Physics - Geometrical Optics Question with Solution | TestHub

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

When the prism is in air, the deviation is:

${\delta }_1=({}_a{n}_g-1)A=\left(\frac{3}{2}-1\right)A=\frac{A}{2}$

When the prism is immersed in the liquid, the relative refractive index of the glass with respect to the liquid is:

${}_l{n}_g=\frac{{}_a{n}_g}{{}_a{n}_l}=\frac{3/2}{4/3}=\frac{9}{8}$

The deviation in the liquid is:

${\delta }_2=({}_l{n}_g-1)A=\left(\frac{9}{8}-1\right)A=\frac{A}{8}$

Taking the ratio of the two deviations:

$\frac{{\delta }_2}{{\delta }_1}=\frac{\frac{A}{8}}{\frac{A}{2}}=\frac{1}{4}$

${\delta }_2=\frac{{\delta }_1}{4}$

Answer: ${\delta }_2=\frac{{\delta }_1}{4}$