Arrange the following steps in correct order to derive Snell's law us…

The correct answer is Option (1) → (D), (A), (B), (C)

Huygens’ principle: every point on a wavefront is a source of secondary wavelets, and the new wavefront is the common tangent to these wavelets.

(D) Consider the incident wave front approaching the interface separating the two media. — Start with the plane wavefront AB meeting the interface at angle of incidence i.

(A) Construct secondary wavelets in both media using different speeds (V₁ and V₂) of light. — In the time \( \tau \) that the other end B takes to reach the surface at C, travelling \( v_1\tau \) in the first medium, the wavelet from A spreads to a radius \( v_2\tau \) in the second medium.

(B) Identify the refracted wave front as the tangent plane to the secondary wavelets in the second medium. — The refracted wavefront CE is the tangent plane from C to this wavelet, at angle of refraction r.

(C) Show the relationship between the angles and velocities to obtain Snell's law. — From the two right-angled triangles with common hypotenuse AC, \( \sin i = v_1\tau/AC \) and \( \sin r = v_2\tau/AC \), which give Snell’s law:

    \( \displaystyle \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = n_{21} \)

So the correct order is (D), (A), (B), (C).

NCERT (2026-27):
Refraction of a Plane Wave – “… We assume a plane wavefront AB propagating in the direction A′A incident on the interface at an angle \( i \) …. … In order to determine the shape of the refracted wavefront, we draw a sphere of radius \( v_2\tau \) from the point A in the second medium (the speed of the wave in the second medium is \( v_2 \)). Let CE represent a tangent plane drawn from the point C on to the sphere. Then, \( \mathrm{AE} = v_2\,\tau \) and CE would represent the refracted wavefront. … Thus we obtain

    \( \displaystyle \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \qquad (10.3) \)”