Question

A point object is kept in air at $40\text{cm}$ in front of an equiconvex lens silvered on one side. Radius of curvature of each side of lens is $20\text{cm}$ and refractive index of lens material is $3/2$. Where is the final image formed from silvered lens?

• A. $60\text{cm}$
• B. $\frac{30}{7}\text{cm}$
• C. $90\text{cm}$
• D. $\frac{40}{7}\text{cm}$

Answer 1

The correct option is D $\frac{40}{7}\text{cm}$


Focal length of mirror$;f_m=\frac{-20}{2}\text{cm}=-10\text{cm}$
Focal length of lens
$\frac{1}{f_l}=[\mu -1][\frac{1}{R_1}-\frac{1}{R_2}]=[\frac{3}{2}-1][\frac{1}{20}-\frac{-1}{20}]$
$\therefore f_l=20\text{cm}$
We know for combination of lens and mirror
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f_m}-\frac{2}{f_l}$
$\frac{1}{v}+\frac{1}{-40}=\frac{1}{-10}-\frac{2}{20}$
$v=\frac{-40}{7}\text{cm}$