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(D) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the prism material relati

Vedclass · Accepted Answer

$\frac{\delta_1}{4}$

Vedclass · Answer

(D) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the prism material relative to the surrounding medium and $A$ is the prism angle.In air,the refractive index of glass relative to air is $_{a}\mu_{g} = \frac{3}{2}$. Thus,$\delta_1 = (_{a}\mu_{g} - 1)A = (\frac{3}{2} - 1)A = \frac{1}{2}A$.When immersed in water,the refractive index of glass relative to water is $_{w}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{w}} = \frac{3/2}{4/3} = \frac{9}{8}$.The new angle of minimum deviation is $\delta_2 = (_{w}\mu_{g} - 1)A = (\frac{9}{8} - 1)A = \frac{1}{8}A$.Comparing $\delta_2$ with $\delta_1$: $\frac{\delta_2}{\delta_1} = \frac{\frac{1}{8}A}{\frac{1}{2}A} = \frac{2}{8} = \frac{1}{4}$.Therefore,$\delta_2 = \frac{\delta_1}{4}$.

The angle of minimum deviation produced by a thin prism in air is $\delta_1$. If it is immersed in water,the angle of minimum deviation is $\left[_{a}\mu_{g}=\frac{3}{2}, _{a}\mu_{w}=\frac{4}{3}\right]$.

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