A ray of light passes from medium 1 into medium 2. The angle of incidence is \(30^\circ\) and the angle of refraction is \(60^\circ\). The refractive index of medium 1 with respect to medium 2 is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\sqrt{3}\)
Snell’s law gives \( n_{21} \), the refractive index of medium 2 with respect to medium 1. The question asks for \( n_{12} \), which is its reciprocal.
\( i=30^\circ \)
\( r=60^\circ \)
\( \begin{aligned} n_{21} &= \frac{\sin i}{\sin r} \\ &= \frac{1/2}{\sqrt{3}/2} \\ &= \frac{1}{\sqrt{3}} \end{aligned} \)
\( \begin{aligned} n_{12} &= \frac{1}{n_{21}} \\ &= \sqrt{3} \end{aligned} \)
From the above, \( n_{12}=\sqrt{3} \). Medium 1 is the denser medium, so the ray bends away from the normal.
NCERT (2026-27):
Refraction – “(ii) The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. … 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 – “If n21 is the refractive index of medium 2 with respect to medium 1 and n12 the refractive index of medium 1 with respect to medium 2, then it should be clear that”
\( \displaystyle n_{12}=\frac{1}{n_{21}} \)\( \qquad (9.11) \)