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What will be the height of the image when an object of height $50 mm$ stands on the axis of a concave mirror of focal length $100 mm$ at a distance of $300 mm$ from the mirror?

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2.5 cm

10 cm

3.4 cm

1.9 cm

answer is A.

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Detailed Solution

The height of the image will be

2.5 cm

when an object of height

50 mm

stands on the axis of a concave mirror of focal length

100 mm

at a distance of

300 mm

from the mirror.
Given,

u = - 300 mm

\Rightarrow - 30 cm

f = - 100 mm

\Rightarrow - 10 cm

h' = 50 mm

\Rightarrow 5 cm

Substituting the given values in mirror formula we get,

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

\Rightarrow \frac{1}{(- 10)} = \frac{1}{(- 30)} + \frac{1}{v}

\Rightarrow \frac{1 - 3}{30} = \frac{1}{v}

\Rightarrow v = - 15 cm

The image formed will be

15 cm

away from the pole of the mirror.
Substituting the value of

v

u

and

h ’

in the magnification formula, we get

\Rightarrow m = \frac{h}{h'} = - \frac{v}{u}

\Rightarrow h = - \frac{v}{u} \times h^{'}

\Rightarrow h = - \frac{(- 15)}{(- 30)} \times 5

\Rightarrow h = - 2.5 cm

Therefore, the height of the image will be

2.5 cm

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