Question

Focal length of an equiconvex lens is $20cm.$ The refractive index of material of the lens is $\mathrm{1.5.}$ Now one of the curved surface is silvered. At what distance from the lens an object is to be placed, so that image coincides with the object?

• A. $-10cm$
• B. $-20cm$
• C. $-30cm$
• D. $-40cm$

Answer 1

The correct option is A $-10cm$

$\begin{array}{c}\text{Equiconvex lens,}\\ |R_1|=|R_2|=R\end{array}$

$\text{refractive index}\mu =1.5$

$\text{focal length}f=20cm$

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{20}=(1.5-1)(\frac{1}{R}-\frac{1}{-R})$

$\begin{array}{cc}\frac{1}{20}& =0.5\times \frac{2}{R}\\ R& =20cm\end{array}$ $\text{if lens is silvered,}$

$\frac{1}{F}=\frac{1}{f_m}-\frac{2}{f_L}$ $\begin{array}{c}{f}_m=\text{mirror}\text{focal length}\\ f_L=\text{lens focal}\text{length}\end{array}$

$\begin{array}{c}F=\text{effective focal}\text{length}\end{array}$

$\begin{array}{cc}\frac{1}{F}& =\frac{1}{-R/2}-\frac{2}{20}\\ \frac{1}{F}& =-\frac{2}{20}-\frac{2}{20}=-\frac{4}{20}=\frac{-1}{5}\\ F& =-5cm\end{array}$

$\begin{array}{c}\text{For image and object to}\\ \text{coincide, object is placed at}2F\\ \text{i.e - 10cm.}\end{array}$