 . 8 cm from the lens, as shown in the figure below....
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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, Work

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