Question

When an equilateral prism is immersed in a liquid A of refractive index $\frac{4}{3}$ the angle of minimum deviation is equal to the angle of the prism. If the same prism is immersed in another liquid B, the angle of minimum deviation is 50% of the angle of the prism, then the refractive index of the liquid B is}

• A. $\frac{5}{3}$
• B. $\sqrt{\frac{5}{3}}$
• C. $\frac{8}{3}$
• D. $\frac{8}{3}$

Hint:

For minimum deviation: \[ \mu= \frac{\sin\left(\frac{A+\delta_m}{2}\right)} {\sin\left(\frac{A}{2}\right)} \] This formula is extremely important for prism-based numerical problems.

Answer 1

The Correct Option is B

Concept: For a prism immersed in a medium, \[ \mu_r= \frac{\sin\left(\frac{A+\delta_m}{2}\right)} {\sin\left(\frac{A}{2}\right)} \] where \(\mu_r\) is the relative refractive index.

Step 1: Use the first condition.
For an equilateral prism, \[ A=60^\circ \] Given, \[ \delta_m=A=60^\circ \] Hence, \[ \mu_r= \frac{\sin60^\circ}{\sin30^\circ} \] \[ =\sqrt3 \] \[ \frac{\mu_p}{4/3}=\sqrt3 \] \[ \mu_p=\frac{4\sqrt3}{3} \]

Step 2: Use the second condition.
Now, \[ \delta_m=\frac{A}{2}=30^\circ \] \[ \mu_r= \frac{\sin45^\circ}{\sin30^\circ} \] \[ =\sqrt2 \] \[ \frac{\mu_p}{\mu_B}=\sqrt2 \] \[ \mu_B= \frac{\frac{4\sqrt3}{3}}{\sqrt2} \] \[ =\sqrt{\frac53} \] \[ \boxed{\sqrt{\frac53}} \]