JEE Advanced 2021 · Paper 1 · Physics Q12

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PhysicsMultiple correct+4 / -2Question 12 of 57

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 $1/n_1$ . Take refractive index of air as 1. Which of the following statement(s) is(are) correct?

A wide horizontal slab: a lower layer of index n1 and an upper layer of index n2, with air above. A ray in medium n1 travels up toward the n1-n2 interface, making angle θ with the normal (dashed vertical line).

The light ray enters air if

n_2 = n_1

The light ray is finally reflected back into the medium of refractive index

n_1

n_2 < n_1

The light ray is finally reflected back into the medium of refractive index

n_1

n_2 > n_1

The light ray is reflected back into the medium of refractive index

n_1

n_2 = 1

Answer and worked solution

Official answer: B, C, D

Key fact. For parallel layers, $n\sin\theta$ is the same in every layer. Here $n_1\sin\theta$ is slightly larger than 1, so the ray can never exist in air (it would need $\sin\theta_{air}>1$ ). Wherever it meets air, it is totally reflected.

Check each option.

(A) If $n_2=n_1$ , the ray reaches the air boundary and is totally reflected. False.
(B) If $n_2<n_1$ , the ray is either totally reflected at the $n_1$ - $n_2$ interface (if $n_2<n_1\sin\theta$ ), or enters $n_2$ and is totally reflected at the air boundary. Either way it ends up back in $n_1$ . True.
(C) If $n_2>n_1$ , the ray enters $n_2$ , is totally reflected at the air boundary, and returns to $n_1$ . True.
(D) If $n_2=1$ , then $n_2<n_1\sin\theta$ , so total internal reflection happens at the first interface. True.

The official key accepts (B, C, D) or (C, D). (B), (C), (D)

Topics:Total internal reflection Refraction at plane surfaces

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JEE Advanced 2021 · Paper 1, Physics question 12