Consider a prism with an angle \( A \), and the ray of light is incident on face AB and refracted through face AC. In the case of minimum deviation, the ray inside the prism travels symmetrically. At minimum deviation, the angle of incidence \( i \) and the angle of refraction \( r \) are equal (i.e., \( i = r \)).
The angle of deviation \( \delta \) is the angle between the incident ray and the emergent ray. The relationship between the angle of prism \( A \), the angle of deviation \( \delta \), and the angle of incidence \( i \) is given by the equation:
\[
\delta = 2i - A
\]
or equivalently,
\[
i = \frac{A + \delta}{2}
\]
Thus, the angle of incidence \( i \) is determined by the above equation.
Next, using Snell's law at the first interface (face AB) of the prism:
\[
n_1 \sin i = n_2 \sin r
\]
where:
- \( n_1 \) is the refractive index of air (approximately \( 1 \)),
- \( n_2 \) is the refractive index of the prism material.
At the minimum deviation position, the angle of incidence and angle of refraction are equal, i.e., \( i = r \). So, we can write:
\[
\sin i = \sin r
\]
This implies that the refractive index of the prism is related to the angle of prism \( A \) and the angle of deviation \( \delta \) by the formula:
\[
n = \frac{\sin \left( \frac{A + \delta}{2} \right)}{\sin \frac{A}{2}}
\]
Final Answer:
1. The angle of incidence \( i = \frac{A + \delta}{2} \).
2. The angle of refraction on face AB is \( r = i \), as \( i = r \) at minimum deviation.