A ray of light travelling in water (refractive index 4/3) falls on a glass surface (refractive index 3/2). The angle of incidence is \(30^\circ\). The angle of refraction in the glass is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \sin^{-1}(4/9) \)
The light goes from water into glass. So Snell’s law needs the refractive index of glass with respect to water, \( n_{gw} \). We find it from the two values given with respect to air. Here \( n_{aw} \) is the reciprocal of \( n_{wa} \), so \( n_{aw}=3/4 \).
\( n_{ga}=3/2 \)
\( n_{wa}=4/3 \)
\( i=30^\circ \)
\( \begin{aligned} n_{gw} &= n_{ga}\times n_{aw} \\ &= \frac{3}{2}\times\frac{3}{4} \\ &= \frac{9}{8} \end{aligned} \)
\( \begin{aligned} \sin r &= \frac{\sin i}{n_{gw}} \\ &= \frac{1/2}{9/8} \\ &= \frac{4}{9} \end{aligned} \)
Hence, the angle of refraction in the glass is \( \sin^{-1}(4/9) \).
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 n32 is the refractive index of medium 3 with respect to medium 2 then n32 = n31 × n12, where n31 is the refractive index of medium 3 with respect to medium 1.”