An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. the height of the image will be ____cm.

An object

4 cm

in height, is placed at

15 cm

in front of a concave mirror of focal length

10 cm

. the height of the image will be

- 8 cm

.
It has been given that,
Object distance,

(u) = - 15 cm

Focal length

(f) = - 10 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.

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

\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 15}

v =

- 30 cm

Now,

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

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

h' = 4 \frac{- 30}{- 15}

h^{'} = - 8 cm

An object $4 cm$ in height, is placed at $15 cm$ in front of a concave mirror of focal length $10 cm$ . the height of the image will be ____ $cm$ .