A plane wavefront travels in water of refractive index 4/3. In the Huygens construction, secondary wavelets are drawn to find the position of this wavefront 1 ns later. The radius of each wavelet is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 22.5 cm
Each secondary wavelet spreads with the speed of the wave in the medium. So its radius after time \( \tau \) is \( v\tau \), with \( v=c/n \) in water.
\( \begin{aligned} v &= \frac{c}{n} \\ &= \frac{3\times10^{8}}{4/3} \\ &= 2.25\times10^{8}\ \mathrm{m\,s^{-1}} \end{aligned} \)
\( \begin{aligned} r &= v\tau \\ &= 2.25\times10^{8}\times1\times10^{-9} \\ &= 0.225\ \mathrm{m} \end{aligned} \)
Hence, each secondary wavelet has a radius of 22.5 cm.
NCERT (2026-27):
Huygens Principle – “Thus, if we wish to determine the shape of the wavefront at t = τ, we draw spheres of radius vτ from each point on the spherical wavefront where v represents the speed of the waves in the medium. …”
Refraction of a plane wave – “… Now, if c represents the speed of light in vacuum, then,”
\( \displaystyle n_1=\frac{c}{v_1} \)\( \qquad (10.4) \)
“… are known as the refractive indices of medium 1 and medium 2, respectively.”