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A glass prism of small angle and of refractive index 1.5 produces a deviation of $D_{1}$ . Now it is immersed in water of refractive index 4/3, the deviation produced is $D_{2}$ . The ratio $D_{2}$ : $D_{1}$ is

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1 : 2

8 : 9

1 : 4

9 : 8

answer is D.

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Detailed Solution

d = A( $μ$ -1)

$\frac{D_{1}}{D_{2}} = \frac{(μ_{g} - 1)}{(\frac{μ_{g}}{μ_{l}} - 1)} = \frac{0.5}{(\frac{9}{8} - 1)} = \frac{0.5}{\frac{1}{8}} = \frac{4}{1}$

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