Physics - Geometrical Optics Question with Solution | TestHub

The deviation produced by a thin prism in air is given by $\delta =({\mu }_g-1)A$.

Substituting ${\mu }_g=\frac{3}{2}$, we get $\delta =\left(\frac{3}{2}-1\right)A=\frac{A}{2}$.

When the prism is immersed in water, the relative refractive index of glass with respect to water is ${\mu }_{gw}=\frac{{\mu }_g}{{\mu }_w}=\frac{3/2}{4/3}=\frac{9}{8}$.

The new angle of minimum deviation is ${\delta }^{\prime }=({\mu }_{gw}-1)A$.

Substituting ${\mu }_{gw}=\frac{9}{8}$, we get ${\delta }^{\prime }=\left(\frac{9}{8}-1\right)A=\frac{A}{8}$.

Comparing the two deviations, $\frac{{\delta }^{\prime }}{\delta }=\frac{A/8}{A/2}=\frac{1}{4}$.

Thus, ${\delta }^{\prime }=\frac{\delta }{4}$.

Answer: $\frac{\delta }{4}$