Question

If the angles of minimum deviation when a prism of angle \(80^\circ\) is placed separately in air and in a liquid are \(40^\circ\) and \(10^\circ\) respectively, then the refractive index of the liquid is

• A. \(\sqrt3\)
• B. \(\sqrt{\dfrac32}\)
• C. \(\sqrt{\dfrac23}\)
• D. \(\sqrt2\)

Hint:

For a prism at minimum deviation, \[ \boxed{ \mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)} {\sin\left(\frac{A}{2}\right)}. } \] When immersed in a liquid, \[ \boxed{ \mu_{\text{relative}} = \frac{\mu_{\text{prism}}}{\mu_{\text{liquid}}}. } \]

Answer 1

The Correct Option is B

Step 1: Find the refractive index of the prism in air. For minimum deviation, \[ \mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)} {\sin\left(\frac{A}{2}\right)}. \] Here, \[ A=80^\circ,\qquad \delta_m=40^\circ. \] Thus, \[ \mu_p = \frac{\sin60^\circ}{\sin40^\circ}. \]

Step 2: Find the relative refractive index in the liquid. When the prism is immersed in the liquid, \[ \mu_{pl} = \frac{\sin\left(\frac{80^\circ+10^\circ}{2}\right)} {\sin40^\circ} = \frac{\sin45^\circ}{\sin40^\circ}. \] Also, \[ \mu_{pl} = \frac{\mu_p}{\mu_l}, \] where \(\mu_l\) is the refractive index of the liquid. Hence, \[ \mu_l = \frac{\mu_p}{\mu_{pl}} = \frac{\sin60^\circ}{\sin45^\circ} = \frac{\frac{\sqrt3}{2}}{\frac1{\sqrt2}} = \sqrt{\frac32}. \]

Step 3: Write the answer. Therefore, \[ \boxed{\sqrt{\frac32}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.