Option A: If $\left(n_1=n_2\right)$,

$n_1\sin \theta =n_2\sin {\theta }_1 ...\left(1\right)$

and $n_2\sin {\theta }_1=1\times \sin r ...\left(2\right)$

From (1) and (2),

$\sin r=n_1\sin \theta$

Now, $\sin \theta >\frac{1}{n_1}$, (given)

$\therefore \sin r>1$ (which is not possible)

and hence light will reflect back from $\left(\text{air}, n_2\right)$ interface

Option B: If $\left(n_2<n_1\right)$,

Snell's law at $A$,

$n_1\sin \theta =n_2\sin {\theta }_1 ...\left(3\right)$

Snell's law at $B$

$n_2\sin {\theta }_1=\sin r ...\left(4\right)$

From $\left(3\right) \text{and} \left(4\right)$,

$n_1\sin \theta =\sin r$

$\therefore \sin r>1$ (not possible)

Hence, light might reflect from $A$ or $B$.

Option C: If $\left(n_2>n_1\right)$,

$n_1\sin \theta =n_2\sin {\theta }_1 ...\left(5\right)$ at $A$

$n_2\sin \theta =1\times \sin r ...\left(6\right)$ at $B$

$\therefore n_1\sin \theta =1\times \sin r$

$\therefore \sin r>1$ (not possible)

$\therefore$ Light will reflect back into medium $n_1$.

Option D: If $\left(n_2=1\right)$,

$n_1\sin \theta =n_2\sin {\theta }_1$

$\Rightarrow \sin {\theta }_1=n_1\sin \theta$

or $\sin {\theta }_1>1$ not possible

Hence, light will reflect back into the medium.