Question 5

Core Geometrical Constraint: Faces $a_1b_1$ and $a_2b_2$ are parallel to each other and perpendicular to the horizontal mirror $M$. When a ray emerges from Prism 1 with an angle of emergence $e_1$, it reflects off the mirror $M$ and strikes Prism 2. Due to the symmetry of reflection and parallel vertical faces, the angle of incidence on Prism 2 will always be equal to the angle of emergence from Prism 1: $e_1 = i_2$ (This is always true).

1. Checking Option (D): If Prism 1 is at minimum deviation, then $i_1 = e_1$ and by Snell's law, $\sin(e_1) = n_1 \sin(A_1/2)$. Since $e_1 = i_2$ (from our core constraint), we can substitute $e_1$ to get: $\sin(i_2) = n_1 \sin(A_1/2)$. Therefore, Option (D) is always true.

2. Checking Option (B): If Prism 2 is at minimum deviation, then $i_2 = e_2$ and $\sin(i_2) = n_2 \sin(A_2/2)$. Since $e_1 = i_2$, we can write $\sin(e_1) = n_2 \sin(A_2/2)$. However, because Prism 1 is not necessarily at minimum deviation, $i_1 \neq e_1$. Thus, $\sin(i_1) = n_2 \sin(A_2/2)$ is NOT always true. Option (B) is incorrect.

3. Checking Option (A): If both prisms are at minimum deviation: For Prism 1: $i_1 = e_1$ and $\sin(e_1) = n_1 \sin(A_1/2)$ For Prism 2: $i_2 = e_2$ and $\sin(i_2) = n_2 \sin(A_2/2)$ Since $e_1 = i_2$, we can equate the two expressions: $n_1 \sin(A_1/2) = n_2 \sin(A_2/2) \implies \frac{n_2}{n_1} = \frac{\sin(A_1/2)}{\sin(A_2/2)}$ Therefore, Option (A) is correct.

4. Checking Option (C): For thin prisms at minimum deviation, the deviation angle is $\delta_m = (n-1)A$. This gives $A_1 = \frac{\delta_{m1}}{n_1-1}$ and $A_2 = \frac{\delta_{m2}}{n_2-1}$. From the geometry of the dashed lines (altitudes/bisectors of the isosceles triangles), the angle $\theta$ between the prisms is: $\theta = \frac{A_1}{2} + \frac{A_2}{2}$ Substituting the values of $A_1$ and $A_2$: $\theta = \frac{\delta_{m1}}{2(n_1-1)} + \frac{\delta_{m2}}{2(n_2-1)}$ Therefore, Option (C) is correct.

Correct Options: (A), (C), (D)