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The ratio of angle of minimum deviation of a thin prism in air and when dipped in water will be $({}_{a}μ_{g} = \frac{3}{2} and {}_{a}μ_{w} = \frac{4}{3})$

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1/8

1/4

3/4

4/1

answer is D.

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Detailed Solution

For a thin prism $δ = (μ - 1) A$

$\frac{δ_{a}}{δ_{w}} = \frac{(μ_{g} - 1) A}{(\frac{μ_{g}}{μ_{w}} - 1) A} = \frac{\frac{3}{2} - 1}{\frac{3 / 2}{4 / 3} - 1} \frac{δ_{a}}{δ_{w}} = \frac{4}{1}$

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