Question

A pin is placed $10$ cm in front of a convex lens of focal length $20$ cm and refractive index $1.5$. The surface of the lens farther away from the pin is silvered and has a radius of curvature of $22$ cm. How far from the lens is the final image formed?

• A. $10cm$
• B. $11cm$
• C. $12cm$
• D. $13cm$

Answer 1

The correct option is B $11cm$

The vet focus of systom $\Rightarrow f_{nt}$

Q we Rusus

$\Rightarrow \frac{1}{{\int }_{\text{vet}}}=\frac{-2}{f_e}+\frac{1}{f_m}$

as $R_m=22\mathrm{con}\therefore f=\frac{22}{2}=12cm$

$\frac{y}{f_{n+1}}=\frac{-2}{20}+\frac{1}{11}\Rightarrow \frac{1}{10}+\frac{1}{11}\to \frac{-11+10}{110}$

$\therefore f_{\mathrm{met}}=-110cm$

by mirror formula $\Rightarrow \frac{1}{y}+\frac{1}{4}-\frac{1}{f}\Rightarrow$

$\Rightarrow \frac{1}{y}+\frac{1}{(-10)}=\frac{-1}{110}\Rightarrow \frac{1}{v}=\frac{1}{110}+\frac{1}{10}$

$\begin{array}{c}\frac{1}{v}=\frac{-10+110}{1100}\Rightarrow \frac{100}{1100}=\frac{1}{11}\\ \therefore v=11cm\end{array}$

Wenee option $B^{\prime }$ is corvect

the final imbge is formed $11cm$ from the lens.