A wide slab consisting of two media of refractive indices n _ 1 and...

Option A: If $\left({n}_{1} = {n}_{2}\right)$,

${n}_{1}sin\theta = {n}_{2}sin{\theta}_{1} . . . \left(1\right)</p><p>and${n}_{2}sin{\theta}_{1} = 1 \times sinr . . . \left(2\right)$</p><p>From (1) and (2),</p><p>$sinr = {n}_{1}sin\theta$</p><p>Now,$sin\theta > \dfrac{1}{{n}_{1}}$, (given)</p><p>$∴ sinr > 1$(which is not possible)</p><p>and hence light will reflect back from$\left(\text{air}, {n}_{2}\right)$interface</p><p><strong>Option B:</strong> If \left({n}_{2} < {n}_{1}\right)$,

Snell's law at A$,</p><p>${n}_{1}sin\theta = {n}_{2}sin{\theta}_{1} . . . \left(3\right)$</p><p>Snell's law at$B$</p><p>${n}_{2}sin{\theta}_{1} = sinr . . . ( 4 )$</p><p>From \left(3\right)\text{and} ( 4 )$,

${n}_{1}sin\theta = sinr$

$∴ sinr > 1$ (not possible)

Hence, light might reflect from A or B$.</p><p><strong>Option C:</strong> If$\left({n}_{2} > {n}_{1}\right),

${n}_{1}sin\theta = {n}_{2}sin{\theta}_{1} . . . \left(5\right)$ at A$</p><p>${n}_{2}sin\theta = 1 \times sinr . . . $\left(6\right)$ at $B$

$∴ {n}_{1}sin\theta = 1 \times sinr$

$∴ sinr > 1$ (not possible)

$∴$ Light will reflect back into medium ${n}_{1}$.

Option D: If \left({n}_{2} = 1\right)$,</p><p>${n}_{1}sin\theta = {n}_{2}sin{\theta}_{1}$</p><p>$⇒ sin{\theta}_{1} = {n}_{1}sin\theta$</p><p>or$sin{\theta}_{1} > 1 not possible

Hence, light will reflect back into the medium.