A plane wavefront in air reaches a flat glass surface. The wavefront makes an angle of \(60^\circ\) with the surface. The refractive index of the glass is \(\sqrt{3}\). The angle that the refracted wavefront makes with the surface is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(30^\circ\)
The angle between the incident wavefront and the surface is the angle of incidence \( i \), since \( \sin i=BC/AC \). In the same way, the refracted wavefront makes the angle of refraction \( r \) with the surface. Snell’s law then gives \( r \).
\( i = 60^\circ \)
\( \begin{aligned} \sin r &= \frac{\sin i}{n} \\ &= \frac{\sqrt{3}/2}{\sqrt{3}} \\ &= \frac{1}{2} \end{aligned} \)
\( r = 30^\circ \)
Hence, the refracted wavefront makes an angle of \( 30^\circ \) with the glass surface.
NCERT (2026-27):
Refraction of a plane wave – “… If we now consider the triangles ABC and AEC, we readily obtain”
\( \displaystyle \sin i=\frac{BC}{AC}=\frac{v_1\tau}{AC} \)\( \qquad (10.1) \)
“and”
\( \displaystyle \sin r=\frac{AE}{AC}=\frac{v_2\tau}{AC} \)\( \qquad (10.2) \)
“where i and r are the angles of incidence and refraction, respectively. …”
Refraction of a plane wave – “… In terms of the refractive indices, … can be written as”
\( \displaystyle n_1\sin i=n_2\sin r \)\( \qquad (10.6) \)
“This is the Snell’s law of refraction. …”