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At what distance should an object be placed from the pole of a converging mirror of focal length $20 cm$ so that the size of the real image formed will be ${\frac{1}{4}}^{th}$ the size of the object?

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125 cm

100 cm

324 cm

129 cm

answer is B.

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Detailed Solution

An object be placed at

100 cm

from the pole of a converging mirror of focal length

20 cm

so that the size of the real image formed will be

{\frac{1}{4}}^{th}

the size of the object.
As the size of the image is

\frac{1}{4}

times than the size of the object, magnification(m) will be

- \frac{1}{4}

(because the image formed is real).
By magnification formula, we get

m = - \frac{v}{u}

\Rightarrow - \frac{1}{4} = - \frac{v}{u}

u = 4 v

Now, substituting the value of

u

in mirror formula for calculating the distance of the object

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\Rightarrow \frac{1}{(- 20)} = \frac{1}{4 v} + \frac{1}{v}

\Rightarrow \frac{1}{(- 20)} = \frac{5}{4 v}

\Rightarrow 4 v = - 100

\Rightarrow v = - 25 cm

Substituting the value of

v

u

, we get

u = 4 v

\Rightarrow u = 4 (- 25)

\Rightarrow u = - 100 cm

So, the object should be placed at

100 cm

from the mirror.

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