A thin prism in air produces the angle of min | vedclass

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

Vedclass · Accepted Answer

$\frac{\delta}{4}$

Vedclass · Answer

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

$A$ thin prism in air produces the angle of minimum deviation $\delta$. If the prism is immersed in water, the angle of minimum deviation for the same ray is (refractive index of water is $4/3$ and that of glass prism is $3/2$)

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