solutions_40

solutions_40 - PHY2061 R. D. Field Chapter 40 Solutions...

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PHY2061 R. D. Field Solutions Chapter 40 Page 1 of 11 Chapter 40 Solutions Problem 1: An object placed 180 cm in front of a spherical mirror produces an image with a magnification m = 0.1 . What happens to the image when the object is moved 150 cm closer to the mirror? Answer: the image increases in size by a factor of 4 Solution: We know that 11 1 pi f m i p += = , and hence cm cm m p f 20 ) 10 1 ( 180 / 1 1 1 1 = = = , where I have used i = -m 1 p 1 , and m 1 = 0.1 . Now, if I move the object 150 cm closer to the mirror the new object position p 2 = 30 cm (180 cm - 150 cm) and the new image position i 2 will be given by cm cm cm p f i 12 1 30 1 20 1 1 1 1 2 2 = = = . Hence the new magnification is 4 . 0 30 ) 12 ( 2 2 2 = = = cm cm p i m , which is four times the original magnification . Problem 2: A converging thin lens has a focal length of 30 cm . For what two object locations ( in cm ) will this lens form an image three times the size of the object? Answer: 20, 40 Solution: We know that 1 f m i p = , and hence

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PHY2061 R. D. Field Solutions Chapter 40 Page 2 of 11 () m f p / 1 1 = . For m = 3 we get cm cm cm p 20 ) 30 ( 3 2 3 / 1 1 ) 30 ( = = = , and for m = -3 we get cm cm cm p 40 ) 30 ( 3 4 3 / 1 1 ) 30 ( = = + = . Problem 3: You decide to take a picture of the moon with your 35 mm camera using its normal lens of 50 mm focal length. Taking the moon&s diameter to be approximately 3.5x10 6 m and its distance from the earth as 3.84x10 8 m , how large ( in mm ) is the moon’s image on the film? Answer: 0.46 Solution: We know that p i m f i p = = + 1 1 1 , and hence f i (because 1/p is very small) and since |m| = h i /h o we have mm m mm m p f h h m h o o i 456 . 0 10 84 . 3 ) 50 )( 10 5 . 3 ( 2 2 | | 2 8 6 = × × = = = . Problem 4: Which of the following types of image of a real, erect object cannot be formed by a concave spherical mirror? (A) Real, erect, enlarged (B) Real, inverted, reduced (C) Virtual. erect, enlarged (D) Enlarged, inverted, and farther away than 2f (E) Virtual, inverted, reduced? Answer: A, E
PHY2061 R. D. Field Solutions Chapter 40 Page 3 of 11 Problem 5: A concave mirror forms a real image which is twice the size of the object. If the object is 20 cm from the mirror, the radius of curvature of the mirror ( in cm) must be approximately: Answer: 27 Solution: We know that 11 1 pi f m i p += = , and hence cm p m p f R 7 . 26 3 4 / 1 1 2 2 = = = = , where I have used i = -mp , f = R/2 , and m = -2 ( since i is positive ) . Problem 6: An erect image formed by a concave mirror ( f = 20 cm ) has a magnification of 2 . Which way and by how much should the object be moved to double the size of the image? Answer: 5 cm further from the mirror Solution: We know that 1 f m i p = , and hence = m f p 1 1 , where I have used i = -mp. Thus, cm cm p 10 2 1 1 20 1 = = , and cm cm p 15 4 1 1 20 2 = = .

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This note was uploaded on 05/31/2011 for the course PHY 2061 taught by Professor Fry during the Spring '08 term at University of Florida.

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solutions_40 - PHY2061 R. D. Field Chapter 40 Solutions...

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