Animated Solution for Physics - Optics: 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 $a_1{b}_1$ and $a_2{b}_2$ are parallel to each other and perpendicular to the mirror $M$. If a ray of light is incident on the face $a_1{c}_1$ and emerges from the face $a_2{c}_2$, then the correct statement(s) is/are:

Imagine a beam of light embarking on a complex journey: entering a prism, refracting, bouncing off a mirror, and then navigating through a second prism. This problem is a beautiful symphony of optics and geometry. Let's break it down step-by-step and see why the math behaves the way it does.

The most crucial observation in this entire setup is purely geometric. Look at the inner faces of the two prisms, $a_1{b}_1$ and $a_2{b}_2$. The problem states they are parallel to each other and perpendicular to the mirror $M$. This means these faces are perfectly vertical, and the mirror is perfectly horizontal.

When the light ray exits the first prism, it makes an angle of emergence $e_1$ with the normal to the face $a_1{b}_1$. Because the face is vertical, this normal is horizontal. The ray then travels down, hits the horizontal mirror, and reflects. By the law of reflection, the angle it makes with the horizontal is preserved. Therefore, when it strikes the vertical face $a_2{b}_2$ of the second prism, the angle of incidence $i_2$ (measured from the horizontal normal) must be exactly equal to $e_1$.

This gives us our unbreakable master equation: $e_1=i_2$.

Now, let's evaluate the options based on the condition of minimum deviation. Recall that at minimum deviation, a ray passes symmetrically through a prism, meaning the angle of incidence equals the angle of emergence ($i=e$).

Evaluating Option A: Suppose both prisms are at minimum deviation. For Prism 1, applying Snell's law at the second interface gives us $\sin e_1=n_1\sin \left(\frac{A_1}{2}\right)$. For Prism 2, applying Snell's law at the first interface gives us $\sin i_2=n_2\sin \left(\frac{A_2}{2}\right)$. Since we know $e_1=i_2$, we can equate the two right-hand sides: Rearranging this beautifully yields $\frac{n_2}{n_1}=\frac{\sin (A_1/2)}{\sin (A_2/2)}$. Thus, Option A is correct.

Evaluating Option B: What if only Prism 2 is at minimum deviation? We still have $\sin i_2=n_2\sin \left(\frac{A_2}{2}\right)$, and our geometric rule $e_1=i_2$ still holds. However, because Prism 1 is not necessarily at minimum deviation, its angle of incidence $i_1$ is not guaranteed to equal $e_1$. Therefore, we cannot substitute $i_1$ into the equation. Option B is a trap and is incorrect.

Evaluating Option D: If only Prism 1 is at minimum deviation, we have $\sin e_1=n_1\sin \left(\frac{A_1}{2}\right)$. Because $e_1=i_2$ is always true, we can simply swap $e_1$ for $i_2$ to get $\sin i_2=n_1\sin \left(\frac{A_1}{2}\right)$. This makes Option D perfectly correct.

Finally, let's look at Option C, which deals with thin prisms. For thin prisms at minimum deviation, the geometry simplifies. By tracing the angles of the extended incident and emergent rays relative to the horizontal, it can be shown that the total angle $\theta$ between them is the sum of half of each prism's angle:

We also know the standard formula for the deviation of a thin prism is ${\delta }_m=(n-1)A$. By rearranging this to solve for the prism angle, we get $A=\frac{{\delta }_m}{n-1}$.

Substituting this expression for both $A_1$ and $A_2$ into our $\theta$ equation gives: This exactly matches the expression in Option C, confirming it is correct.

In conclusion, by carefully combining geometric constraints with the laws of refraction, we successfully navigated this complex optical system!