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An object of 1 cm high produces a real image $1.5 cm$ high when placed at a distance of $15 cm$ from a concave mirror. What is the focal length of the mirror?

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- 9 cm

- 10 cm

+ 9 cm

+ 16 cm

answer is A.

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Detailed Solution

An object of 1 cm high produces a real image

1.5 cm

high when placed at a distance of

15 cm

from a concave mirror. The focal length of the mirror

f = - 9 cm

Given:
Height of the object

h = + 1 cm

Distance of the object from the mirror,

u = - 15

We know

m = \frac{h^{'}}{h}

= - \frac{v}{u}

hence

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

= - \frac{1.5}{+ 1} \times (- 15)

= - 22.5 cm

.
Thus, the real and inverted image of the object is formed at a distance of

22.5 cm

in front of the concave mirror.
Now,

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

= \frac{1}{- 22.5} + \frac{1}{- 15}

= \frac{- 2 - 3}{45}

= - \frac{5}{45}

= - \frac{1}{9}

So,

f = - 9 cm

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