Minimum Deviation in Prism

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

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 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.