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The distance between an object and its real image produced by a converging lens is 0.72 m.The magnification is 2. What will be the magnification when the object is moved by 0.04 m towards the lens ?

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answer is B.

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Detailed Solution

Given, m = 2 and the image formed is real v=2u and v+u=0.72

we get u=0.24 and v=0.48
$For a real image \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\Rightarrow f = \frac{0.72 \times 2}{3} = 0.16 m$

$present u = 0.2 m \Rightarrow magnification (m) = \frac{f}{u - f} = \frac{0.16}{0.2 - 0.16} = 4$

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