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A converging lens forms a five fold magnified image of an object. The screen is moved towards the object by a distance d=0.5 m, and the lens is shifted so that the image has the same size as the object. Find the power of lens and the initial distance between the object and the screen. |
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Answer» image and object are of equal size. Hence, `|v|=|u|`. From the two figures, `6x=2y+d` or `6x-2y=0.5`……(i)` Using the lens formula for both the cases, `(1)/(5X)-1/-x=1/f` or `6/(5x)=1/f`......(ii) `1/y-1/(-y)=1/f` or `2/y=1/f`......(iii) SOLVING these three equations, we GET `x=0.1875 m and f=0.15625 m` Therefore, initial distance between the object and the screen `=6x=1.125 m` Power of the lens, `P=1/f` `=1/0.15625=6.4D` |
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