Question

A biconvex thin lens is prepared from glass of refractive index $\frac{3}{2}$. The two bounding surfaces have equal radii of $25\text{cm}$ each. One of the surface is silvered from outside to make it reflecting. Where should an object be placed before this lens so that the image coincides with the object ?

• A. $25\text{cm}$
• B. $6.25\text{cm}$
• C. $12.5\text{cm}$
• D. $11.25\text{cm}$

Answer 1

The correct option is C $12.5\text{cm}$

Equivalent focal length of this system which behaves like a mirror is given by formula (taking rightwards direction as positive),


Refraction through surface $1$,

$\frac{{\mu }_2}{v_1}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}....\left(1\right)$

For the reflection at the surface $2$,

$\frac{1}{v_2}+\frac{1}{v_1}=\frac{2}{R_2}....\left(2\right)$

Again refraction at the surface $1$,

$\frac{{\mu }_1}{-v}-\frac{{\mu }_2}{v_2}=\frac{{\mu }_2-{\mu }_1}{R_1}....\left(3\right)$

From $\left(1\right)$,$\left(2\right)$ and $\left(3\right)$,

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}=\frac{2\left(\frac{{\mu }_2}{{\mu }_1}\right)}{R_2}-\frac{2(\frac{{\mu }_2}{{\mu }_1}-1)}{R_1}$

Here, $R_1=+25\text{cm},R_2=-25\text{cm},{\mu }_1=1$ and ${\mu }_2=\frac{3}{2}$

As image coincides with object, hence, $u=v=-x$ (say)

Substituting in mirror formula, we have

$\frac{1}{-x}-\frac{1}{x}=\frac{2\left(\frac{3}{2}\right)}{-25}-\frac{2(\frac{3}{2}-1)}{25}$

$\Rightarrow \frac{2}{x}=\frac{3}{25}+\frac{1}{25}$

$\Rightarrow \frac{2}{x}=\frac{4}{25}$

$\therefore x=12.5\text{cm}$

Hence, the object should be palced at a disance $12.5\text{cm}$ in front of the silvered surface.