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 from the silvered lens should an object be placed to obtain a real image magnified three times? $({\mu }_g=1.5)$

• 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 that,
Focal length of biconvex lens, $f_l=40\text{cm}$
Radius of curvature of non silvered lens, $R_1=60\text{cm}$
Refractive index of lens, ${\mu }_g=1.5$

Let $x$ is the magnitude of radius of the curvature of the silvered surface.
Sign convention: Direction along the incident ray is taken as positive.


From lens maker's formula:
$\frac{1}{f_l}=(\frac{{\mu }_g}{{\mu }_0}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\Rightarrow \frac{1}{40}=(\frac{1.5}{1}-1)(\frac{1}{+60}-\frac{1}{-x})$
$\Rightarrow \frac{1}{40}=\frac{1}{2}(\frac{1}{60}+\frac{1}{x})$
$\Rightarrow x=\frac{60\times 20}{60-20}=30\text{cm}$
$\Rightarrow x=R_2=30\text{cm}$
Now
$f_m=$ focal length of mirror
$f_m=\frac{-R}{2}=-15\text{cm}$

The silvered lens will behave like a converging mirror with focal length $F$ because it is a combination of a converging lens and a converging mirror
$\frac{1}{F}=\frac{-2}{f_l}+\frac{1}{f_m}$


Where $f_l=40\text{cm}$
$\frac{1}{F}=\frac{-2}{40}+\frac{1}{-15}$
$\Rightarrow F=\frac{-60}{7}\text{cm}$
For the concave mirror
Magnification $=-3,F=-60/7\text{cm}$
$M=\frac{-f}{u-f}$
$\Rightarrow -3=\frac{+60/7}{u+60/7}$
$\Rightarrow -3u-3\times \frac{60}{7}=\frac{60}{7}$
$\Rightarrow -3u=\frac{60}{7}+\frac{180}{7}$
$\Rightarrow -3u=\frac{240}{7}$
$\Rightarrow u=\frac{-80}{7}\text{cm}$
$\Rightarrow u=-11.42\text{cm}$

Why this question ? Note: Entire setup acts like a mirror so the magnification formula of the mirror is to be used.