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A point object O is placed on the principle axis of a convex lens of focal length 10 cm at 12 cm from the lens. When object is displaced 1 mm along the principle axis magnitude of displacement of image is x1. When the lens is displaced by 1 mm perpendicular to the principal axis displacement of image is x2 in magnitude. Find the value of x1/x2 . |
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Answer» U = -12 f = -10 1/v - 1/u = 1/f 1/v + 1/12 = 1/10 1/v = 1/60 \(-\frac{1}{v^2}\)dv or \(\frac{1}{u^2}\)du du = \(\frac{v^2}{u^2}\)du \(\frac{v}{u}\)= same path △u1 = △u2 = 1mm △u1 = △v2 △v2 = vf - vi x2 = △v2 + 1mm x2 = 2mm \(\frac{x_1}{x_2}=\frac{1}{2}\) Given u = -10 f = 12 \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) \(\frac{1}{v}-(\frac{1}{-10})=\frac{1}{12}\) \(\frac{1}{v}=\frac{1}{12}-\frac{1}{10}\) \(\frac{1}{v}=\frac{10-12}{120}\) \(\frac{1}{v}=\frac{-2}{120}\) \(\frac{1}{v}=\frac{-1}{60}\) \(-\frac{1}{v^2}\) dv + \(\frac{1}{u^2}\) du du = \(\frac{v^2}{u^2}\) du \(\frac{v}{u}\) = same path △u1 = △u2 = 1mm △u1 = △v2 △v2 = vf - vi x2 = △v2 + 1mm x2 = 2mm \(\frac{x_1}{x_2}=\frac{1}{2}\) |
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