Minimum Angle of Deviation in an Equilateral Prism

To find the minimum angle of deviation for a ray of light passing through an equilateral prism, we follow these steps:

1. Refractive Index of the Prism Material ($\mu$): The angle of an equilateral prism is $A = 60^\circ$. The speed of light in vacuum is $c = 3 \times 10^8\text{ m/s}$, and the velocity of light in the prism material is $v = 2.12 \times 10^8\text{ m/s}$.

The refractive index $\mu$ of the material is given by: $\mu = \frac{c}{v} = \frac{3 \times 10^8\text{ m/s}}{2.12 \times 10^8\text{ m/s}} \approx 1.415 \approx \sqrt{2}$

2. Prism Formula for Minimum Deviation ($\delta_m$): The refractive index of a prism is related to the prism angle $A$ and the minimum angle of deviation $\delta_m$ by the formula: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

Substituting $A = 60^\circ$ and $\mu = \sqrt{2}$: $\sqrt{2} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}$

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

Since $\sin(30^\circ) = \frac{1}{2}$: $\sin\left(\frac{60^\circ + \delta_m}{2}\right) = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$

Taking the inverse sine on both sides: $\frac{60^\circ + \delta_m}{2} = 45^\circ$

$60^\circ + \delta_m = 90^\circ$

$\delta_m = 90^\circ - 60^\circ = 30^\circ$

Thus, the minimum angle of deviation is $30^\circ$.

Correct Option: B