Question

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 $\frac1 { n _{1}}$ Take refractive index of air as $1$ Which of the following statement(s) is(are) correct?

• A. The light ray enters air if $n_{2}=n_{1}$
• B. The light ray is finally reflected back into the medium of refractive index $n _{1}$ if $n _{2}< n _{1}$
• C. The light ray is finally reflected back into the medium of refractive index $n_{1}$ if $n_{2}>n_{1}$
• D. The light ray is reflected back into the medium of refractive index $n_{1}$ if $n_{2}=1$

Answer 1

The Correct Option is D

Given:

• Refractive indices of two media: \( n_1 \) and \( n_2 \)
• Light ray incident from \( n_1 \) to \( n_2 \) at an angle \( \theta \)
• \( \sin \theta \) is slightly greater than \( \dfrac{1}{n_1} \)
• Refractive index of air = 1

Condition: \( \sin\theta > \dfrac{1}{n_1} \Rightarrow \theta > \theta_c \) (Critical angle for total internal reflection from \( n_1 \) to air)

If the ray hits the interface between \( n_1 \) and air at an angle greater than the critical angle, it will be totally internally reflected.

However, in this setup, the ray is incident from \( n_1 \) to \( n_2 \). For total internal reflection to happen at the top interface (between \( n_2 \) and air), \( n_2 \) must be such that total internal reflection condition is satisfied there.

Now consider the case: if \( n_2 = 1 \) (i.e., equal to the refractive index of air), then the interface between \( n_2 \) and air becomes ineffective in bending the light — it behaves as if it is just air.

As the ray travels from \( n_1 \) through \( n_2 \) and reaches the top (air interface), the total internal reflection happens at the first interface itself — from \( n_1 \) to \( n_2 \), if \( n_2 = 1 \), since:

\[ \text{If } \sin\theta > \frac{n_2}{n_1}, \text{ total internal reflection occurs} \]

With \( n_2 = 1 \), and \( \sin\theta > \frac{1}{n_1} \), this condition is satisfied.

✓ Hence, the light ray is reflected back into medium of index \( n_1 \) if \( n_2 = 1 \)

Correct Answer: Option D