Derive the expression $\delta = i + e - A$ for a prism.

(N/A) Consider a prism $ABC$ with angle $A$. A monochromatic light ray $PQ$ is incident on the face $AB$ at an angle of incidence $i$.
At point $Q$, the ray refracts and travels along $QR$ inside the prism, making angles of refraction $r_1$ and $r_2$ at faces $AB$ and $AC$ respectively.
The ray emerges from face $AC$ at point $R$ at an angle of emergence $e$.
Let the incident ray $PQ$ be extended forward and the emergent ray $RS$ be extended backward to meet at point $M$. The angle between these two rays is the angle of deviation $\delta$.
In $\triangle MQR$, the exterior angle $\delta = \angle MQR + \angle MRQ$.
Since $\angle MQR = i - r_1$ and $\angle MRQ = e - r_2$, we have $\delta = (i - r_1) + (e - r_2) = (i + e) - (r_1 + r_2)$.
In the quadrilateral $AQNR$ (where $N$ is the intersection of normals at $Q$ and $R$), $\angle A + \angle QNR = 180^{\circ}$. Also, in $\triangle QNR$, $r_1 + r_2 + \angle QNR = 180^{\circ}$.
Comparing these, we get $A = r_1 + r_2$.
Substituting this into the deviation equation, we get $\delta = i + e - A$.