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(A) For a thin prism,the angle of minimum deviation $\delta$ is given by the formula $\delta = A(\mu - 1)$,where $A$ is the prism angle and $\mu$ is t

Vedclass · Accepted Answer

$4: 1$

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(A) For a thin prism,the angle of minimum deviation $\delta$ is given by the formula $\delta = A(\mu - 1)$,where $A$ is the prism angle and $\mu$ is the refractive index of the prism material relative to the surrounding medium.$1$. When the prism is in air:The refractive index of the prism with respect to air is $\mu_1 = \frac{3}{2}$.Therefore,the angle of minimum deviation is $\delta_1 = A(\frac{3}{2} - 1) = A(\frac{1}{2}) = \frac{A}{2}$.$2$. When the prism is in water:The refractive index of water is $\mu_w = \frac{4}{3}$. The refractive index of the prism with respect to water is $\mu_2 = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.Therefore,the angle of minimum deviation is $\delta_2 = A(\frac{9}{8} - 1) = A(\frac{1}{8}) = \frac{A}{8}$.$3$. The ratio of the angles of minimum deviation is:$\frac{\delta_1}{\delta_2} = \frac{A/2}{A/8} = \frac{8}{2} = \frac{4}{1}$.Thus,the ratio is $4: 1$.

$A$ small-angled prism is made of a material of refractive index $\frac{3}{2}$. The ratio of the angles of minimum deviation when the prism is placed in air and in water of refractive index $\frac{4}{3}$ is:

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