Geometrical Optics

If both the prisms are at minimum deviation condition, then $\dfrac{n_2}{n_1} = \dfrac{\sin\left(\dfrac{A_1}{2}\right)}{\sin\left(\dfrac{A_2}{2}\right)}$.

If prism 2 is at minimum deviation condition, then $\sin i_1 = n_2 \sin \left(\dfrac{A_2}{2}\right)$ is always true.

If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation $\delta_{m1}$ and $\delta_{m2}$, respectively, then $\theta = \dfrac{\delta_{m1}}{2(n_1 - 1)} + \dfrac{\delta_{m2}}{2(n_2 - 1)}$.

If prism 1 is at minimum deviation condition, then $\sin i_2 = n_1 \sin \left(\dfrac{A_1}{2}\right)$ is always true.