An object of 3 cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 4.5 cm high. The formation of the image will be?

An object of

3 cm

high is placed at a distance of

8 cm

from a concave mirror which produces a virtual image

4.5 cm

high. The formation of the image will be behind the mirror (virtual and erect).
Given,
Distance of the object from the mirror,

u = - 8 cm

Height of the object

{, h}_{0} = 3 cm

Height of the image,

h_{i} = 4.5 cm

Since ho and hi are given

m = \frac{h_{i}}{h_{o}}

\frac{h_{i}}{h_{o}} = - \frac{v}{u}

\frac{4.5}{3} = - \frac{v}{(- 8)}

3 v = 4.5 \times 8

v = + 12 cm

Since

v = + 12 cm,

the image is formed behind the mirror.
The object is placed between the pole and the focal length (

f = 24 cm

and

u = 8 cm

). Hence, the formation of the image will be behind the mirror.

An object of $3 cm$ high is placed at a distance of $8 cm$ from a concave mirror which produces a virtual image $4.5 cm$ high. The formation of the image will be?