Question

The angle of minimum deviation for a ${75}^0$ prism of dense glass is found to be ${45}^0$ when in air and ${15}^0$ when immersed in certain liquid. The refractive index of the liquid is

• A. $\sqrt{\frac{2}{3}}$
• B. $\frac{3}{2}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{3}$

Answer 1

The correct option is C $\sqrt{\frac{3}{2}}$

Refractive index of prism when the prism is in aix $a{\mu }_g=\frac{\mu q}{{\mu }_a}=\frac{\sin \left(\frac{A+B_m}{2}\right)}{\sin A/2}$-{1} Refractive index of prism $l{\mu }_g=\frac{\mu g}{u_1}=\frac{\sin \left(\frac{A+D_m^{\prime }}{2}\right)}{\sin \frac{A}{2}}$-{2}

Divide (1) and (2)

$\Rightarrow \frac{{\mu }_l}{{\mu }_a}=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A+D_m^{\prime }}{2}\right)}$

$A=75,D_m={45}^{\circ },0^{\prime }m={15}^{\circ },{\mu }_a=1$

Put all. these values.

$\begin{array}{cc}{\mu }_l& =\frac{\sin \left(\frac{75+45}{2}\right)}{\sin \left(\frac{75+15}{2}\right)}\\ \Rightarrow & {\mu }_e=\frac{\sin 60}{\sin 45}=\frac{\sqrt{3}/2}{1/\sqrt{2}}\\ \Rightarrow & {\mu }_l=\sqrt{\frac{3}{2}}\\ \text{Hence,}& \left(c\right)\text{option is correct.}\end{array}$