The light rays travel from a glass slab with refractive index \( \mu_1 \) to the second slab with refractive index \( \mu_2 \). The refractive index of the second slab with respect to the first one is:
(Assume that the two slabs have the same thickness)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( {}_{1}\mu_{2} = \mu_2/\mu_1 \)
The refractive index of medium 2 with respect to medium 1 is the ratio of the speed of light in medium 1 to that in medium 2. Since \( \mu = c/v \), this becomes a ratio of the absolute indices.
\( \displaystyle {}_{1}\mu_{2} = \frac{v_1}{v_2} = \frac{c/\mu_1}{c/\mu_2} = \frac{\mu_2}{\mu_1} \)
Option (1) \( {}_{1}\mu_{2} = \mu_1/\mu_2 \) — Incorrect. \( \mu_1/\mu_2 \) is the index of medium 1 with respect to medium 2, the inverse of the one asked.
Option (2) \( {}_{1}\mu_{2} = \mu_2/\mu_1 \) — Correct. The index of the second slab with respect to the first is \( {}_{1}\mu_{2} = \mu_2/\mu_1 \).
Option (3) \( {}_{1}\mu_{2} = \mu_1\times\mu_2 \) — Incorrect. The relative index is a ratio; the product has no such meaning.
Option (4) \( {}_{2}\mu_{1} = \mu_1/\mu_2 \) — Incorrect. The relation \( \mu_1/\mu_2 \) is true, but for \( {}_{2}\mu_{1} \), the first slab with respect to the second; the question asks for \( {}_{1}\mu_{2} \).
NCERT (2026-27):
Refraction – “… We have
\( \displaystyle \frac{\sin i}{\sin r} = n_{21} \qquad (9.10) \)
where n21 is a constant, called the refractive index of the second medium with respect to the first medium.”
Refraction of a Plane Wave – “… Now, if c represents the speed of light in vacuum, then,
\( \displaystyle n_1 = \frac{c}{v_1} \qquad (10.4) \)
and
\( \displaystyle n_2 = \frac{c}{v_2} \qquad (10.5) \)
are known as the refractive indices of medium 1 and medium 2, respectively. In terms of the refractive indices, Eq. (10.3) can be written as
\( \displaystyle n_1 \sin i = n_2 \sin r \qquad (10.6) \)”