For a prism of prism angle θ = 60 ° , the refractive indices of the...

Given, ${n}_{1} = {n}_{2} = n = \dfrac{3}{2}$

In case of minimum deviation,

${r}_{m} = \dfrac{A}{2} = \dfrac{60 °}{2} = 30 °$

and $i = e$

Applying expression for minimum deviation,

n = \dfrac{\sin \left(\dfrac{{\delta}_\min + A}{2}\right)}{\sin \left(\dfrac{A}{2}\right)} ⇒ \dfrac{3}{2} = \dfrac{\sin ( i )}{\sin \left(\dfrac{60 °}{2}\right)} ⇒ \sin \left(i\right) = \dfrac{3}{4}, \cos \left(i\right) = \dfrac{\sqrt{7}}{4}

If ${n}_{1} = n = \dfrac{3}{2}$ and {n}_{2} = n + ∆ n:

Angle of emergence, $e = i + \Deltae$

Applying Snell's law, we get,

$1 \times sini = nsin30 °$

Then,

\left(n + \Deltan\right)\sin 30 ° = \left(1\right)\sin \left(i + \Deltae\right)$</p><p>Solving both equations we get,$\dfrac{1}{2}\left(\Deltan\right) = \sin \left(i + \Deltae\right) - sini$</p><p>$\dfrac{\Deltan}{2} = \dfrac{\sin \left(i + \Deltae\right) - sini}{\Deltae} \times \Deltae$</p><p>Now using the definition of derivative,</p><p>$\dfrac{d\left(sini\right)}{di} = cosi$, </p><p>For small change,$∆ e \approx ∆ i$.</p><p>Therefore,</p><p>$\dfrac{\Deltan}{2} = cosi\Deltae ⇒ \dfrac{\Deltan}{2} = \dfrac{\sqrt{7}}{4}\Deltae ⇒ \dfrac{\Deltan}{\Deltae} = \dfrac{\sqrt{7}}{2} = \dfrac{2 . 64}{2}$</p><p>$\dfrac{\Deltan}{\Deltae} = 1 . 34 > 1 ⇒ \Deltan > \Deltae$,</p><p>Hence, option (A) is incorrect.</p><p>$\Deltan = \left(1 . 34\right)\Delta e ⇒ \Delta e ∝ \Deltan$</p><p>Hence, option (B) is correct.</p><p>$\dfrac{\Deltan}{\Deltae} = \dfrac{\sqrt{7}}{2}\text{If}\Deltan = 2 . 8 \times {10}^{- 3}\text{,} ⇒ \dfrac{2 . 8 \times {10}^{- 3}}{\Deltae} = \dfrac{\sqrt{7}}{2}$</p><p>$⇒ \Deltae = \dfrac{5 . 6 \times {10}^{- 3}}{\sqrt{7}} = \sqrt{7} \times 0 . 8 \times {10}^{- 3}$</p><p>$⇒ \Deltae = \left(2 . 64 \times 0 . 8\right) \times {10}^{- 3} = 2 . 11 \times {10}^{- 3}rad

Option (C) is correct.