Physics - Geometrical Optics Question with Solution | TestHub

For a thin prism, the angle of minimum deviation is given by $\delta =(n_{rel}-1)A$, where $n_{rel}$ is the refractive index of the prism relative to the surrounding medium and $A$ is the angle of the prism.

When the prism is in air, the deviation ${\delta }_1$ is:
${\delta }_1=({}^a{n}_g-1)A$
Substituting ${}^a{n}_g=\frac{3}{2}$:
${\delta }_1=\left(\frac{3}{2}-1\right)A=\frac{A}{2}$

When the prism is immersed in the liquid, the relative refractive index of the glass with respect to the liquid is:
${}^{\ell }{n}_g=\frac{{}^a{n}_g}{{}^a{n}_{\ell }}=\frac{\frac{3}{2}}{\frac{1}{3}}=\frac{9}{2}$

The new deviation ${\delta }_2$ in the liquid is:
${\delta }_2=({}^{\ell }{n}_g-1)A$
${\delta }_2=\left(\frac{9}{2}-1\right)A=\frac{7A}{2}$

Taking the ratio of ${\delta }_2$ to ${\delta }_1$:
$\frac{{\delta }_2}{{\delta }_1}=\frac{\frac{7A}{2}}{\frac{A}{2}}=7$

${\delta }_2=7{\delta }_1$

Answer: ${\delta }_2=7{\delta }_1$