Question

RAY OPTICS
Q.1 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

Answer:

Given

Radius of curvature, R = 160 cm

Focal length, f = R/2 = 80 cm

Distance of the image, v = 70cm (sign is positive as virtual image)

By applying mirror's formula

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

$\mathsf{\implies\frac{1}{u} = \frac{1}{f} - \frac{1}{v}}$

$\mathsf{\implies\frac{1}{u} = \frac{1}{80} - (\frac{1}{70})}$

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

$\mathsf{\implies\frac{1}{u} = \frac{-10}{5600}}$

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

$\fbox{\mathsf{ u = -560 cm}}$

magnification = -v/u

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

$\mathsf{ \implies\: m = \frac{1}{8}}$

$\fbox{\mathsf{ m = + 0.125 }}$