Question

With the help of a neat diagram, explain the reflection of Light on a plane-reflecting surface.

Answer 1

Consider a plane wavefront AB of monochromatic light incident obliquely at an angle on a plane mirror MN as shown in the figure.

MN : plane mirror, RA and QC : Incident rays, AP and CS : Normals to MN, AB : Incident wavefront, CE : Reflected wavefront, i : Angle of incidence, r : Angle of reflection

The wave front AB touches the reflecting surface MN at A at time t = 0.

Let v be the speed of light in the medium. lf T is the time taken by the incident wavefront to travel from B to C.

Then BC = v T.

During this time, secondary wave originating from A covers the same distance, so that the secondary spherical wavelet has a radius vT at time T. Drow a tangent CE to the wavelet.

From figure,

∠RAP = ∠i = angle of incidence

∠PAE = ∠r = angle of redlection

In ΔABC and Δ AEC

AC is common side

AE = BE

∠ABC = ∠AEC

= 90°

∴ ΔABC and ΔAEC are congruent.

∴ ∠ACE = ∠BAC

= ∠i    ....(i)

Also as AE is perpendicular to CE and AP is perpendicular to AC

∠ACE = ∠PAE

= ∠r     ...(ii)

∴ From equations (i) and (ii),

∠ i = ∠ r

As a result, the angle of incidence and the angle of reflection are identical. The first rule of reflection is this.

Additionally, it is clear from the figure that incident and reflected rays both reside in the same plane but on opposing sides of the normal. The second low of reflection is at hand.

Thus, the laws of reflection of light ore deduced.