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.

Velocity of light in vacuum, \(c = 3 \times 10^8\) m/s

Velocity of light in the prism, \(v = 2.12 \times 10^8\) m/s

Refractive index of the prism material, \(\mu = \dfrac{c}{v} = \dfrac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 \approx \sqrt{2}\)

For an equilateral prism, the angle of the prism \(A = 60^\circ\).

Using the prism formula:

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

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

\(\sqrt{2} = \dfrac{\sin\left(30^\circ + \dfrac{\delta_m}{2}\right)}{\dfrac{1}{2}}\)

\(\sin\left(30^\circ + \dfrac{\delta_m}{2}\right) = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}\)

\(30^\circ + \dfrac{\delta_m}{2} = 45^\circ\)

\(\dfrac{\delta_m}{2} = 15^\circ\)

\(\delta_m = 30^\circ\)

Answer: \(30\)