JEE Challenger
More from Ray Optics and Optical Instruments
PhyHardAdv 2021 • Paper 1

Total Internal Reflection and Refraction in Two Layered Slab

A wide slab comprising two media with refractive indices n1n_1 and n2n_2 is kept in air as illustrated in the figure. A light ray is incident from medium n1n_1 to medium n2n_2 at an angle θ\theta, such that sin⁡θ\sin \theta is marginally greater than 1/n11/n_1. Assuming the refractive index of air to be 1, identify the correct statement(s):

Question Diagram 1

Options

A

The light ray enters air if n2=n1n_2 = n_1

B

The light ray is finally reflected back into the medium of refractive index n1n_1 if n2<n1n_2 < n_1

Correct
C

The light ray is finally reflected back into the medium of refractive index n1n_1 if n2>n1n_2 > n_1

Correct
D

The light ray is reflected back into the medium of refractive index n1n_1 if n2=1n_2 = 1

Correct

Topics & Concepts

Ray Optics and Optical InstrumentsReflection and Refraction

Step-by-Step Solution

To determine the correct statement(s), we analyze the propagation of the light ray using Snell's Law at each interface.

1. Reflection and Refraction Analysis

Let θ\theta be the angle of incidence in medium n1n_1, θ2\theta_2 be the angle of refraction in medium n2n_2, and θair\theta_{\text{air}} be the angle of refraction in air.

Applying Snell's Law at the interface between medium n1n_1 and medium n2n_2: n1sin⁡θ=n2sin⁡θ2n_1 \sin \theta = n_2 \sin \theta_2

Applying Snell's Law at the interface between medium n2n_2 and air (nair=1n_{\text{air}} = 1): n2sin⁡θ2=1⋅sin⁡θairn_2 \sin \theta_2 = 1 \cdot \sin \theta_{\text{air}}

Combining these two relations yields: sin⁡θair=n1sin⁡θ\sin \theta_{\text{air}} = n_1 \sin \theta

We are given that sin⁡θ>1n1\sin \theta > \frac{1}{n_1}, which means: n1sin⁡θ>1n_1 \sin \theta > 1

Therefore: sin⁡θair>1\sin \theta_{\text{air}} > 1

Since the sine of a real angle cannot exceed 11, the light ray cannot enter the air under any condition. Thus, total internal reflection (TIR) must occur at one of the boundaries, sending the light ray back into medium n1n_1.

2. Analysis of the Options

  • Option (A): If n2=n1n_2 = n_1, the light ray enters medium n2n_2 without bending (θ2=θ\theta_2 = \theta). At the n2n_2-air interface, sin⁡θair=n1sin⁡θ>1\sin \theta_{\text{air}} = n_1 \sin \theta > 1, so total internal reflection occurs at the n2n_2-air interface. The ray is reflected back into medium n2n_2 and subsequently enters medium n1n_1. Thus, the light ray does not enter air.
    (Option A is incorrect)

  • Option (B): If n2<n1n_2 < n_1, there are two cases:

    1. If n2<n1sin⁡θn_2 < n_1 \sin \theta, then sin⁡θ2=n1sin⁡θn2>1\sin \theta_2 = \frac{n_1 \sin \theta}{n_2} > 1. Total internal reflection occurs directly at the n1n_1-n2n_2 interface, sending the light back into n1n_1.
    2. If n2≥n1sin⁡θn_2 \ge n_1 \sin \theta, the ray enters n2n_2. However, at the n2n_2-air interface, n2sin⁡θ2=n1sin⁡θ>1n_2 \sin \theta_2 = n_1 \sin \theta > 1, so total internal reflection occurs at the n2n_2-air interface. The ray then travels back and re-enters medium n1n_1.
      In both scenarios, the light ray is finally reflected back into medium n1n_1.
      (Option B is correct)

  • Option (C): If n2>n1n_2 > n_1, since sin⁡θ≤1\sin \theta \le 1, we have n2>n1sin⁡θn_2 > n_1 \sin \theta. Thus, sin⁡θ2=n1sin⁡θn2<1\sin \theta_2 = \frac{n_1 \sin \theta}{n_2} < 1, and the ray refracts into medium n2n_2. At the n2n_2-air interface, the product n2sin⁡θ2=n1sin⁡θ>1n_2 \sin \theta_2 = n_1 \sin \theta > 1, causing total internal reflection. The reflected ray travels back through medium n2n_2 and returns into medium n1n_1.
    (Option C is correct)

  • Option (D): If n2=1n_2 = 1, then sin⁡θ2=n1sin⁡θ1>1\sin \theta_2 = \frac{n_1 \sin \theta}{1} > 1. Consequently, total internal reflection occurs immediately at the n1n_1-n2n_2 interface, and the light ray is reflected back into medium n1n_1.
    (Option D is correct)

Conclusion

The correct options are B, C, and D.

PreviousNext