Refraction Through Two Isosceles Prisms and Plane Mirror

To analyze the given optical setup, we first use geometric optics to find the relationship between the angles of light propagation through the two prisms and the plane mirror $M$.

1. Geometric Relation Between Prisms 1 and 2

• The faces $a_1b_1$ and $a_2b_2$ are given to be parallel to each other and perpendicular to the plane mirror $M$.
• Taking the plane of mirror $M$ to be horizontal, the normals to the faces $a_1b_1$ and $a_2b_2$ are both horizontal.
• Let a ray of light emerge from face $a_1b_1$ of Prism 1 at an angle of emergence $e_1$ with respect to the normal of face $a_1b_1$. The ray thus makes an angle $e_1$ with the horizontal.
• The ray reflects off the horizontal plane mirror $M$. By the law of reflection, the angle of the reflected ray with the horizontal remains $e_1$.
• Since the normal to face $a_2b_2$ of Prism 2 is also horizontal, the angle of incidence $i_2$ at face $a_2b_2$ must be equal to the angle of emergence $e_1$ from face $a_1b_1$: $i_2 = e_1$

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2. Analysis of Option A

For Prism 1 at minimum deviation condition:

• The angle of refraction at the internal surface is $r_1 = r_1' = \frac{A_1}{2}$.
• Applying Snell's law at the emerging face $a_1b_1$: $\sin e_1 = n_1 \sin\left(\frac{A_1}{2}\right)$

For Prism 2 at minimum deviation condition:

• The angle of refraction at the internal surface is $r_2 = r_2' = \frac{A_2}{2}$.
• Applying Snell's law at the incident face $a_2b_2$: $\sin i_2 = n_2 \sin\left(\frac{A_2}{2}\right)$

Since $i_2 = e_1$, we have $\sin i_2 = \sin e_1$, which yields: $n_1 \sin\left(\frac{A_1}{2}\right) = n_2 \sin\left(\frac{A_2}{2}\right)$ $\implies \frac{n_2}{n_1} = \frac{\sin\left(\frac{A_1}{2}\right)}{\sin\left(\frac{A_2}{2}\right)}$

Thus, Option A is correct.

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3. Analysis of Option D

If Prism 1 is at minimum deviation condition, we have: $\sin e_1 = n_1 \sin\left(\frac{A_1}{2}\right)$

Using the relation $i_2 = e_1$, it directly follows that: $\sin i_2 = n_1 \sin\left(\frac{A_1}{2}\right)$

This relation holds universally whenever Prism 1 is at minimum deviation, regardless of the parameters or state of Prism 2.

Thus, Option D is correct.

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4. Analysis of Option B

If Prism 2 is at minimum deviation condition, we have: $\sin i_2 = n_2 \sin\left(\frac{A_2}{2}\right) \implies \sin e_1 = n_2 \sin\left(\frac{A_2}{2}\right)$

For $\sin i_1 = n_2 \sin\left(\frac{A_2}{2}\right)$ to hold, we would require $i_1 = e_1$, which is only true if Prism 1 is also at minimum deviation condition. Since Prism 1 is not specified to be at minimum deviation, this statement is not always true.

Thus, Option B is incorrect.

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5. Analysis of Option C

• The face $a_1b_1$ is vertical and forms an angle $A_1$ with face $a_1c_1$.
• The face $a_2b_2$ is vertical and forms an angle $A_2$ with face $a_2c_2$.
• Since $a_1b_1 \parallel a_2b_2$, the angle $\theta$ between the extended surfaces $a_1c_1$ and $a_2c_2$ is given by: $\theta = A_1 + A_2$

For a full thin isosceles prism with minimum deviation angle $\delta_{m}$, the angle of deviation is related to the half-angle by $\delta_{m} = 2(n-1)A$, which gives: $A_1 = \frac{\delta_{m1}}{2(n_1 - 1)} \quad \text{and} \quad A_2 = \frac{\delta_{m2}}{2(n_2 - 1)}$

Substituting $A_1$ and $A_2$ into the expression for $\theta$: $\theta = \frac{\delta_{m1}}{2(n_1-1)} + \frac{\delta_{m2}}{2(n_2-1)}$

Thus, Option C is correct.

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Conclusion

The correct options are A, C, and D.