The angle of deviation through a triangular prism is defined as the angle between the incident ray and the emerging ray (angle $\delta$ ). It can be shown that when the angle of incidence $i$ is equal to the angle of refraction $r^{\prime}$ for the emerging ray, the angle of deviation is at a minimum. Show that the minimum deviation angle ( $\delta_{\min }=D$ ) is related to the prism angle $A$ and the index of refraction $n$, by
$$
n=\frac{\sin \frac{1}{2}(A+D)}{\sin \frac{1}{2} A}
$$
[Hint: For an isosceles triangular prism, the minimum angle of deviation occurs when the ray inside the prism is parallel to the base, as shown in the figure.]