Question

A pin is placed $10cm$ in front of a convex lens of focal length $20cm,$ made of material having refractive index $\mathrm{1.5.}$ The surface of the lens farther away from the pin is silvered and has a radius of curvature $22cm.$ Determine the position of the final image.

• A. $11cm$ in front
• B. $21cm$ in front
• C. $15cm$ in front
• D. $31cm$ in front

Answer 1

Answer: (A)

$11cm$ in front

$\text{In this case,}$

$\begin{array}{c}\text{for a silvered lens,}\\ \frac{1}{F}=-(\frac{2}{f_{\ell }}-\frac{1}{f_m})\end{array}$

$\begin{array}{cc}\text{where}& F=\text{net focal lenght of lens}\\ & f_{\ell }=\text{focal length of lens}\\ & f_m=\text{focal length of mirror}\end{array}$

$f_L=20cm$

$\begin{array}{cc}{f}_m& =-\frac{R}{2}=-11cm\\ \frac{1}{F}& =-(2\times \frac{1}{20}-\frac{1}{-11})\\ \frac{1}{F}& =-(\frac{1}{10}+\frac{1}{11})\\ \frac{1}{F}& =-\left(\frac{11+10}{110}\right)=-\frac{21}{110}\\ F& =\frac{-110}{21}=-5.23cm\\ \frac{1}{F}=\frac{1}{v}+\frac{1}{u}& \Rightarrow \frac{1}{v}=\frac{1}{F}-\frac{1}{u}\Rightarrow v=-11cm\end{array}$