Deviation Angle Sensitivity in Split Refractive Index Prism

To determine which of the statements are correct, we analyze the refraction of light through the two-part prism step-by-step.

Step 1: Minimum Deviation Condition in the Symmetric Prism

For a uniform prism of apex angle $\theta = 60^\circ$ and refractive index $n_1 = n_2 = n = 1.5$ undergoing minimum deviation:

• The angle of refraction at the first surface is: $r_1 = \frac{\theta}{2} = \frac{60^\circ}{2} = 30^\circ$
• Using Snell's law at the first surface: $\sin i = n \sin r_1 = 1.5 \sin 30^\circ = 1.5 \times 0.5 = 0.75$
• The cosine of the angle of incidence $i$ is: $\cos i = \sqrt{1 - \sin^2 i} = \sqrt{1 - (0.75)^2} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$

Under the condition of minimum deviation, the ray inside the prism travels perpendicular to the angle bisector (the vertical interface separating the two media).

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Step 2: Refraction across the Interface and Second Surface

The interface between the two halves of the prism lies along the bisector of the apex angle $\theta$. Since the light ray inside the first half is perpendicular to this interface:

1. The angle of incidence at the internal boundary is $0^\circ$ (normal incidence).
2. The ray passes straight into the second medium without any deviation at the internal boundary.
3. Therefore, inside the right half of the prism, the ray continues along the same direction and strikes the second surface at an angle of incidence: $r_2 = 30^\circ$

For the second half with refractive index $n_2 = n + \Delta n$, Snell's law at the emergent surface gives: $n_2 \sin r_2 = 1 \cdot \sin e$ $(n + \Delta n) \sin 30^\circ = \sin(i + \Delta e)$

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Step 3: Relation between $\Delta e$ and $\Delta n$

Expanding the right-hand side using small angle approximation for $\Delta e$ ($\Delta e \ll 1$): $n \sin 30^\circ + \Delta n \sin 30^\circ \approx \sin i + \cos i \cdot \Delta e$

Since $n \sin 30^\circ = \sin i$, the equation simplifies to: $\Delta n \sin 30^\circ = \cos i \cdot \Delta e$

$\Delta e = \frac{\sin 30^\circ}{\cos i} \Delta n = \frac{1/2}{\sqrt{7}/4} \Delta n = \frac{2}{\sqrt{7}} \Delta n$

Using the numerical value of $\sqrt{7} \approx 2.646$: $\Delta e \approx \frac{2}{2.646} \Delta n \approx 0.756 \, \Delta n$

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Step 4: Verification of the Options

1. Option A: $\Delta e \approx 0.756 \, \Delta n < \Delta n$. Thus, $\Delta e$ is less than $\Delta n$, making Statement A incorrect.

2. Option B: Since $\Delta e = \left(\frac{2}{\sqrt{7}}\right) \Delta n$, $\Delta e$ is directly proportional to $\Delta n$. Thus, Statement B is correct.

3. Options C and D: For $\Delta n = 2.8 \times 10^{-3}$: $\Delta e = \frac{2}{\sqrt{7}} \times 2.8 \times 10^{-3} = \frac{5.6}{2.64575} \times 10^{-3} \text{ rad} \approx 2.116 \times 10^{-3} \text{ rad} = 2.116 \text{ mrad}$

   Since $2.116\text{ mrad}$ lies between $2.0\text{ mrad}$ and $3.0\text{ mrad}$:

   • Statement C is correct.
   • Statement D is incorrect.

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Conclusion

The correct options are B and C.