Physics - Geometrical Optics Question with Solution | TestHub

For refraction at first surface,

$u=-7.5 \mathrm{c}\mathrm{m}, R_1=2.5 \mathrm{c}\mathrm{m}$

${\mu }_1=1, {\mu }_2=\frac{4}{3}$

$\therefore \frac{{\mu }_2}{v^{\prime }}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}$

$\therefore \frac{4/3}{v\prime }-\frac{1}{\left(-7.5\right)}=\frac{4/3-1}{2.5}$

$\Rightarrow \frac{4}{3v^{\prime }}=\frac{1}{7.5}-\frac{1}{7.5}=0$

$\Rightarrow v^{\prime }=\infty$

It means the ray is parallel within the sphere.

For refraction at second surface,

$u=-\infty , {\mu }_1=\frac{4}{3}$

${\mu }_2=1, R_2=-2.5 \mathrm{c}\mathrm{m}$

$\therefore \frac{{\mu }_2}{v^{\prime }}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_2}$gives

$\Rightarrow \frac{1}{v^{\prime }}-\frac{4/3}{\left(-\infty \right)}=\frac{1-\left(4/3\right)}{\left(-2.5\right)}$

$\Rightarrow v=+ 7.5 \mathrm{c}\mathrm{m}$

i.e. Image is formed at distance$7.5 \mathrm{c}\mathrm{m}$to the right of$P_2$. Hence, distance of final image$P$from centre$C=7.5+2.5=10 \mathrm{c}\mathrm{m}.$