Convex Lens Algebra
Date: 11/09/98 at 07:33:54 From: Karl Pudniks Subject: Convex lens algebra A focused image of an object produced by a convex lens and is projected onto a screen at 3x magnification. Then the screen is moved 20cm, and the object moved a certain distance until another focused image is produced at 2.5x magnification. What is the focal length of the lens? Could you please try to explain the use of the formula 1/u + 1/v = 1/f, as this is all we've been taught. I'm having trouble with the algebra. I keep ending up with u = u or something similar. The closest I get is two unknowns, two equations, but I can't do the last step. Thanks a lot, Kurt
Date: 11/10/98 at 08:28:25 From: Doctor Rick Subject: Re: Convex lens algebra Hi, Kurt. I wish you had told us what two equations you got, so I could help you with the last step! The trickiest part of this problem for me was that it isn't clear whether the screen was moved closer to the lens or farther from it, so I had to try it both ways and see which one gave a solution. If you look at each position separately, you can find the screen distance in terms of the focal length f. Use the equation you gave, plus the fact that the ratio of image distance to object distance is equal to the magnification. That is, if the magnification is 3, then v = 3u where u is the object distance and v is the image distance (the screen). That means that you can write: 1 1 1 ------- + --- = --- (v_1)/3 v_1 f where v_1 is the screen distance at the first position. Solve this for v_1 in terms of f. Do the same for screen distance v_2, with a magnification of 2.5. You can use these two equations to get an equation for v_2 in terms of v_1. Now you can use the information that v_2 = v_1 + 20 cm - or is it v_2 = v_1 - 20 cm? Try it both ways. Substitute for v_2 from the previous equation, and you have an equation for v_1. The answer should be positive. Now go back to the equation for v_1 in terms of f, and solve it for f in terms of v_1. Plug in your solution for v_1 and you have the focal length. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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