Physics - Geometrical Optics Question with Solution | TestHub

For the refraction at the left surface, $u=-\infty , R=+15 \mathrm{cm}$.

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{1.5}{v_1}-\frac{1}{-\infty }=\frac{1.5-1}{+15}$

$\Rightarrow v_1=45 \mathrm{cm}$

For the refraction at the right surface, $u=45-30=15 \mathrm{cm}, R=-15 \mathrm{cm}$

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{1}{v_2}-\frac{1.5}{15}=\frac{1-1.5}{-15}$

$\Rightarrow v_2=7.5 \mathrm{cm}$

Therefore, the distance from the centre where the beam will converge is,

$15+7.5=22.5 \mathrm{cm}=225 \mathrm{mm}$