Question

Consider two isosceles prisms 1 and 2 with prism angles $A_1$ and $A_2$ and refractive indices $n_1$ and $n_2$, respectively, as shown in the figure. The faces $a_1b_1$ and $a_2b_2$ are parallel to each other and perpendicular to the mirror $M$. If a ray of light is incident on the face $a_1c_1$ and emerges from the face $a_2c_2$, then the correct statement(s) is/are:

• A. If both the prisms are at minimum deviation condition, then $\frac{n_2}{n_1} = \sin \left( \frac{A_1}{2} \right) / \sin \left( \frac{A_2}{2} \right)$.
• B. If prism 2 is at minimum deviation condition, then $\sin i_1 = n_2 \sin \left( \frac{A_2}{2} \right)$ is always true.
• C. 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 $\theta = \frac{\delta_{m1}}{2(n_1-1)} + \frac{\delta_{m2}}{2(n_2-1)}$.
• D. If prism 1 is at minimum deviation condition, then $\sin i_2 = n_1 \sin \left( \frac{A_1}{2} \right)$ is always true.

Hint:

In optics problems with mirrors and multiple prisms, track the angle of the ray relative to the normals of parallel faces. Symmetry often reduces complex trigonometric equations to simple equalities.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The setup involves two prisms separated by a mirror. Because faces $a_1b_1$ and $a_2b_2$ are parallel and perpendicular to the mirror, the geometry of reflection implies that the angle of emergence from prism 1 ($e_1$) is equal to the angle of incidence on prism 2 ($i_2$).
Step 2: Key Formula or Approach:

Snell's Law at prism interfaces.

Minimum deviation condition: $r = A/2$ and $i = e$.

Relation from reflection: $i_2 = e_1$ (due to symmetry of parallel faces and perpendicular mirror).

Step 3: Detailed Explanation:

Checking (A) and (D): If prism 1 is at minimum deviation, then $\sin i_1 = \sin e_1 = n_1 \sin(A_1/2)$. Since $i_2 = e_1$, we have $\sin i_2 = n_1 \sin(A_1/2)$. Thus (D) is correct.
If prism 2 is also at minimum deviation, $\sin i_2 = n_2 \sin(A_2/2)$.
Equating both: $n_1 \sin(A_1/2) = n_2 \sin(A_2/2) \implies \frac{n_2}{n_1} = \sin(A_1/2) / \sin(A_2/2)$. Thus (A) is correct.

Checking (C): For thin prisms, $\delta_m = (n-1)A \implies A = \frac{\delta_m}{n-1}$. The angle $\theta$ in the figure is the angle between the two normals or the combined wedge angle. From geometry, $\theta = \frac{A_1}{2} + \frac{A_2}{2}$.
Substituting $A_1$ and $A_2$: $\theta = \frac{\delta_{m1}}{2(n_1-1)} + \frac{\delta_{m2}}{2(n_2-1)}$. Thus (C) is correct.

Step 4: Final Answer:
The correct statements are (A), (C), and (D).