If a thin prism of glass is dipped into water | vedclass

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(B) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$.When the prism is in air,the deviation is $\delta_a = (_a\mu_g -

Vedclass · Accepted Answer

$\frac{1}{4}$

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(B) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$.When the prism is in air,the deviation is $\delta_a = (_a\mu_g - 1)A$.When the prism is in water,the refractive index of glass with respect to water is $_w\mu_g = \frac{_a\mu_g}{_a\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.The deviation in water is $\delta_w = (_w\mu_g - 1)A = \left( \frac{9}{8} - 1 \right)A = \frac{1}{8}A$.Comparing the two,$\frac{\delta_w}{\delta_a} = \frac{(1/8)A}{(1/2)A} = \frac{1}{4}$.Therefore,the new deviation is $\delta_w = \frac{\delta_a}{4}$.

If a thin prism of glass is dipped into water,then the minimum deviation (with respect to air) of light produced by the prism will be: $\left( {_a{\mu _g} = \frac{3}{2} \text{ and } _a{\mu _w} = \frac{4}{3}} \right)$

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