Question

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of image.

Answer 1

Given,

​Distance of the object from the mirror = $u=-27cm$.

Height of the object = $h_o=+7cm$.

Focal length of the mirror = $f=-18cm$.

We have to find the distance of the image $v$ and height of the image $h_i$.

Using the mirror formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$, we get

$\frac{1}{-18}=\frac{1}{v}-\frac{1}{-27}$

$\Rightarrow \frac{1}{v}=\frac{1}{27}-\frac{1}{18}=\frac{2-3}{54}=-\frac{1}{54}$

$\Rightarrow v=-54cm$.

Thus, the distance of the image $v$ is 54 cm.

Now, using the magnification formula $m=\frac{h_i}{h_o}=-\frac{v}{u}$, we get

$h_i=(-\frac{v}{u})\times h_o$

$h_i=(-\frac{-54}{-27})\times 7$

$h_i=-14cm$.

Therefore, the height of the image $h_i$ is 14 cm.

Again, using the magnification formula $m=-\frac{v}{u}$, we get

$m=(-\frac{-54}{-27})=-2$.

As the magnification produced by the mirror is negative and more than 1, the image is real, inverted and large in size.