Q.Using Huygens' geometrical construction, show that when a plane wavefront is reflected at a plane surface, the angle of incidence is equal to the angle of reflection.

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:

$i = r$

Also, since AB, AA′, and the reflected wavefront A′C all lie in the plane of the diagram containing the normal, the incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane. This proves both laws of reflection.

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 $D$

Hence triangles $ABC$ and $ADC$ are congruent, so $\angle BAC = \angle DCA$. Since $\angle BAC$ is the angle of incidence $i$ and $\angle DCA$ is the angle of reflection $r$:

$i = r$

Also, the incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane, which is the second law of reflection.