Question

One surface of a biconvex lens having focal length $40\text{cm}$ is silvered. The radius of the curvature of the other surface is $60\text{cm}$. At what distance the silvered lens should an object be placed to obtain a real image magnified three times ?
$(\text{Refractive index of lens,}{\mu }_L=1.5)$

• A. $10\text{cm}$
• B. $11.4\text{cm}$
• C. $12\text{cm}$
• D. $12.4\text{cm}$

Answer 1

The correct option is B $11.4\text{cm}$

Given:
$f_L=40\text{cm};R_1=60\text{cm}$


Let, $x=$ magnitude of radius of the curvature of the silvered surface.

From lens maker's formula:

$\frac{1}{f_L}=(\frac{{\mu }_L}{{\mu }_m}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

where, ${\mu }_L\to$RI of lens $=1.5$
${\mu }_m\to$RI of medium $=1$

$\Rightarrow \frac{1}{40}=(\frac{1.5}{1}-1)(\frac{1}{+60}-\frac{1}{-x})$

$\Rightarrow \frac{1}{40}=\frac{1}{2}(\frac{1}{60}+\frac{1}{x})$

$x=\frac{60\times 20}{60-20}=30\text{cm}$

The silvered lens will behave like a converging mirror with focal length $F_e$ because it is a combination of a converging lens and a converging mirror.

$\frac{-1}{F_e}=\frac{1}{f_L}+\frac{-1}{f_m}+\frac{1}{f_L}.....\left(1\right)$

where, $f_L=40\text{cm}$
$f_m$ = focal length of mirror$=\frac{-30}{2}=-15\text{cm}$

Putting the value of above data in Eq.(1)

$\Rightarrow \frac{-1}{F_e}=\frac{2}{40}-\frac{1}{-15}\Rightarrow F_e=\frac{-60}{7}\text{cm}$

$\because$ Focal length of the mirror is negative means the mirror is concave mirror.


For the concave mirror
Magnification $=-3,F_e=-\frac{60}{7}\text{cm}$

$M=\frac{f}{f-u}$

$\Rightarrow -3=\frac{-\frac{60}{7}}{-\frac{60}{7}-u}$

$3u+\frac{180}{7}=-\frac{60}{7}$

$\Rightarrow u=\frac{-80}{7}\text{cm}=-11.42\text{cm}$
Hence, option $\left(b\right)$ is correct.

$\begin{array}{c}\text{Why this Question}?\\ \text{CAUTION: Entire setup acts like a mirror so the magnification}\\ \text{formula of the mirror is to be used}\end{array}$