Geometrical Optics

Concept:The geometry of the parallel vertical faces and the mirror forces the angle of emergence from prism 1 to equal the angle of incidence on prism 2.Explanation:At minimum deviation for a prism, $r = \frac{A}{2}$ and Snell’s law gives $\sin i = n \sin\frac{A}{2}$.Faces $a_1b_1$ and $a_2b_2$ are vertical and parallel, and perpendicular to the horizontal mirror $M$.The ray emerges from prism 1 at angle $e_1$ (with horizontal), reflects off $M$ (angle unchanged), and enters prism 2 at the same angle, so $i_2 = e_1$.If both prisms are at minimum deviation: $\sin e_1 = n_1\sin\frac{A_1}{2}$ and $\sin i_2 = n_2\sin\frac{A_2}{2}$.Since $\sin i_2 = \sin e_1$, we get $n_1\sin\frac{A_1}{2} = n_2\sin\frac{A_2}{2}$, i.e. $\frac{n_2}{n_1} = \frac{\sin(A_1/2)}{\sin(A_2/2)}$. Hence option A is correct.If only prism 2 is at minimum deviation, $\sin i_2 = n_2\sin\frac{A_2}{2}$ but $i_2 = e_1$ depends on prism 1; $\sin i_1$ does not equal that expression generally. Option B is false.For thin prisms ($A \ll 1$), $\delta_m = (n-1)A$ so $A = \frac{\delta_m}{n-1}$. The angle $\theta$ between outer faces is $\frac{A_1}{2} + \frac{A_2}{2}$. Substituting gives $\theta = \frac{\delta_{m1}}{2(n_1-1)} + \frac{\delta_{m2}}{2(n_2-1)}$. Option C is correct.If only prism 1 is at minimum deviation, $\sin e_1 = n_1\sin\frac{A_1}{2}$ and $i_2 = e_1$, so $\sin i_2 = n_1\sin\frac{A_1}{2}$ holds always. Option D is correct.Answer:Options A, C, D are correct.