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

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.
It states that when a completely plane polarised light beam is incident on an analyzer, the intensity of the emergent light varies as the square of the cosine of the angle between the plane of transmission of the analyzer and the polarizer.
I = I0cos2θ
Law: Rayleigh’s Criterion
According to Rayleigh’s criterion, two objects are just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other.
Law: Malus’ Law
I2 = I1cos2θ
It gives the intensity of plane polarized light after passing through a second polarizer, where θ is the angle between the axes of the two polarizers.
Law: Huygens' Principle
"Each point on a wavefront acts as a secondary source of light emitting secondary light waves called wavelets in all directions which travel with the speed of light in the medium. The new wavefront can be obtained by taking the envelope of these secondary wavelets travelling in the forward direction and is thus, the envelope of the secondary wavelets in forward direction. The wavelets travelling in the backward direction are in effective".
Law: Laws of Reflection
First Law of Reflection:
i = r
The angle of incidence is equal to the angle of reflection.
Second Law of Reflection:
The incident ray, reflected ray, and the normal at the point of incidence lie in the same plane.
Law: Snell’s Law of Refraction
n1 sin i = n2 sin r
\[\frac{\sin i}{\sin r}=\frac{v_1}{v_2}\]