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At what distance should a screen be placed to get a sharp image when an object of $5 cm$ in size is placed at a distance of $20 cm$ from a converging mirror of a focal length of $15 cm$ ?

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25 cm

60 cm

34 cm

19 cm

answer is B.

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Detailed Solution

The screen should be placed at

60 cm

to get a sharp image when an object of

5 cm

in size is placed at a distance of

20 cm

from a converging mirror of a focal length of

15 cm

.
Given,

f = - 15 cm

(negative : as the mirror is converging)

u = - 20 cm

Substituting the given values in mirror formula we get,

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

\Rightarrow \frac{1}{(- 15)} = \frac{1}{v} + \frac{1}{- 20}

\Rightarrow \frac{1}{v} = \frac{3 - 4}{60}

\Rightarrow \frac{1}{v} = - \frac{1}{60}

\Rightarrow v = - 60 cm

The negative sign indicates that the image will be formed on the same side as that of the object. So, the screen should be placed at

60 cm

in front of the mirror.
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