Question

An object is placed exactly midway between a concave mirror of radius of curvature $40$cm and a convex mirror of radius of curvature $30$cm. The mirrors face each other and are $50$cm apart. Determine the nature and position of the image formed by successive reflections first at the concave mirror and then at the convex mirror.

Answer 1

Given: An object is placed exactly midway between a concave mirror of radius of curvature 40cm and a convex mirror of radius of curvature 30cm. The mirrors face each other and are 50cm apart.

To find the nature and position of the image formed by successive reflections first at the concave mirror and then at the convex mirror.

Solution:

For concave mirror,

As per the given condition,

object distance, $u=25cm$, as the mirrors are 50cm apart.

Radius of curvature, $R=40cm$

So the focal length, $f=\frac{R}{2}=\frac{40}{2}=20cm$

Applying lens formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}⟹\frac{1}{v}=\frac{1}{f}-\frac{1}{u}⟹\frac{1}{v}=\frac{1}{20}-\frac{1}{25}⟹\frac{1}{v}=\frac{5-4}{100}⟹v=100cm$

Therefore the image is 100cm behind the concave mirror.

Magnification, $m_1=-\frac{v}{u}=-\frac{100}{25}=-4$

Hence the image formed by the concave mirror is inverted, virtual and enlarged.

This becomes object for convex mirror, so

object distance, $u^{\prime }=-(100-50)=-50cm$

Radius of curvature of convex mirror, $r=30cm$

So the focal length, $f^{\prime }=\frac{r}{2}=\frac{30}{2}=-15cm$

Applying lens formula, we get

$\frac{1}{f^{\prime }}=\frac{1}{v^{\prime }}+\frac{1}{u^{\prime }}⟹\frac{1}{v^{\prime }}=\frac{1}{f^{\prime }}-\frac{1}{u^{\prime }}⟹\frac{1}{v^{\prime }}=\frac{1}{-15}-\frac{1}{-50}⟹\frac{1}{v^{\prime }}=\frac{-10+3}{150}⟹v=-21.4cm$

Hence the position of the final image is 21.4 behind the convex mirror.

Magnification, $m_2=-\frac{v^{\prime }}{u^{\prime }}=-\frac{-21.4}{-50}=-0.43$

So the nature of the final image formed is inverted of inverted, i.e., upright.