Question

With the help of a diagram, show how a plane wave is reflected from a surface. Hence, verify the law of reflection.

Draw the reflected wave front for a plane wave front incident on a plane reflecting surface. Hence, verify the laws of reflection using Huygen’s principle.

A plane wave is incident on a reflecting surface. Using Huygens principle, show how it is reflected from the surface. Hence, verify the law of reflection.

Answer 1

According to the laws of reflection:

• At the point of incidence, the incident rays, reflected rays, and normal to the reflecting surface all lie in the same plane.
• On opposing sides of the normal are the incident and reflected rays.
• The angle of incidence and the angle of reflection are the same. i.e., ∠i = ∠r.

Explanation:

                Reflection of light

XY: Plane reflecting surface

AB: Plane wavefront

RB₁: Reflecting wavefront

A₁M, B₁N: Normal to the plane

∠AA₁M = ∠BB₁N = ∠i = Angle of incidence

∠TA₁M = ∠QB₁N = ∠r = Angle of reflection

A plane wavefront AB is advancing obliquely towards the plane reflecting surface XY. The AA₁ and BB₁ are incident rays.

When ‘A’ reaches XY at A₁, then the ray at ‘B’ reaches point ‘P’, and it has to cover the distance PB₁ to reach the reflecting surface XY.

Let ‘t’ be the time required to cover the distance PB₁. During this time interval, secondary wavelets are emitted from A₁ and will spread over a hemisphere of radius A₁R, in the same medium. The distance covered by secondary wavelets to reach from A₁ to R in time t is the same as the distance covered by primary waves to reach from P to B₁. Thus, A₁R = PB₁ = ct.

All other rays between AA₁ and BB₁ will reach XY after A₁ and before B₁. Hence, they will also emit secondary wavelets of decreasing radii.

The surface touching all such hemispheres is RB₁ which is the reflected wavefront, bounded by reflected rays A₁R and B₁Q.

Draw A₁M ⊥ XY and B₁N ⊥ XY.

Thus, the angle of incidence is ∠AA₁M = ∠BB₁N = i, and the angle of reflection is ∠MA₁R = ∠NB₁Q = r.

∠RA₁B₁ = 90 − r

∠PB₁A₁ = 90 − i

In ΔA₁RB₁ and ΔA₁PB₁

∠A₁RB₁ = ∠A₁PB₁

A₁R = PB₁    ....(Reflected waves travel an equal distance in the same medium in equal time.)

A₁B₁ = A₁B₁     ....(Common side)

∴ ΔA₁RB₁ ≅ ΔA₁PB₁

∴ ∠RA₁B₁ = ∠PB₁A₁

∴ 90 − r = 90 − i

∴ i = r

Also from the figure, it is clear that incident rays, reflected rays, and normal lie in the same plane.

This explains the laws of reflection of light from a plane reflecting surface on the basis of Huygen’s wave theory.

Frequency, wavelength, and speed of light do not change after reflection. If reflection takes place from a denser medium, then the phase changes by π radians.

Answer 2

AB = Incident wavefront

CD = Reflected wavefront

XY = Reflecting surface

If c be the speed of light and t be the time taken by light to go from B to C or A to D or E to G through F, then

t = `(EF)/C + (FG)/C`

= `(AF sin i)/C + (FC sin r)/C`

= `(AC sin r + AF(sin i - sin r))/C`

For rays of light from different parts of the incident wavefront, the values of AF are different. But light from different points of the incident wavefront should take the same time to reach the corresponding points on the reflected wavefront.

So, ‘t’ should not depend upon AF.

This is possible only if sin i – sin r = 0.

i.e., sin i = sin r

⇒ i = r

Hence proved.