Consider two isosceles prisms 1 and 2 with prism angles A₁ and A₂ and refractive indices n₁ and n₂, respectively, as shown in the figure. The faces a₁b₁ and a₂b₂ are parallel to each other and perpendicular to the mirror M. If a ray of light is incident on the face a₁c₁ and emerges from the face a₂c₂, then the correct statement(s) is/are:

Correct Options:

1. If both the prisms are at minimum deviation condition, then $\frac{n_2}{n_1}=\frac{\sin \left(\frac{A_1}{2}\right)}{\sin \left(\frac{A_2}{2}\right)}$.

2. If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation δ_m1 and δ_m2, respectively, then $\theta =\frac{{\mathrm{\delta ;}}_{m1}}{2(n_1-1)}+\frac{{\mathrm{\delta ;}}_{m2}}{2(n_2-1)}$.

3. If prism 1 is at minimum deviation condition, then $\sin i_2=n_1\sin \left(\frac{A_1}{2}\right)$ is always true.

Analysis of the Given Setup:

From the given figure:

We have two isosceles prisms with apex angles A₁ and A₂ and refractive indices n₁ and n₂. The base faces a₁b₁ and a₂b₂ are perpendicular to the plane mirror M. The extensions of the base faces a₁b₁ and a₂b₂ intersect at an angle θ at the top.

Since a₁b₁ and a₂b₂ are both perpendicular to mirror M, the normals to the surfaces a₁b₁ and a₂b₂ are parallel to the mirror M (horizontal).

The ray emerges from face a₁b₁ at an angle of emergence e₁ relative to the normal of a₁b₁. The angle made by the ray leaving prism 1 with the mirror M is equal to the angle with the base a₁b₁, which is:

${\alpha }_1={90}^{\circ }-e_1$

Upon reflection at the plane mirror M, the angle of incidence equals the angle of reflection. Thus, the ray strikes the face a₂b₂ of Prism 2 at an angle of incidence i₂ with its normal such that:

$i_2=e_1$

Step-by-step Verification of Statements:

1. Condition when Prism 1 is at Minimum Deviation:
When Prism 1 is at minimum deviation condition:
• The ray inside Prism 1 passes symmetrically, so the angle of refraction at the first face is $r_1=\frac{A_1}{2}$.
• The angle of incidence equals the angle of emergence, i.e., $i_1=e_1$.
• Applying Snell's law at the emerging face a₁b₁ of Prism 1:

$1\cdot \sin e_1=n_1\sin r_2=n_1\sin \left(\frac{A_1}{2}\right)$

Since $i_2=e_1$, we directly obtain:

$\sin i_2=\sin e_1=n_1\sin \left(\frac{A_1}{2}\right)$

Therefore, the statement "If prism 1 is at minimum deviation condition, then $\sin i_2=n_1\sin \left(\frac{A_1}{2}\right)$ is always true" is correct.

2. Condition when both prisms are at Minimum Deviation:
If Prism 2 is also at minimum deviation condition, then its angle of incidence i₂ must satisfy:

$\sin i_2=n_2\sin \left(\frac{A_2}{2}\right)$

Equating both expressions for $\sin i_2$:

$n_1\sin \left(\frac{A_1}{2}\right)=n_2\sin \left(\frac{A_2}{2}\right)$

Rearranging the terms yields:

$\frac{n_2}{n_1}=\frac{\sin \left(\frac{A_1}{2}\right)}{\sin \left(\frac{A_2}{2}\right)}$

Therefore, the statement "If both the prisms are at minimum deviation condition, then $\frac{n_2}{n_1}=\frac{\sin \left(\frac{A_1}{2}\right)}{\sin \left(\frac{A_2}{2}\right)}$" is correct.

3. Condition for Thin Prisms at Minimum Deviation:
For a thin prism of angle A and refractive index n at minimum deviation:

${\mathrm{\delta ;}}_m=(n-1)A\Rightarrow A=\frac{{\mathrm{\delta ;}}_m}{n-1}$

Thus:
$A_1=\frac{{\mathrm{\delta ;}}_{m1}}{n_1-1}$ and $A_2=\frac{{\mathrm{\delta ;}}_{m2}}{n_2-1}$

From the geometry of the triangle formed by extending faces a₁b₁ and a₂b₂:
The base angle of prism 1 is ${90}^{\circ }-\frac{A_1}{2}$ and for prism 2 is ${90}^{\circ }-\frac{A_2}{2}$.
The interior angles of the triangle with apex angle θ are $\frac{A_1}{2}$ and $\frac{A_2}{2}$.
Summing the angles in the top triangle gives:

$\theta =\frac{A_1}{2}$ + $\frac{A_2}{2}$

Substituting $A_1$ and $A_2$:

$\theta =\frac{{\mathrm{\delta ;}}_{m1}}{2(n_1-1)}+\frac{{\mathrm{\delta ;}}_{m2}}{2(n_2-1)}$

Thus, this statement is also correct.