Question

A biconvex lens, $f_1=20cm$, is placed 5 cm in front of a convex mirror, $f_2=15cm$. An object of length 2 cm is placed at a distance 10 cm from the lens. If overall magnification is X/3. Then find out the value of X?

Answer 1

Given: A biconvex lens, $f_1=20cm$, is placed 5 cm in front of a convex mirror, $f_2=15cm$. An object of length 2 cm is placed at a distance 10 cm from the lens. If overall magnification is $\frac{X}{3}$.

To find the value of X

Solution:

From lens formula

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

Therefore magnification of first image, $m_1=\frac{v}{u}=+2$

The first image will act as the object for convex mirror. Object distance for mirror is $u^{\prime }=(20+5)=25cm$.

From lens equation,

$\frac{1}{v^{\prime }}+\frac{1}{-25}=\frac{1}{+15}⟹v^{\prime }=\frac{75}{8}cm$

Therefore the second magnification will be, $m_2=\frac{v^{\prime }}{u^{\prime }}=\frac{\frac{75}{8}}{-25}=\frac{3}{8}$

The second image acts as object for the lens. The object distance for the second refraction at the lens, $u^{\prime \prime }=\frac{75}{8}+5=\frac{115}{8}$

From lens equation,

$\frac{1}{v^{\prime \prime }}-\frac{1}{+\frac{115}{8}}=\frac{1}{-20}⟹v^{\prime \prime }=\frac{460}{9}cm$

So final magnification $m_3=\frac{v^{\prime \prime }}{u^{\prime \prime }}=\frac{\frac{460}{9}}{\frac{115}{8}}=\frac{32}{9}$

Therefore, overall magnification will be

$m=m_1\times m_2\times m_3⟹m=2\times \frac{3}{8}\times \frac{32}{9}⟹m=\frac{8}{3}$

Hence $X=8$