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Moving Object, Stationary Lens

A convex lens...

Question

A convex lens forms a real image with magnification $m_{1}$ on a screen.Now,the screen is moved by a distance x and the object is also moved so as to obtain a real image with magnification $m_{2} (> m_{1})$ on the screen.Then,the focal length of the lens is

A

(\frac{m_{1}}{m_{2}}) x

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B

(\frac{m_{2}}{m_{1}}) x

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C

x (m_{2} - m_{1})

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D

(\frac{x}{(m_{2} - m_{1})})

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Solution

The correct option is D $(\frac{x}{(m_{2} - m_{1})})$
The magnification in case of displacement method is given by,
$m = \frac{v}{f} - 1$
For the first case, v = v, and the second case, $v = v + x$
Solving for f gives, $f = \frac{x}{m_{2} - m_{1}}$
where, $m_{1}$ and $m_{2}$ are the respective magnifications.

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