Question

One of the surfaces of thin convex lens of focal length 12 cm is silvered and a object is placed at 12 cm in front of lens as shown in figure. If refractive index of lens is 1.5, then distance of image from lens in cm is

Answer 1

Combination behaves as a mirror of focal length $f_{eq}$

Here $f_1=f_3=12cm$ (given)

and $f_2=\frac{R}{2}$

$\frac{1}{12}=(1.5-1)\frac{2}{R}\Rightarrow R=12cm$ and $f_2=6cm$

$\frac{1}{f_{eq}}=\frac{1}{12}+\frac{1}{6}+\frac{1}{12}=\frac{1}{3}\Rightarrow f_{eq}=3cm$

By mirror formula $\Rightarrow \frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$-\frac{1}{3}=\frac{1}{v}-\frac{1}{12}\Rightarrow \frac{1}{v}=\frac{1}{12}-\frac{1}{3}=\frac{1-4}{12}=\frac{-3}{12}cm$

$v=-4cm$