Question

Two spherical mirrors $M_1$ and $M_2$ are on optical Axis and at a separation of $50cm$. A point object $O$ is placed between Mirrors on optical Axis. Find Location and Nature of its Image after two successive Reflection at $M_1$ and $M_2$.

Answer 1

Given: Two spherical mirrors $M_1$ and $M_2$ are on optical Axis and at a separation of 50 cm. A point object O is placed between Mirrors on optical Axis.

To find Location and Nature of its Image after two successive Reflection at $M_1$ and $M_2$.

Solution:

From figure it is clear that $M_1$ is concave mirror, and $M_2$ is convex mirror

For concave mirror,

As per the given condition,

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

the focal length, $f=20cm$

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 infront of 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, $u^{\prime }=-(100-50)=-50cm$

the focal length, $f^{\prime }=-30cm$

Applying lens formula, we get

$\frac{1}{f^{\prime }}=\frac{1}{v^{\prime }}+\frac{1}{u^{\prime }}⟹\frac{1}{v^{\prime }}=\frac{1}{f^{\prime }}-\frac{1}{u^{\prime }}⟹\frac{1}{v^{\prime }}=\frac{1}{-30}-\frac{1}{-50}⟹\frac{1}{v^{\prime }}=\frac{-5+3}{150}⟹v=-75cm$

Hence the position of the final image is 75 behind the convex mirror.

Magnification, $m_2=-\frac{v^{\prime }}{u^{\prime }}=-\frac{-75}{-50}=-1.5$

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