A plane wavefront passes from water into a transparent liquid. The speed of light in water is \(2.25\times10^{8}\ \mathrm{m\,s^{-1}}\). In the Huygens construction, one edge of the wavefront moves 10 cm in water to reach the surface. In the same time, the secondary wavelet from the other edge grows to a radius of 8 cm in the liquid. The refractive index of the liquid is (speed of light in vacuum \(=3\times10^{8}\ \mathrm{m\,s^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.67
In the Huygens construction, the edge in water and the wavelet in the liquid move for the same time \( \tau \). So the two distances are in the ratio of the two speeds. The refractive index then follows from \( n=c/v \).
\( \begin{aligned} BC &= v_1\tau=10\ \mathrm{cm} \\ AE &= v_2\tau=8\ \mathrm{cm} \end{aligned} \)
\( \begin{aligned} v_2 &= v_1\times\frac{AE}{BC} \\ &= 2.25\times10^{8}\times\frac{8}{10} \\ &= 1.8\times10^{8}\ \mathrm{m\,s^{-1}} \end{aligned} \)
\( \begin{aligned} n_2 &= \frac{c}{v_2} \\ &= \frac{3\times10^{8}}{1.8\times10^{8}} \\ &= 1.67 \end{aligned} \)
Hence, the refractive index of the liquid is about 1.67.
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 – “… we draw a sphere of radius v2τ from the point A in the second medium … Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2τ and CE would represent the refracted wavefront. …”
Refraction of a plane wave – “… Now, if c represents the speed of light in vacuum, then,”
\( \displaystyle n_1=\frac{c}{v_1} \)\( \qquad (10.4) \)
“and”
\( \displaystyle n_2=\frac{c}{v_2} \)\( \qquad (10.5) \)
“are known as the refractive indices of medium 1 and medium 2, respectively. …”