Physics - Geometrical Optics Question with Solution | TestHub

The effective focal length of the silvered lens is given by

$\frac{1}{F}=-\frac{2}{f}+\frac{1}{f_m}=-\frac{2}{15}+\frac{1}{\infty }=-\frac{2}{15},$which gives

$F=-\frac{15}{2}cm.$The silvered lens behaves like a concave mirror. Using the spherical mirror formula$\frac{1}{v}+\frac{1}{u}=\frac{1}{F},$we have$\left(\because u=-20 cm \mathrm{a}\mathrm{n}\mathrm{d} F=-\frac{15}{2}cm\right)$

$\frac{1}{v}+\frac{1}{-20}=\frac{2}{-15}$

Which gives$v=-12 cm.$The negative sing indicates that the image is formed to the left of the lens.