Question

A thin glass prism of n = 1.5 is immersed in water of n = 1.33. The ratio of deviation of the ray water to that in air for the same prism is:

• A. 1 : 2
• B. 1 : 4
• C. 1 : 3
• D. 1 : 8

Answer 1

Answer: (B)

Angle of deviation ${\delta }_{air}$ is given by when the prism is in air ${\delta }_{air}=(n-1)A=\left(\frac{3}{2}-1\right)A=\frac{A}{2}$ Again when the prism is immered in water then angle of deviation ${\delta }_{water}$ ${\delta }_{water}=({}_g{n}_w-1)A$ Here ${}_g{n}_w$ is the refractive index of glass with respect to water so, ${}_g{n}_w=\frac{n_g}{n_w}=\frac{3\text{/}2}{4\text{/}3}=\frac{9}{8}$ So, ${\delta }_{water}=\left(\frac{9}{8}-1\right)A=\frac{1}{8}A$ Hence, required ratio is $\frac{\frac{1}{8}A}{\frac{1}{2}A}=1:4.$