A ray of light from air passes through an equilateral triangular glass prism and undergoes minimum deviation when the angle of incidence is $\frac{3}{4}$ of the angle of the prism. The speed of light in prism is

$2.12 × 10^8 m/s$

The correct answer is Option (2) → $2.12 × 10^8 m/s$

Let the angle of the prism be $A$ and the angle of incidence at minimum deviation be $i$. Given $i = \frac{3}{4}A$.

At minimum deviation, the angle of incidence $i$ equals the angle of emergence $e$: $i = e$. Also, the angle of deviation $D_{\text{min}} = 2i - A$.

Given the prism is equilateral, so $A = 60^\circ$.

$i = \frac{3}{4} \cdot 60^\circ = 45^\circ$.

$D_{\text{min}} = 2i - A = 2(45^\circ) - 60^\circ = 90^\circ - 60^\circ = 30^\circ$.

Using the prism formula: $n = \frac{\sin\frac{A + D_{\text{min}}}{2}}{\sin\frac{A}{2}}$

$n = \frac{\sin\frac{60^\circ + 30^\circ}{2}}{\sin 30^\circ} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}$

Speed of light in prism: $v = \frac{c}{n} = \frac{3 \times 10^8}{\sqrt{2}} \approx 2.12 \times 10^8\ \text{m/s}$

Final Answer: $v \approx 2.12 \times 10^8 m/s$