Question

An extended object is placed at point $O,10\text{cm}$ in front of a convex lens $L_1$ and a concave lens $L_2$ is placed $10\text{cm}$ behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are $20\text{cm}$. The refractive index of both the lenses is $1.5$. The total magnification of this lens system is

• A. $0.4$
• B. $0.8$
• C. $1.3$
• D. $1.6$

Answer 1

The correct option is B $0.8$

Focal length of convex lens $\left(f_1\right)$ is given by
$\frac{1}{f_1}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$
$=(1.5-1)[\frac{1}{20}-\left(\frac{1}{-20}\right)]$
$\Rightarrow \frac{1}{f_1}=\frac{1}{20}\Rightarrow f_1=+20\text{cm}$
Similarly, focal length of concave lens $\left(f_2\right)$ is given as
$\frac{1}{f_2}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$
$=(1.5-1)[-\frac{1}{20}-\frac{1}{20}]$
$\Rightarrow \frac{1}{f_2}=-\frac{1}{20}\Rightarrow f_2=-20\text{cm}$
Using lens formula for convex lens, we have
$\frac{1}{v_1}-\frac{1}{-10}=\frac{1}{20}$
$v_1=-20\text{cm}$
$m_1=\frac{v_1}{u_1}=\frac{(-20)}{-10}=2$
Again, applying lens formula for concave lens, we have
$\frac{1}{v_2}-\frac{1}{-30}=\frac{1}{-20}$
$\frac{1}{v_2}=-\frac{1}{20}-\frac{1}{30}=\frac{-3-2}{60}$
$v_2=-12\text{cm}$
$m_2=\frac{v_2}{u_2}=\frac{(-12)}{-30}=\frac{2}{5}$
Therefore, overall magnification is given by
$m=m_1\times m_2=2\times \frac{2}{5}=0.8$