Question

An equiconvex lens, $f_1=10cm$, is placed 40 cm in front of a concave mirror, $f_2=7.50cm$, as shown in fig. An object 2 cm high is placed 20 cm to the left of the lens. If overall magnification is X/1000. Then find out the value of X?

Answer 1

Given: An equiconvex lens, $f_1=10cm$, is placed $40cm$ in front of a concave mirror, $f_2=7.50cm$. An object $2cm$ high is placed $20cm$ to the left of the lens. If overall magnification is $\frac{X}{1000}$.

To find the value of X

Solution:

From lens formula

$\frac{1}{v}-\frac{1}{-20}=\frac{1}{10}⟹v=+20cm$

Therefore magnification of first image, $m_1=\frac{v}{u}=-1$

The image formed by the equiconvex lens right will be real, inverted and of same size as the object.

The first image will act as the object for concave mirror. Object distance for mirror is $(40-20)cm$.

From mirror equation,

$\frac{1}{v^{\prime }}+\frac{1}{-20}=\frac{1}{-7.5}⟹v^{\prime }=-12cm$

Therefore the second magnification will be, $m_2=\frac{v^{\prime }}{u^{\prime }}=\frac{-12}{-20}=0.6$

The second image acts as object for the lens. The object distance for the second refraction at the lens, $u^{\prime \prime }=+28cm$

From lens equation,

$\frac{1}{v^{\prime \prime }}-\frac{1}{+28}=\frac{1}{-10}⟹v^{\prime \prime }=-15.6cm$

So final magnification $m_3=\frac{v^{\prime \prime }}{u^{\prime \prime }}=\frac{-15.6}{+28}=-0.556$

Therefore, overall magnification will be

$m=m_1\times m_2\times m_3⟹m=(-1)\left(0.6\right)(-0.556)=-0.333⟹m=\frac{333}{1000}$

Hence $X=333$