Question

The ratio of angle of minimum deviation produced by a thin prism $(\mu =\frac{3}{2})$ in air to that in liquid of refractive index $\frac{9}{7}$ is :

• A. $\frac{1}{3}$
• B. $3$
• C. $\frac{1}{4}$
• D. $4$

Answer 1

The correct option is B $3$

Given,

${}_a{\mu }_g=\frac{3}{2}$

${}_a{\mu }_l=\frac{9}{7}$

We know $\delta =\left(\mu -1\right)A$

For air, ${\delta }_a=\left(\frac{3}{2}-1\right)A=\frac{A}{2}$

${}_l{\mu }_g=\frac{{}_a{\mu }_g}{{}_a{\mu }_l}=\frac{3/2}{9/7}=\frac{7}{6}$

For liquid, ${\delta }_l=\left(\frac{7}{6}-1\right)A=\frac{A}{6}$

Hence, $\frac{{\delta }_a}{{\delta }_l}=\frac{A/2}{A/6}=3$

Therefore the correct answer is option (B).