Question

An object 2 cm in height is placed at a distance
of 16 cm from a concave mirror. If the focal
length of the mirror is 9.6 cm., find the image
distance, nature and size of the image.​

Answer 1

$\huge \fcolorbox{red}{pink}{Solution :)}$

According to sign convention ,

• Height of object (h) = 2 cm
• Object distance (u) = - 16 cm
• Focal length (f) = - 9.6 cm

We know that , the formula of mirror is given by

$\large \sf \fbox{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

Substitute the known values , we get

$\sf \hookrightarrow - \frac{1}{9.6} = \frac{1}{v} + ( - \frac{1}{16} ) \\ \\ \sf \hookrightarrow \frac{1}{v} = \frac{1}{16} - \frac{1}{9.6} \\ \\ \sf \hookrightarrow \frac{1}{v} = \frac{96 - 160}{1536} \\ \\ \sf \hookrightarrow v = - \frac{1536}{64} \\ \\ \sf \hookrightarrow v = - 24 \: \: cm$

As v is negative , a real image is formed 24 cm from the mirror on the same side as the object

We know that , the formula of linear magnification for spherical mirror is given by

$\mathtt{ \large \fbox{Magnification = - \frac{v}{u} }}$

Substitute the known values , we get

$\sf \hookrightarrow M = -( \frac{ - 24}{ - 16} ) \\ \\ \sf \hookrightarrow M = - 1.5$

Hence , the image is magnified , real and inverted