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