Q.Using Huygens' principles, prove the laws of reflection of light.

Constructing secondary wavelets from each point of the incident wavefront as it strikes a reflecting surface, and finding the envelope of these wavelets after equal time, geometrically forces the angle of reflection to equal the angle of incidence.

Let $XY$ be a plane reflecting surface, and $AB$ a plane wavefront incident on it at angle $i$ to the surface normal, with A having just touched the surface while B is still travelling toward it, a distance $BC = vt$ away (v = speed of light, t = time for the wave to travel from B to C).

By Huygens' principle, while the disturbance from B travels to C, the point A (already on the surface) also sends out a secondary wavelet into the same medium, of radius $AA' = vt$ (equal path since same speed, same time). The reflected wavefront is the common tangent (envelope) $A'C$ to all such secondary wavelets from every point between A and C.

In right triangles ABC and AA′C:

• $BC = AA' = vt$ (equal radii)
• $AC$ is common

So triangles ABC and AA′C are congruent (RHS), giving $\angle BAC = \angle A'CA$, i.e. the angle of incidence equals the angle of reflection:

$i = r$ …

A wavefront is a surface of constant phase; applying Huygens' construction to a wavefront reflecting off a plane surface shows that the angle of incidence equals the angle of reflection, and the incident ray, reflected ray and normal are coplanar.

a) Wavefront: A wavefront is the locus (surface) of all the points of a medium that are vibrating in the same phase at a given instant. It is perpendicular to the direction of propagation (ray) at every point.

b) Laws of reflection via Huygens' principle:

Let $AB$ be a plane wavefront incident on reflecting surface $MN$ at angle $i$ to the normal, travelling with speed $v$. When point $A$ of the wavefront touches the surface, point $B$ still has to travel distance $BC = vt$ to reach the surface (say at point $C$).

By Huygens' principle, while $B$ travels to $C$, the secondary wavelet from $A$ spreads out and travels the same distance $AD = vt$ (since both travel in the same medium with the same speed) in the same time. The new wavefront is the tangent plane $CD$ from the reflecting surface.

In right triangles $ABC$ and $ADC$:

• $BC = AD = vt$
• $AC$ is common
• Both are right angles at $B$ and $D$ …