A plane wavefront travelling in air strikes a flat glass surface at an angle of incidence of \(30^\circ\). One edge of the wavefront touches the surface 0.5 ns before the other edge. The distance between the two points where the edges touch the surface is (speed of light in air \(=3\times10^{8}\ \mathrm{m\,s^{-1}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30 cm
Let edge A touch the surface first. In the time \( \tau \), the other edge B moves along its ray to the point C on the surface, so \( BC=v_1\tau \). In the right triangle ABC, the angle at A is the angle of incidence, so \( \sin i=BC/AC \).
\( \tau = 0.5\ \mathrm{ns}=0.5\times10^{-9}\ \mathrm{s} \)
\( \begin{aligned} BC &= v_1\tau \\ &= 3\times10^{8}\times0.5\times10^{-9} \\ &= 0.15\ \mathrm{m} \end{aligned} \)
\( \begin{aligned} AC &= \frac{BC}{\sin i} \\ &= \frac{0.15}{\sin 30^\circ} \\ &= 0.30\ \mathrm{m} \end{aligned} \)
Hence, the two points of contact are 30 cm apart.
NCERT (2026-27):
Refraction of a plane wave – “… Let τ be the time taken by the wavefront to travel the distance BC. Thus,”
\( \displaystyle BC = v_1\tau \)
Refraction of a plane wave – “… If we now consider the triangles ABC and AEC, we readily obtain”
\( \displaystyle \sin i=\frac{BC}{AC}=\frac{v_1\tau}{AC} \)\( \qquad (10.1) \)