Question

The angle of prism is equal to its minimum deviation angle (\( \delta_m \)). In minimum deviation condition, calculate the following: (i) The angle of incidence
(ii) The angle of refraction
(iii) The refractive index of the material of the prism.

Hint:

In minimum deviation condition, the refracted ray is symmetric inside the prism, and refractive index can be calculated using the formula \( n = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \).

Answer 1

i. The angle of incidence Step 1: At minimum deviation condition, the angle of incidence \( i \) is related to the prism angle \( A \) and minimum deviation angle \( \delta_m \) by: \[ i = A + \frac{\delta_m}{2} \] Step 2: Given \( A = \delta_m \), we substitute: \[ i = A + \frac{A}{2} = \frac{3A}{2} \] \[ \boxed{i = \frac{3A}{2}} \] ii. The angle of refraction Step 1: At minimum deviation, the refracted ray passes symmetrically through the prism, meaning: \[ r = \frac{A}{2} \] \[ \boxed{r = \frac{A}{2}} \] iii. The refractive index of the material of the prism Step 1: The refractive index \( n \) of the prism material is given by: \[ n = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] Step 2: Since \( A = \delta_m \): \[ n = \frac{\sin \left( \frac{A + A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] \[ n = \frac{\sin A}{\sin \left( \frac{A}{2} \right)} \] \[ \boxed{n = \frac{\sin A}{\sin \left( \frac{A}{2} \right)}} \]