Step-by-step Derivations

1. Analyze the given condition and its implication for the final interface.

The problem states that a ray of light is incident from medium $n_1$ to $n_2$ at an angle θ, and $\sin{\theta}$ is slightly larger than $1/n_1$. We can write this as: $\sin{\theta} > \frac{1}{n_1} \implies n_1 \sin{\theta} > 1$

Let's analyze the path of the light ray. The ray first travels from medium $n_1$ to $n_2$, and then from $n_2$ to air (refractive index = 1). Let the angle of refraction at the $n_1$-$n_2$ interface be r. According to Snell's law: $n_1 \sin{\theta} = n_2 \sin{r}$

The ray then strikes the $n_2$-air interface. Since the interfaces of the slab are parallel, the angle of incidence at this second interface is also r. Let's assume the ray emerges into the air at an angle e. Applying Snell's law at the $n_2$-air interface: $n_2 \sin{r} = 1 \cdot \sin{e}$

Combining the two Snell's law equations, we get: $n_1 \sin{\theta} = \sin{e}$

Using the given condition $n_1 \sin{\theta} > 1$, we find: $\sin{e} > 1$ This is physically impossible for a real angle e. This crucial result implies that the light ray can never emerge into the air from the $n_2$ medium. It must undergo Total Internal Reflection (TIR) at the $n_2$-air interface, provided it reaches that interface.

After TIR at the $n_2$-air interface, the ray is reflected back into medium $n_2$ and travels towards the $n_2$-$n_1$ interface. By the principle of reversibility, it will re-enter medium $n_1$. Therefore, if the ray enters medium $n_2$, it is ultimately reflected back into medium $n_1$.

Now we must also consider the possibility of TIR at the first interface ($n_1$ to $n_2$). This can only happen if $n_1 > n_2$.

2. Evaluate each option based on the analysis.

• A: The light ray enters air if n_2 = n_1 If $n_2 = n_1$, the slab is effectively a single medium of refractive index $n_1$. The light ray travels undeviated to the $n_1$-air interface. The angle of incidence at this interface is θ. The critical angle for the $n_1$-air interface, $\theta_c$, is given by $\sin{\theta_c} = 1/n_1$. The problem states $\sin{\theta} > 1/n_1$, which means $\theta > \theta_c$. Therefore, TIR occurs at the $n_1$-air interface, and the ray is reflected back into medium $n_1$. It does not enter the air. Thus, statement A is incorrect.

• B: The light ray is finally reflected back into the medium of refractive index n_1 if n_2 < n_1 In this case, the ray travels from a denser medium ($n_1$) to a rarer medium ($n_2$). The critical angle for this interface is $\theta_{c12}$, where $\sin{\theta_{c12}} = n_2/n_1$.

  • Case 1: If $\sin{\theta} > n_2/n_1$ (i.e., $\theta > \theta_{c12}$), TIR occurs at the first ($n_1$-$n_2$) interface. The ray is immediately reflected back into medium $n_1$.
  • Case 2: If $\sin{\theta} \le n_2/n_1$, the ray refracts into medium $n_2$. As established in our initial analysis, this ray will then undergo TIR at the $n_2$-air interface and subsequently re-enter medium $n_1$. In both possible scenarios, the ray is finally reflected back into medium $n_1$. Thus, statement B is correct.

• C: The light ray is finally reflected back into the medium of refractive index n_1 if n_2 > n_1 In this case, the ray travels from a rarer medium ($n_1$) to a denser medium ($n_2$). TIR cannot occur at the first interface, so the ray will always enter medium $n_2$. As established in our initial analysis, the ray will then undergo TIR at the $n_2$-air interface and be reflected back, eventually re-entering medium $n_1$. Thus, statement C is correct.

• D: The light ray is reflected back into the medium of refractive index n_1 if n_2 = 1 This is a special case of option B where $n_2 < n_1$ (assuming $n_1 > 1$, which is necessary for the condition $\sin{\theta} > 1/n_1$ to be possible). The first interface is between medium $n_1$ and air ($n_2 = 1$). The critical angle $\theta_c$ is given by $\sin{\theta_c} = n_2/n_1 = 1/n_1$. The given condition is $\sin{\theta} > 1/n_1$, which means $\theta > \theta_c$. Therefore, TIR occurs at the very first interface, and the ray is reflected back into medium $n_1$. Thus, statement D is correct.

Conclusion

Based on the step-by-step analysis, options B, C, and D are correct.