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CE-19-SS2005 - The absolute value of is 100 for the image...

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CLASS EXERCISE #19 - 20 July 2005 You are supposed to make a slide projector using a single convergent lens. The projector projects an image of a slide (which is placed near the lens, in the projector). Suppose you want the slide to be magnified 100 times (meaning | | 100) and appear right-side up on a screen 5.0 m away from 7 œ the projector lens. Determine (in this suggested order): (a) the image distance (including the correct sign), 3 (b) the magnification (including its correct sign), 7 (c) the object distance (including its correct sign), and 9 (d) the approximate focal length of the lens you need. (e) Also, determine whether the slide (the object) needs to be upside-down or right-side up to give an image that is right-side up. S ince the image is to appear on a screen 5.0 m away from the lens, 3 œ  5.0 m. This must be positive since the image is real.
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Unformatted text preview: The absolute value of is 100, for the image to be magnified 100 times. Is 7 7 3 9 7 œ 3Î9 100 or 100? Since and must be positive, must be negative, so the value of must be 100. 7 7 œ Since 5.0 m 100 0.050 m or 5.0 7 œ 3Î9ß 9 œ 3Î7 œ Ð ÑÎÐ Ñ œ cm. This is positive, as we expect, since the slide is a real object. Then we can calculate the focal length from: 1 1 1 1 10 0 m 1 0.050 m Î0 œ Î3 Î9 œ ÎÐ Þ Ñ ÎÐ Ñ 0.100/m 20.0/m 20.1/m œ œ so 1 m/20.1 0.050 m or 5.0 cm (but just a little less). 0 œ œ Note that the slide must be placed just beyond the focal length: is just a 9 little greater than . Since is negative, the image is inverted, so for the image to 7 be right-side up, the slide must be upside-down in the projector....
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