exam4sol - PHY2049 Spring 2011 Profs. P. Avery, S....

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Unformatted text preview: PHY2049 Spring 2011 Profs. P. Avery, S. Hershfield Final Exam Solution 1. A small insect starts walking away from a concave mirror along its central axis. Which statement is true? Answer: The image is initially erect, then flips upside down Solution: For a concave mirror the object distance, p , and image distance, i , are related via 1 /p + 1 /i = 1 /f with the focal length, f , being positive. Near the mirror p is small so 1 /p > 1 /f and the image distance is negative (virtual image). Consequently, the magnification, m =- i/p , is positive, and the image is erect. Far from the mirror, 1 /p < 1 /f , and the image distance is positive (real image). In this case the magnification, m =- i/p , is negative, and the image is inverted. 2. A concave shaving mirror has a radius of curvature of 30 cm. It is positioned so that the (upright) image of a man’s face is 3 times the size of the face. How far is the mirror from the face (in cm)? Answer: 10 Solution: Because the radius of curvature is 30 cm, the focal length is f = r/ 2 = 15 cm. An upright image three times the object size corresponds to a magnification of m =- i/p = 3. This implies that i =- 3 p and 1 /p +1 /i = 1 /p- 1 / (3 p ) = 2 / 3 p = 1 /f . The object distance is (2 / 3) f = 10 cm. 3. An object is placed 30 cm to the left of a diverging lens of focal length- 15 cm. A converging lens of focal length 30 cm is placed 30 cm to the right of the first lens. Where is the final image located relative to lens 2, and what is the overall magnification m ? Answer: 120 cm to the right; m =- 1 . Solution: This is a compound lens problem. Find the image of the first lens, and use that as the object for the second lens. For the first lens we have 1 30 + 1 i 1 = 1- 15 . The image relative to the first lens is -10 cm, which is 10 cm to the left of the first lens and 40 cm to the left of the second lens. Thus, the object distance for the second lens is 40 cm: 1 40 + 1 i 2 = 1 30 . The final image is i 2 = 120 cm to the right of the second lens, and the overall magnification is m = (- i 1 /p 1 )(- i 2 /p 2 ) = (10 / 30)(- 120 / 40) =- 1. 4. One of the two slits in a Young’s experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result: Answer: the dark fringes get brighter and the bright ones get darker Solution: If one of the slits in Young’s experiment is covered up, then there will be no interference pattern. Partially covering up a slit reduces the interference: dark fringes get brighter and bright ones get darker. 5. A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. The intensities of wavelengths 480 nm and 800 nm and no wavelengths in between are to be maximized in the reflected beam. (Take the index of refraction of air to be one.) The thickness of the film is: Answer: 400 nm Solution:...
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This note was uploaded on 07/31/2011 for the course PHY 2049 taught by Professor Any during the Summer '08 term at University of Florida.

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exam4sol - PHY2049 Spring 2011 Profs. P. Avery, S....

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