Refractive Index of Equilateral Prism with Given Minimum Deviation

To find the refractive index of the prism, we use the prism formula:

$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

where:

• $\mu$ is the refractive index of the prism,
• $A$ is the angle of the prism,
• $\delta_m$ is the angle of minimum deviation.

For an equilateral prism, the angle of the prism is: $A = 60^\circ$

According to the given problem, the angle of minimum deviation is equal to half of the angle of the prism: $\delta_m = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ$

Substitute the values of $A$ and $\delta_m$ into the prism formula:

$\mu = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$

Using the trigonometric values $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\sin(30^\circ) = \frac{1}{2}$:

$\mu = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Thus, the refractive index of the prism is $\sqrt{2}$.

Correct Option: C ($\sqrt{2}$)