Question

A ray of light is incident on face AB of a prism ABC with angle of prism \( A \) and emerges out from face AC. The prism is set in the position of minimum deviation with angle of deviation \( \delta \). Find: \begin{enumerate} \item the angle of incidence and \item the angle of refraction on face AB. \end{enumerate}

Hint:

At the minimum deviation position, the incident and refracted angles are equal in a prism. The refractive index of the prism can be found using the relation involving the prism angle \( A \) and the deviation angle \( \delta \).

Answer 1

The deviation angle \( \delta \) for a prism with prism angle \( A \) at minimum deviation, where the internal ray is symmetrical, is related to the angle of incidence \( i \) by \( \delta = 2i - A \). This can be rearranged to \( i = \frac{A + \delta}{2} \). Using Snell's law, \( n_1 \sin i = n_2 \sin r \), where \( n_1 \) is the refractive index of the surrounding medium (air, \( \approx 1 \)) and \( n_2 \) is the refractive index of the prism material. At minimum deviation, \( i = r \). Therefore, the refractive index \( n \) of the prism material is given by \( n = \frac{\sin \left( \frac{A + \delta}{2} \right)}{\sin \frac{A}{2}} \). Final Answer: 1. The angle of incidence \( i \) is given by \( i = \frac{A + \delta}{2} \). 2. The angle of refraction on face AB, denoted as \( r \), is equal to the angle of incidence \( i \) at minimum deviation.