A wide slab consisting of two media of refractive indices $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ is placed in air as shown in the figure. A ray of light is incident from medium $\mathrm{n}_{1}$ to $\mathrm{n}_{2}$ at an angle $\theta$, where $\sin\theta$ is slightly larger than $1/\mathrm{n}_{1}$. Take refractive index of air as 1. Which of the following statement(s) is(are) correct?

(B) The light ray is finally reflected back into the medium of refractive index $\mathrm{n}_{1}$ if $\mathrm{n}_{2} \mathrm{n}_{1}$; (D) The light ray is reflected back into the medium of refractive index $\mathrm{n}_{1}$ if $\mathrm{n}_{2}=1$

(A) The light ray enters air if $\mathrm{n}_{2}=\mathrm{n}_{1}$

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Geometrical Optics

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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}

n_{2}

at an angle

θ

, where

sin θ

is slightly larger than

1/ 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 $n_{2} < n_{1}$ ✓ Correct
(C)The light ray is finally reflected back into the medium of refractive index $n_{1}$ if $n_{2} > n_{1}$ ✓ Correct
(D)The light ray is reflected back into the medium of refractive index $n_{1}$ if $n_{2} = 1$ ✓ Correct

Answer: (B), (C), (D)

Solution

Given:

sin θ_{1} > 1/ n_{1}

.

**Case (A):

n_{1} = n_{2}

**. Then

θ_{2} = θ_{1}

. At the

n_{2}

-air interface:

n_{2} sin θ_{2} = sin θ_{3}

, so

sin θ_{3} = n_{1} sin θ_{1} > 1

. This is impossible, so TIR occurs and the ray cannot enter air. ✗

**Case (B):

n_{2} < n_{1}

**. From Snell's law:

sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1} > \frac{n _{1}}{n _{2}} \cdot \frac{1}{n _{1}} = \frac{1}{n _{2}}

. At the

n_{2}

-air interface,

sin θ_{3} = n_{2} sin θ_{2} > 1

, so TIR occurs. The ray reflects back and can enter medium 1. ✓

**Case (C):

n_{2} > n_{1}

**. Then

sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1}

. At the

n_{2}

-air interface,

sin θ_{3} = n_{2} sin θ_{2} = n_{1} sin θ_{1} > 1

, so TIR occurs. The ray reflects back and enters medium 1. ✓

**Case (D):

n_{2} = 1

**. Then

sin θ_{2} = n_{1} sin θ_{1} > 1

, which is impossible. TIR occurs at the first interface itself, and the ray is reflected back into medium 1. ✓

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Geometrical Optics

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