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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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 = = .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 11

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online