An object 2 cm high is placed at a distance of 16 cm from a concave mirror which produces a real image 3 cm high. The focal length of the mirror is -[[1]]cm.

An object

2 cm

high is placed at a distance of

16 cm

from a concave mirror which produces a real image

3 cm

high. Focal length of mirror is

- 9.6 cm

.
Given,
Object height,

h = 2 cm

Image height,

h' = - 3 cm

(since the real image is inverted).

m = \frac{h'}{h}

= \frac{- v}{u}

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

v = - 24 cm

Applying mirror formula,

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

Where

v

is the distance of the image from the mirror,

u

is the distance of the object from the mirror, and

f

is the focal length of the mirror.

\Rightarrow \frac{1}{- 16} + \frac{1}{- 24} = \frac{1}{f}

\Rightarrow - \frac{3 + 2}{48} = \frac{1}{f}

f = - 9.6 cm

An object $2 cm$ high is placed at a distance of $16 cm$ from a concave mirror which produces a real image $3 cm$ high. The focal length of the mirror is $-$ [[1]] $cm$ .