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IIT-JEE 2010 Paper 1 Offline

Numerical

The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio $${{{m_{25}}} \over {{m_{50}}}}$$ is __________.

Your Input ________

Answer

Correct Answer is 6

Explanation

$${1 \over v} - {1 \over u} = {1 \over f}$$

or, $${u \over v} - 1 = {u \over f}$$

or, $${u \over v} = \left( {{{u + f} \over f}} \right)$$

$$\therefore$$ $$m = {v \over u} = \left( {{f \over {u + f}}} \right)$$

$${{{m_{25}}} \over {{m_{50}}}} = {{\left( {{{20} \over { - 25 + 20}}} \right)} \over {\left( {{{20} \over { - 50 + 20}}} \right)}} = 6$$

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