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In an optics experiment, with the position of the object fixed, studen varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45^(@) with the x - axis meets the experimental curve at P. The coordinates of P will be : |
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Answer» (2f, 2f) One has a take that u is negative again for calculation, it effectively come to `(1)/(v) + (1)/(u) = (1)/(f)` If u = radius of curvature, 2f, v = 2f i.e.,`(1)/(2f) + (1)/(2f) = (1)/(f)` v and u have the same value when the object is at the centre of curvature. According thereal and virtual system, u is + ve and v is also + ve as both REAL if u = v, u = 2f = radius of curvature `therefore (1)/(v) + (1)/(u) = (1)/(f) rArr (1)/(2f) + (1)/(2f) = (1)/(f)`. |
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