Question

Find the angle of minimum deviation for an equilateral prism made of refractive index \( 1.732 \). What is the angle of incidence for the deviation?

Hint:

For the minimum deviation in a prism, the angle of incidence is equal to the angle of emergence, and the total deviation is minimized.

Answer 1

Step 1: Formula for minimum deviation.
The angle of minimum deviation \( D_{\text{min}} \) for a prism is given by: \[ D_{\text{min}} = 2i - A \] where \( i \) is the angle of incidence, and \( A \) is the angle of the prism.
Step 2: Formula for the refractive index.
The refractive index \( n \) of the material of the prism is related to the minimum deviation by the equation: \[ n = \frac{\sin \left( \frac{A + D_{\text{min}}}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] Substitute \( n = 1.732 \) and \( A = 60^\circ \): \[ 1.732 = \frac{\sin \left( \frac{60^\circ + D_{\text{min}}}{2} \right)}{\sin \left( \frac{60^\circ}{2} \right)} \] Using the known values: \[ \sin \left( 30^\circ \right) = 0.5 \] Solving for \( D_{\text{min}} \), we get: \[ D_{\text{min}} = 30^\circ \]
Step 3: Conclusion.
The angle of minimum deviation is \( 30^\circ \). The angle of incidence for the deviation is \( 30^\circ \).