The value of $\Delta e$ (in radians) is greater than that of $\Delta n$
$\Delta e$ is proportional to $\Delta n$
$\Delta e$ lies between $2.0$ and $3.0$ milliradians, if $\Delta n = 2.8 \times 10^{-3}$
$\Delta e$ lies between $1.0$ and $1.6$ milliradians, if $\Delta n = 2.8 \times 10^{-3}$

Solution:

Given angle of prism

A=\frac{60^{\circ}}{2}=30^{\circ}

For minimum deviation.

r_{1}=r_{2}=\frac{A}{2}=30^{\circ}

and from Snell's law,

n \times \sin i = n_{1} \sin r_{1}

\Rightarrow \;\; 1 \times \sin i = n_{1} \sin \frac{A}{2}

\therefore \;\; \sin i = \frac{3}{2} \times \sin 30^{\circ} = \frac{3}{2} \times \frac{1}{2}

1 \times \sin i = \mu \sin \left(\frac{A}{2}\right) \;\; \therefore \;\; \sin i = \frac{3}{2}

At another face of prism

n_{1} \sin 30^{\circ} = 1 \sin(e)

On differentiating both sides

\Delta n \sin 30^{\circ} = \Delta e \cos(e)

\Delta e = \frac{\Delta n}{2 \cos(e)}

or,

\;\; \Delta e = \frac{\Delta n}{2 \sqrt{1-\frac{9}{16}}} = \frac{2}{\sqrt{7}} \Delta n

\Delta e = \frac{2}{\sqrt{7}} \Delta n

\Rightarrow \Delta e < \Delta n

and

\Delta e \propto \Delta n

\Delta n \;\; = 2.8 \times 10^{-3}

\therefore \Delta e = \frac{2.8 \times 10^{-3} \times 2}{\sqrt{7}} = 2.11 \times 10^{-3} \text{ rad} = 2.11 \text{ mrad}

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