A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is_

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A $10 mm$ long awl pin is placed vertically in front of a concave mirror. A $5 mm$ long image of the awl pin is formed at $30 cm$ in front of the mirror. The focal length of this mirror is__cm?

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- 30

- 20

- 40

- 60

answer is B.

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Detailed Solution

Object height,

h = 10 mm

, Image height,

H = - 5 mm

, image distance

v = - 30 cm

Magnification is equal to,

M = \frac{H}{h}

Here, M is magnification, H is Image height and h is Object height.

M = \frac{- 5}{10}

= \frac{- 1}{2}

Also, magnification is equal to,

M = - \frac{v}{u}

Here, M is magnification, v is image distance and u is object distance.

M = - \frac{(- 30)}{u}

- \frac{1}{2} = \frac{30}{u}

u = - 60 cm

The object distance is equal to

u = - 60 cm

.
The mirror formula is equal to,

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

Here, f is focal length, v is image distance and u is object distance.

\frac{1}{f} = \frac{1}{- 30} + \frac{1}{- 60}

\frac{1}{f} = - \frac{1}{30} - \frac{1}{60}

\frac{1}{f} = - \frac{1}{30} - \frac{1}{60}

\frac{1}{f} = \frac{- 2 - 1}{60}

= \frac{- 3}{60}

= \frac{- 1}{20}

f = - 20 cm

The focus of the concave mirror is

f = - 20 cm

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