Light goes from medium 1 into medium 2, and \(n_{21}\) is less than 1. Snell’s law then sets an upper limit on the angle of incidence for which refraction can take place. This upper limit exists because
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → the value of \(\sin r\) cannot exceed 1
Snell’s law gives \( \sin i=n_{21}\sin r \). When \( n_{21} \) is less than 1, the angle of refraction is larger than the angle of incidence. So \( r \) reaches \(90^\circ\) first. At that point \( \sin i=n_{21} \), and this angle of incidence is the critical angle.
Option (1) the angle of refraction cannot exceed the angle of incidence — Incorrect. Here medium 2 is the rarer medium. The ray bends away from the normal, so the angle of refraction is larger than the angle of incidence.
Option (2) \(n_{21}\) depends on the angle of incidence — Incorrect. \( n_{21} \) is a property of the pair of media. It does not change with the angle of incidence.
Option (3) the value of \(\sin r\) cannot exceed 1 — Correct. The largest value of \( \sin r \) is 1, at \( r=90^\circ \). So \( \sin i \) can be at most \( n_{21} \). A larger angle of incidence cannot satisfy the law, and no refraction takes place.
Option (4) some light is always reflected at the surface — Incorrect. Partial reflection takes place at every angle of incidence. It does not put any limit on refraction.
NCERT (2026-27):
Total Internal Reflection – “… if the relative refractive index of the refracting medium is less than one then, since the maximum value of sin r is unity, there is an upper limit to the value of sin i for which the law can be satisfied, that is, i = ic such that”
\( \displaystyle \sin i_c=n_{21} \qquad (9.12) \)