According to Huygens principle, every point on AB is a source of secondary wavelets. Let the secondary wavelets from B strike M1M2 at A’ in t seconds.
Therefore, BA’ = c × t (1)
Where c is the velocity of light in the medium, the secondary wavelets from A will travel the same distance c × t in the same time. Therefore, with A as the centre and c × t as radius, draw an arc B’ so that
AB’ = c × t (2)
From A’ draw a tangent plane A’B’ touching the spherical arc tangentially at B’. Therefore, A’B’ is the secondary wavefront after t seconds. This would advance in the direction of rays 1’, 2’, 3’, which are the corresponding reflected rays perpendicular to A’B’.
For A’B’ to be true reflected wavefront, secondary wavelets starting from any other point D on the incident wavefront AB, must reach the point D’ on A’B’, after reflection at P, and that too, in the same time as the secondary wavelets take to go from B to A’. For this,
DP + PD’ = BA’ (1)
To prove it, draw PN ⊥ BA’
Therefore, DP = BN
In Δs ABA’ and AB’A’ = 90°
And AA’ is common, therefore Δs are congruent.
Therefore, ∠BAA’ = ∠B’A’A (2)
As PN || AB, therefore, ∠ BAA’ = ∠NPA’ (3)
From (2) and (3),
∠BAA’ = ∠B’A’A
= ∠D’A’P +∠ NPA’ (4)
Now, in Δs NPA’ and D’PA’
∠NPA’ = ∠D’PA” (already proved)
∠PNA’ = ∠PD’A’ = 90° and PA’ is common.
Therefore, Δs are congruent.
Therefore, NA’ = PD’
As DP = BN, therefore,
BN + NA’ = DP + PD’
Or BA’ = DP + PD’ which is the necessary condition as per eqn. (1).
Hence, A’B’ is the true reflected wavefront. In fig.
Angle of incident, i = ∠B’AA’
And angle of reflection, r = ∠B’A’A
From (2), ∠B’AA’ = ∠B’A’A i.e. i = r, which is the law of reflection.
Further, the incident wavefront AB, the reflecting surface M1M2 and the reflected wavefront A’B’ are the perpendicular to the plane of the paper. Therefore, incident ray, normal to the mirror M1M2 and reflected ray all lie in the plane of the paper. This is the second law of reflection.
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