A ray of light strikes an air-glass interface at an angle of incidence of \( 60^\circ \) and is refracted at an angle of refraction r. On increasing the angle of incidence, what happens to the angle of refraction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → It increases.
By Snell’s law, \( \sin i/\sin r=n_{21} \). Here the refractive index \( n_{21} \) of glass with respect to air is a constant greater than 1. It does not depend on the angle of incidence. Hence \( \sin r=\sin i/n_{21} \).
Option (1) It increases. — Correct. For angles between \( 0^\circ \) and \( 90^\circ \), a larger \( i \) means a larger \( \sin i \). With \( n_{21} \) fixed, \( \sin r \) grows in the same ratio, so the angle of refraction increases, while staying smaller than \( i \).
Option (2) It remains the same. — Incorrect. If \( r \) stayed the same while \( \sin i \) grew, the ratio \( \sin i/\sin r \) would change. But for the air-glass pair, this ratio is a constant.
Option (3) It decreases. — Incorrect. A smaller \( r \) with a larger \( i \) would raise the ratio \( \sin i/\sin r \) even more, against Snell’s law.
Option (4) It becomes \( 60^\circ \). — Incorrect. Light entering glass from air bends towards the normal, so \( r \) is always less than \( i \). Even at the largest angle of incidence, \( 90^\circ \), \( \sin r=1/n_{21} \), which for glass (\( n_{21} \) about 1.5) gives \( r \) of only about \( 42^\circ \).
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. … We note that n21 is a characteristic of the pair of media (and also depends on the wavelength of light), but is independent of the angle of incidence.”