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

Options

A. - 30 cm
B. -20 cm
C. - 40 cm
D. - 60 cm

Answer

B

Explanation

Here the size of object h = 10 mm = 1 cm Size of the image h. = 5 mm = 0.5 cm
Distance of image from the pole of mirror v = -30 cm
Let distance of object = u
Let focal length = f
Using mirror formula
`1/v + 1/u = 1/f`
Therefore, the value of u is needed for that magnification formula will be used.
Magnification (m)= `(h.)/(h) = - v/u`
`1/f = 1/v + 1/u`
Putting all the values in mirror formula:
`(1)/(- 30) + (1)/(- 60) = 1/f`
`implies (2+1)/(-60)=1/f`
`1/f=(3)/(-60) implies f= -20 cm`
.