Question

On the basis of Huygen's Wave theory of light, show that angle of reflection is equal to angle of incidence. You must draw a labelled diagram for this derivation.

Answer 1

Let SS' be the section of a plane reflecting surface and AB that of a plane wave front striking at A Let v be the velocity of light and seconds the time for the edge B of the wave front to reach the surface at A'.
According to Huygens' principle, such point on the wave front acts as a surface of secondary wavelets. In the absence of SS', the wave front AB would have advanced to the position A'D after is time t where A'D is parallel to AB. But at the presence of SS', as the wave front advances, the points on SS' successively struck by the wave front become the surface of secondary spherical wavelets. Thus, after a time t when the wave front strikes the point A', the secondary wavelet form A has acquired a radius
$A{B}^{\prime }=B{A}^{\prime }=AD=vt$,
By the time incident wave front strike A', the reflected ray has already covered a distance AB' in the medium.
$A{B}^{\prime }=A^{\prime }B=vt$
From triangle ABA' and A'B'A
$\angle B=\angle B^{\prime }={90}^o$

A'A common $A{B}^{\prime }=A^{\prime }B=vt$
Therefore, the two triangles are congruent
$\therefore \angle BA{A}^{\prime }=\angle B^{\prime }{A}^{\prime }{A}^{\prime }$
or $\angle i=\angle r$
that is, the angle of incidence is equal to the angle of reflection. This is the second law of reflection. Since AB, A'S' and SS' are in the plane of the paper, they will also be in the same plane. Therefore, the incidence ray, the reflected ray and the normal at the point of incidence are all in the same plane. This is the first law of reflection.