A ray of light passing through an equilateral prism is having velocity $$2.12 \times 10^8$$ m/s in the prism material, then the minimum angle of deviation is _______ degrees.

1. Given parameters:

Angle of an equilateral prism: $$A = 60^\circ$$
Speed of light in the prism: $$v = 2.12 \times 10^8 \text{ m/s}$$
Speed of light in vacuum: $$c = 3 \times 10^8 \text{ m/s}$$

2. Refractive index calculation:

The refractive index $$\mu$$ of the prism is given by:

$$$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 \approx \sqrt{2}$$$

3. Minimum angle of deviation calculation:

Using the prism formula:

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

Substitute $$A = 60^\circ$$ and $$\mu = \sqrt{2}$$ into the formula:

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

Since $$\sin(30^\circ) = \frac{1}{2}$$:

$$$\sqrt{2} \cdot \frac{1}{2} = \sin\left(\frac{60^\circ + \delta_m}{2}\right)$$$

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

Since $$\sin(45^\circ) = \frac{1}{\sqrt{2}}$$:

$$$\frac{60^\circ + \delta_m}{2} = 45^\circ$$$

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

$$$\delta_m = 30^\circ$$$

The minimum angle of deviation is 30 degrees.