Question

A converging beam of light falls on one surface of the biconcave lens whose other surface is silvered. After reflection from the silvered lens, the beam converges to a point $24\text{cm}$ in front of the lens. The focal length of the lens is $30\text{cm}$ and the silvered surface has a radius of curvature equal to $50\text{cm}$. Where will the beam of light converge if the lens is removed from its path?

• A. $4.5\text{cm}$ behind the lens
• B. $+6.74\text{cm}$ behind the lens
• C. $+6.7\text{cm}$ in front of the lens
• D. $3.3\text{cm}$ in front of the lens

Answer 1

The correct option is B $+6.74\text{cm}$ behind the lens

Given that,
Focal length of the lens, $f_l=-30\text{cm}$
Focal length of the mirror, $f_m=\frac{R}{2}=\frac{50}{2}=25\text{cm}$

The silvered lens is a combination of diverging lens and a diverging mirror. Hence, it will behave like a convex mirror of focal length $F_e$.

For equivalent convex mirror
$\Rightarrow \frac{-1}{F_e}=\frac{1}{f_l}+\frac{-1}{f_m}+\frac{1}{f_l}$

$\Rightarrow \frac{-1}{F_e}=\frac{1}{-30}+\frac{-1}{25}+\frac{1}{-30}$

$\Rightarrow F_e=\frac{75}{8}\text{cm}$


Let $u$ be the distance from the lens at which beam of light converge if the lens is removed from its path.

For equivalent convex mirror
$v=-24\text{cm};F_e=\frac{75}{8}\text{cm}$

using mirror formula,

$u=\frac{F_{e}v}{v-F_e}=\frac{\left(\frac{75}{8}\right)\times (-24)}{(-24)-\left(\frac{75}{8}\right)}$

$\Rightarrow u=+\frac{24\times 25}{89}\Rightarrow u=+6.74\text{cm}$

Hence, the beam will converge at a point $+6.74\text{cm}$ behind the lens if it is removed.

$\begin{array}{c}\text{Why this Question}?\\ \text{Silvered mirror questions can be analysed as the combination of lens and a mirror.}\end{array}$