Step-by-step Solution

The problem asks us to analyze the change in the angle of emergence ($\\\Delta e$) when the refractive index of the right half of a prism is slightly changed. The angle of incidence is fixed to a value that would cause minimum deviation if the prism were homogeneous.

1. Analyze the reference case (homogeneous prism)

First, consider the case where the prism is homogeneous with refractive index $n_1 = n_2 = n = 1.5$. The prism angle is $A = \theta = 60^\circ$. The angle of incidence $i$ is chosen for minimum deviation. For minimum deviation:

• The ray passes symmetrically through the prism, so the angle of incidence equals the angle of emergence ($i=e$).
• The angles of refraction at the first surface ($r_1$) and incidence at the second surface ($r_2$) are equal ($r_1 = r_2$).
• The sum of these angles is equal to the prism angle: $r_1 + r_2 = A$.

From these conditions: $2r_1 = A = 60^\circ \implies r_1 = 30^\circ$ So, $r_1 = r_2 = 30^\circ$.

Now, we use Snell's law at the first surface to find the angle of incidence $i$: $1 \cdot \sin i = n \cdot \sin r_1$ $\sin i = 1.5 \cdot \sin 30^\circ = 1.5 \cdot 0.5 = 0.75$ In this reference case, the angle of emergence $e$ is equal to $i$, so $\sin e = 0.75$.

2. Analyze the modified case (non-homogeneous prism)

Now, the refractive indices are $n_1 = n = 1.5$ and $n_2 = n + \Delta n$. The angle of incidence $i$ is kept the same, so $\sin i = 0.75$.

Let's trace the ray through this composite prism:

• Refraction at the first surface (Air to $n_1$): The angle of incidence is $i$. The angle of refraction is $r_1$. Snell's law gives: $1 \cdot \sin i = n_1 \cdot \sin r_1$ $0.75 = 1.5 \cdot \sin r_1$ $\sin r_1 = 0.5 \implies r_1 = 30^\circ$ The path of the ray in the first half of the prism is unchanged.

• Angle of incidence at the second surface: From the geometry of the prism, the angle of incidence at the second surface, $r_2$, is given by: $r_2 = A - r_1 = 60^\circ - 30^\circ = 30^\circ$

• Refraction at the second surface ($n_2$ to Air): Let the new angle of emergence be $e'$. Snell's law at this surface is: $n_2 \cdot \sin r_2 = 1 \cdot \sin e'$ $(n + \Delta n) \cdot \sin 30^\circ = \sin e'$ $(1.5 + \Delta n) \cdot 0.5 = \sin e'$ $\sin e' = 0.75 + 0.5 \Delta n$

3. Calculate the change in emergence angle, $\Delta e$

The original angle of emergence was $e$, where $\sin e = 0.75$. The new angle is $e' = e + \Delta e$. We have $\sin(e + \Delta e) = 0.75 + 0.5 \Delta n$. Since $\Delta n$ is small, $\Delta e$ will also be small. We can use the first-order Taylor expansion for $\sin(e + \Delta e)$: $\sin(e + \Delta e) \approx \sin e + (\cos e) \Delta e$ Substituting this into our equation: $\sin e + (\cos e) \Delta e = 0.75 + 0.5 \Delta n$ Since $\sin e = 0.75$, the equation simplifies to: $(\cos e) \Delta e = 0.5 \Delta n$ We need to find $\cos e$. Using $\sin^2 e + \cos^2 e = 1$: $\cos e = \sqrt{1 - \sin^2 e} = \sqrt{1 - (0.75)^2} = \sqrt{1 - (3/4)^2} = \sqrt{1 - 9/16} = \sqrt{7/16} = \frac{\sqrt{7}}{4}$ Now we can find the relation between $\Delta e$ and $\Delta n$: $\left(\frac{\sqrt{7}}{4}\right) \Delta e = 0.5 \Delta n$ $\Delta e = \frac{0.5 \times 4}{\sqrt{7}} \Delta n = \frac{2}{\sqrt{7}} \Delta n$ The angle $\Delta e$ is in radians.

4. Evaluate the given options

We have the relationship $\Delta e = \frac{2}{\sqrt{7}} \Delta n$. Let's evaluate the constant factor: $\frac{2}{\sqrt{7}} \approx \frac{2}{2.6457} \approx 0.7559$.

• A: The value of $\Delta e$ (in radians) is greater than that of $\Delta n$ The proportionality constant is $\frac{2}{\sqrt{7}} \approx 0.7559$, which is less than 1. Therefore, $\Delta e < \Delta n$. Option A is incorrect.

• B: $\Delta e$ is proportional to $\Delta n$ The derived relationship is $\Delta e = k \Delta n$, where $k = \frac{2}{\sqrt{7}}$ is a constant. Thus, $\Delta e$ is directly proportional to $\Delta n$. Option B is correct.

• C: $\Delta e$ lies between 2.0 and 3.0 milliradians, if $\Delta n = 2.8 \times 10^{-3}$ Let's calculate $\Delta e$ for the given $\Delta n$: $\Delta e = \frac{2}{\sqrt{7}} (2.8 \times 10^{-3}) \text{ radians}$ $\Delta e \approx 0.7559 \times 2.8 \times 10^{-3} \approx 2.1166 \times 10^{-3} \text{ radians}$ This is equal to 2.1166 milliradians. This value lies between 2.0 and 3.0 milliradians. Option C is correct.

• D: $\Delta e$ lies between 1.0 and 1.6 milliradians, if $\Delta n = 2.8 \times 10^{-3}$ Our calculated value is 2.1166 milliradians, which is not in the range [1.0, 1.6]. Option D is incorrect.

Final correct options are B and C.