Question

An object lies on the principal axis of a concave mirror with radius of curvature $160$cm. Its vertical image appears at a distance $70$cm from it. Determine the position of the object and also the magnification.

Answer 1

Given:

Radius of curvature of the mirror: $R=160cm$

Thus, the focal length of the morror: $f=80cm$ ( as $f=\frac{R}{2}$)

Image distance: $v=70cm$ (sign is positive as virtual image)

As

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ (Mirror Formula)

$\frac{1}{80}=\frac{1}{70}+\frac{1}{u}$

$\frac{1}{u}=\frac{1}{80}-\frac{1}{70}$

$\frac{1}{u}=\frac{70-80}{5600}$

$\frac{1}{u}=\frac{-1}{560}$

$u=-560cm$

As magnification $m=-\frac{v}{u}$

$m=-\frac{70}{(-560)}$

$m=\frac{1}{8}=0.125$

Hence the object is placed at $560cm$ from the mirror to the right and the magnification is $0.125$.