For a prism of prism angle θ = 60 ° , the refractive indices of the left half and the right half…

Given, $n_1=n_2=n=\frac{3}{2}$

In case of minimum deviation,

$r_m=\frac{A}{2}=\frac{60°}{2}=30°$

and $i=e$

Applying expression for minimum deviation,

$$\begin{gathered} n=\frac{\sin \left(\frac{{\delta }_{\min }+A}{2}\right)}{\sin \left(\frac{A}{2}\right)} \\ \Rightarrow \frac{3}{2}=\frac{\sin \left(i\right)}{\sin \left(\frac{60°}{2}\right)} \\ \Rightarrow \sin \left(i\right)=\frac{3}{4}, \cos \left(i\right)=\frac{\sqrt{7}}{4} \end{gathered}$$

If $n_1=n=\frac{3}{2}$ and $n_2=n+∆n$:

Angle of emergence, $\mathrm{e}=i+\Delta e$

Applying Snell's law, we get,

$1\times \sin i=n\sin 30°$

Then,

$\left(n+\mathrm{\Delta }n\right)\sin 30°=\left(1\right)\sin \left(i+\Delta e\right)$

Solving both equations we get,$\frac{1}{2}\left(\mathrm{\Delta }n\right)=\sin \left(i+\mathrm{\Delta }e\right)-\sin i$

$\frac{\mathrm{\Delta }n}{2}=\frac{\sin \left(i+\mathrm{\Delta }e\right)-\sin i}{\mathrm{\Delta }e}\times \mathrm{\Delta }e$

Now using the definition of derivative,

$\frac{\mathrm{d}\left(\sin i\right)}{\mathrm{d}i}=\cos i$, $$

For small change, $∆e\approx ∆i$.

Therefore,

$$\begin{gathered} \frac{\mathrm{\Delta }n}{2}=\cos i\Delta e \\ \Rightarrow \frac{\mathrm{\Delta }n}{2}=\frac{\sqrt{7}}{4}\mathrm{\Delta }e \\ \Rightarrow \frac{\mathrm{\Delta }n}{\mathrm{\Delta }e}=\frac{\sqrt{7}}{2}=\frac{2.64}{2} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{\Delta }n}{\mathrm{\Delta }e}=1.34>1 \\ \Rightarrow \mathrm{\Delta }n>\mathrm{\Delta }e \end{gathered}$$

Hence, option (A) is incorrect.

$$\begin{gathered} \mathrm{\Delta }n=\left(1.34\right)\mathrm{\Delta e} \\ \Rightarrow \mathrm{\Delta e}\propto \mathrm{\Delta }n \end{gathered}$$

Hence, option (B) is correct.

$$\begin{gathered} \frac{\mathrm{\Delta }n}{\mathrm{\Delta }e}=\frac{\sqrt{7}}{2} \\ \text{If} \mathrm{\Delta }n=2.8\times {10}^{-3}\text{,} \\ \Rightarrow \frac{2.8\times {10}^{-3}}{\mathrm{\Delta }e}=\frac{\sqrt{7}}{2} \end{gathered}$$

$\Rightarrow \mathrm{\Delta }e=\frac{5.6\times {10}^{-3}}{\sqrt{7}}=\sqrt{7}\times 0.8\times {10}^{-3}$

$\Rightarrow \mathrm{\Delta }e=\left(2.64\times 0.8\right)\times {10}^{-3}=2.11\times {10}^{-3} \mathrm{rad}$

Option (C) is correct.