Question

find the size, nature and position of image formed when an object of size 2cm is placed at distance of 20cmfrom a concave mirror of focal length 15cm​

Answer 1

HELLO THERE!

Mirror formula:

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

For a concave mirror, according to the sign convention, u (object distance) and f (focal length), both are negative. So, we have:

$-\frac{1}{15} = \frac{1}{v} - \frac{1}{20}\\\\\\\implies \frac{1}{v} = \frac{1}{20} - \frac{1}{15}\\\\\\\implies \frac{1}{v} = -\frac{1}{60}\\\\\\\implies v = -60 cm$

So, the image distance from the pole is 60 cm.

Magnification:

$m = -\frac{v}{u} = -[\frac{-60}{-20}] = -3$

Also,

$m = -\frac{h_{I}}{h_{o}}\\\\\\\implies -3 = -\frac{h_{I}}{2}\\\\\\\implies h_{I} = 6 cm$

So, height of the image is 6 cm. A minus sign in the magnification denotes that the image is Real, and magnified as well. It is formed beyond the centre of curvature.

Thanks!