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Unformatted text preview: Golshani (ag5873) Homework 02 Yao (57045) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Consider a concave mirror with radius R . An upright object is placed between the interval R 2 and R . R R /2 The image is 1. virtual, upright, same size. 2. virtual, inverted, reduced. 3. virtual, upright, reduced. 4. real, upright, enlarged. 5. real, upright, same size. 6. virtual, upright, enlarged. 7. virtual, inverted, enlarged. 8. real, upright, reduced. 9. real, inverted, enlarged. correct 10. real, inverted, reduced. Explanation: R R /2 Note: f = R 2 , where f is positive for a concave mirror by convention. From the mirror equation 1 p + 1 q = 1 f . Solving for q we have q = fp p- f . Since p > f , the image is real and inverted. The absolute value of the magnification is | M | = q p = f p- f . Since f < p < 2 f , recall parenleftbigg f = R 2 parenrightbigg , | M | > 1, or the image is enlarged. This is also confirmed by the ray diagram shown. 002 10.0 points A concave mirror with a radius of curvature of 0 . 9 m is illuminated by a candle located on the symmetry axis 2 . 9 m from the mirror. Where is the image of the candle? Correct answer: 0 . 532653 m. 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 = 0 . 45 m and p = 2 . 9 m . From the mirror equation 1 p + 1 q = 1 f = 2 R f = R 2 q = bracketleftbigg 1 f- 1 p bracketrightbigg 1 = bracketleftbigg 1 (0 . 45 m)- 1 (2 . 9 m) bracketrightbigg 1 = . 532653 m . Golshani (ag5873) Homework 02 Yao (57045) 2 003 10.0 points A concave mirror has a focal length of 51 cm. What is the position of the resulting image if the image is inverted and 6 times smaller than the object?...
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This note was uploaded on 09/27/2010 for the course PHY 58650 taught by Professor Yao during the Fall '10 term at University of Texas at Austin.
- Fall '10