(N/A) Consider a prism $ABC$ with prism angle $A$. $A$ monochromatic light ray $PQ$ is incident on the face $AB$ at an angle of incidence $i$. It refracts along $QR$ with an angle of refraction $r_1$. At face $AC$,it is incident at an angle $r_2$ and emerges as $RS$ at an angle of emergence $e$.
$1$. In the quadrilateral $AQNR$ (where $N$ is the intersection of normals at $Q$ and $R$),the sum of opposite angles is $180^{\circ}$:
$\angle A + \angle QNR = 180^{\circ}$ --- $(1)$
$2$. In $\triangle QNR$,the sum of angles is $180^{\circ}$:
$r_1 + r_2 + \angle QNR = 180^{\circ}$ --- $(2)$
Comparing $(1)$ and $(2)$,we get:
$A = r_1 + r_2$ --- $(3)$
$3$. The total deviation $\delta$ is the angle between the incident ray $PQ$ extended and the emergent ray $RS$ extended. In $\triangle DQR$ (where $D$ is the intersection of extended rays):
$\delta = (i - r_1) + (e - r_2)$
$\delta = (i + e) - (r_1 + r_2)$
Substituting $(3)$ into this equation:
$\delta = i + e - A$
Thus,the final relation is:
$i + e = A + \delta$