Question

An object 4 cm in size is placed at a distance of 25 cm from a concave mirror of focal length 15 cm. At what distance from the mirror should the screen be kept so that a sharp image of the object can be obtained on screen? Find the nature and height of the image.

Answer 1

Given,

Focal length, $f=15cm$

Object distance, $u=25cm$

Object height, $h_0=4cm$

From mirror formula,

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

$\frac{1}{-15}=\frac{1}{-25}+\frac{1}{v}$

$\Rightarrow \frac{1}{v}=\frac{1}{-15}+\frac{1}{25}$

$v=-37.5cm$

magnification, $m=-\frac{v}{u}=\frac{h_i}{h_0}$

$\Rightarrow m=-\frac{(-37.5)}{-25}=\frac{h_i}{4}$

$h_i=-\frac{37.5\times 4}{25}$

$h_i=-6cm$

To get a sharp image, $v=37.5cm$

The image is real, inverted and magnified