Question

The minimum deviation angle obtained by a prism of refracting angle $60^\circ$ is $30^\circ$. Calculate the refractive index of the material of prism.

Hint:

At the position of minimum deviation, the refracted ray inside the prism stays parallel to the base of the prism (if it is an isosceles or equilateral prism).

Answer 1

Step 1: Identifying Given Data.
From the problem, we have the following parameters:

Angle of the prism (Refracting angle), $A = 60^\circ$
Angle of minimum deviation, $\delta_m = 30^\circ$

Step 2: Applying the Prism Formula.
The refractive index ($\mu$) of the material of a prism is given by the prism formula: $$\mu = \frac{\sin\left( \frac{A + \delta_m}{2} \right)}{\sin\left( \frac{A}{2} \right)}$$
Step 3: Calculation.
Substituting the given values into the formula: $$\mu = \frac{\sin\left( \frac{60^\circ + 30^\circ}{2} \right)}{\sin\left( \frac{60^\circ}{2} \right)}$$ $$\mu = \frac{\sin(45^\circ)}{\sin(30^\circ)}$$ Using trigonometric values ($\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\sin 30^\circ = \frac{1}{2}$): $$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$$ The refractive index of the material of the prism is approximately $1.414$.