Physics - Geometrical Optics Question with Solution | TestHub

As$\frac{1}{f}=(\mathrm{\mu }-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\therefore \frac{1}{20}=(1.5-1)\left(\frac{1}{\infty }-\frac{1}{R}\right)$

$\frac{1}{20}=\frac{-1}{2R},R=-10$cm

Refraction from rarer to denser medium

$-\frac{{\mathrm{\mu }}_1}{u}+\frac{{\mathrm{\mu }}_2}{v}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{R},$where$u=\infty ,v=f$

$\therefore 0+\frac{1.5}{f}=\frac{1.5-1}{10}=\frac{1}{20},f=30 \mathrm{c}\mathrm{m}$