A wide slab consisting of two media of refractive indices $$n_1$$ and $$n_2$$ is placed in air as shown in the figure. A ray of light is incident from medium $$n_1$$ to $$n_2$$ at an angle $$\theta$$, where $$\sin \theta$$ is slightly larger than $$\frac{1}{n_1}$$. Take refractive index of air as 1. Which of the following statement(s) is(are) correct?

Let the refractive indices be arranged as
air (top, $$n = 1$$)  |  medium $$n_1$$  |  medium $$n_2$$  |  air (bottom, $$n = 1$$).

The ray starts inside $$n_1$$ and strikes the $$n_1\!-\!n_2$$ interface at an angle $$\theta$$ that satisfies

$$\sin\theta = \frac{1}{n_1} + \varepsilon,$$

where $$\varepsilon \gt 0$$ is a very small positive number. Because $$\sin\theta \gt \dfrac{1}{n_1}$$, the ray would have suffered total internal reflection (TIR) if the medium outside $$n_1$$ had been air. We now examine what really happens in the presence of the $$n_2$$ layer.

Step 1: Refraction at the $$n_1\!-\!n_2$$ interface
Applying Snell’s law

$$n_1\sin\theta = n_2\sin\phi \quad\Longrightarrow\quad \sin\phi = \frac{n_1}{n_2}\sin\theta.$$

Substituting $$\sin\theta = \dfrac{1}{n_1} + \varepsilon$$ gives

$$\sin\phi = \frac{1}{n_2} + \frac{n_1}{n_2}\,\varepsilon. \qquad -(1)$$

The second term in (1) is positive, so

$$\sin\phi \gt \frac{1}{n_2}.$$

Step 2: Incidence at the $$n_2\!-\!{\rm air}$$ interface
The critical angle for the $$n_2\!-\!{\rm air}$$ interface is defined by $$\sin\phi_c = \frac{1}{n_2}.$$ Because $$\sin\phi \gt \dfrac{1}{n_2}$$ from (1), the ray is definitely above the critical angle and therefore undergoes total internal reflection at the lower surface. After reflection, it travels upward in medium $$n_2$$ with the same angle $$\phi$$.

Step 3: Returning to the $$n_2\!-\!n_1$$ interface
When the ray coming from $$n_2$$ meets the $$n_2\!-\!n_1$$ boundary, its angle of incidence is again $$\phi$$. Whether $$n_2$$ is larger, equal to, or smaller than $$n_1$$, Snell’s law gives

$$n_2\sin\phi = n_1\sin\theta,$$

which has a real solution for $$\sin\theta$$ because $$\sin\theta \le 1$$. Therefore total internal reflection cannot occur here; the ray always refracts back into medium $$n_1$$.

Step 4: Meeting the upper $$n_1\!-\!{\rm air}$$ interface
Inside $$n_1$$ the ray still makes the same angle $$\theta$$ with the normal. Since $$\sin\theta = \dfrac{1}{n_1} + \varepsilon \gt \dfrac{1}{n_1},$$ the ray is again above the critical angle for the $$n_1\!-\!{\rm air}$$ surface and is totally internally reflected. Hence it remains trapped inside the slab, always coming back into the $$n_1$$ region.

Now check each statement:

Option A: If $$n_2 = n_1$$ the two internal media are identical, but the analysis above still shows TIR at the lower air surface and again at the upper air surface. The ray never reaches air. False.

Option B: For $$n_2 \lt n_1$$, all steps above remain valid (because (1) was derived without assuming any ordering of $$n_1,n_2$$). The ray finally returns to medium $$n_1$$. True.

Option C: For $$n_2 \gt n_1$$ the same conclusion follows; the ray refracts back into $$n_1$$ after the lower TIR and is again reflected at the top surface. True.

Option D: If $$n_2 = 1$$, the first interface itself is $$n_1\!-\!{\rm air}$$. Because $$\sin\theta \gt \dfrac{1}{n_1}$$, the ray is totally internally reflected immediately, i.e. it stays in $$n_1$$. True.

Hence the correct statements are:
Option B, Option C and Option D.

Final Answer: Option B, Option C, Option D