Light goes from medium 1 to medium 2. List-I gives the refractive index \(n_{12}\) of medium 1 with respect to medium 2. Match it with the critical angle in List-II.
| List-I (\(n_{12}\)) | List-II (Critical angle) |
|---|---|
| (A) \(2\) | (I) \(60^\circ\) |
| (B) \(\sqrt{2}\) | (II) Total internal reflection is not possible |
| (C) \(\frac{2}{\sqrt{3}}\) | (III) \(45^\circ\) |
| (D) \(\frac{3}{4}\) | (IV) \(30^\circ\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
When medium 1 is denser, \( \sin i_c=1/n_{12} \). This needs \( n_{12} \) greater than 1. If \( n_{12} \) is less than 1, medium 1 is the rarer medium.
(A) \(2\) — This matches (IV) \(30^\circ\), because \( \sin i_c=1/2 \).
(B) \(\sqrt{2}\) — This matches (III) \(45^\circ\), because \( \sin i_c=1/\sqrt{2} \).
(C) \(\frac{2}{\sqrt{3}}\) — This matches (I) \(60^\circ\), because \( \sin i_c=\sqrt{3}/2 \).
(D) \(\frac{3}{4}\) — This matches (II). Here medium 1 is rarer than medium 2. The ray bends towards the normal and is always refracted, so there is no critical angle.
NCERT (2026-27):
Total Internal Reflection – “The refractive index of denser medium 1 with respect to rarer medium 2 will be n12 = 1/sin ic. …”
Refraction – “… if n21 > 1, r < i, i.e., the refracted ray bends towards the normal. In such a case medium 2 is said to be optically denser (or denser, in short) than medium 1. …”
Summary – “2. The critical angle of incidence ic for a ray incident from a denser to rarer medium, is that angle for which the angle of refraction is 90°. …”