Question

A light ray going through a prism with the angle of prism 60°, is found to have minimum deviation of 30°. What is the refractive index of the prism material?

Hint:

The refractive index of a prism material can be found using the angle of the prism and the angle of minimum deviation in the formula.

Answer 1

Step 1: Formula for Refractive Index in Terms of Angle of Minimum Deviation.
The refractive index \( n \) of a prism can be found using the formula: \[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \] Where:
- \( A = 60^\circ \) is the angle of the prism,
- \( D = 30^\circ \) is the angle of minimum deviation.
Step 2: Substituting the Given Values.
Substituting the values of \( A \) and \( D \) into the formula: \[ n = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(45^\cir}{\sin(30^\cir} \]
Step 3: Calculation.
Using known values: \[ \sin(45^\cir = \frac{\sqrt{2}}{2}, \quad \sin(30^\cir = \frac{1}{2} \] \[ n = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2} \approx 1.414 \]
Final Answer:
The refractive index of the prism material is \( \boxed{1.414} \).