Reflection of Light at a Plane Surface

The diagram to draw

Draw a plane reflecting surface $XY$ (the mirror), horizontal, with the plane of the paper as the plane of the figure. A plane wavefront $AB$ advances obliquely towards $XY$ with speed $v$, bounded by the incident rays $AA_{1}$ and $BB_{1}$. Mark the point $A_{1}$ where the wavefront first touches the mirror, and the point $P$ on the incident wavefront directly "above" $B_{1}$, so that $B$ still has the distance $PB_{1}$ to travel. Draw the hemispherical wavelet of radius $A_{1}R$ centred at $A_{1}$, and the reflected wavefront $RB_{1}$ tangential to it, bounded by the reflected rays $A_{1}R$ and $B_{1}Q$. Drop the normals $A_{1}M\perp XY$ and $B_{1}N\perp XY$.

Derivation of the laws of reflection using Huygens' principle

• A plane wavefront $AB$ is advancing obliquely towards the plane reflecting surface $XY$ with velocity $v$. $AA_{1}$ and $BB_{1}$ are the incident rays.
• When $A$ reaches $XY$ at $A_{1}$, the ray at $B$ has reached the point $P$ and still has to cover the distance $PB_{1}$ to reach the reflecting surface $XY$.
• Let $t$ be the time required to cover the distance $PB_{1}$. During this interval, secondary wavelets are emitted from $A_{1}$ and spread over a hemisphere of radius $A_{1}R$ in the SAME medium. The distance covered by the secondary wavelet from $A_{1}$ to $R$ in time $t$ is the same as the distance covered by the primary wave from $P$ to $B_{1}$:

$$A_{1}R=PB_{1}=vt$$

• All the other rays between $AA_{1}$ and $BB_{1}$ reach $XY$ after $A_{1}$ and before $B_{1}$, so they emit secondary wavelets of gradually decreasing radii.
• The surface touching all such hemispheres is $RB_{1}$, which is the reflected wavefront, bounded by the reflected rays $A_{1}R$ and $B_{1}Q$.
• Draw $A_{1}M\perp XY$ and $B_{1}N\perp XY$. Then the angle of incidence is $\angle AA_{1}M=\angle BB_{1}N=i$ and the angle of reflection is $\angle MA_{1}R=\angle NB_{1}Q=r$.

$$\therefore\ \angle RA_{1}B_{1}=90-r\qquad\text{and}\qquad \angle PB_{1}A_{1}=90-i$$

• In $\triangle A_{1}RB_{1}$ and $\triangle A_{1}PB_{1}$:

$$\angle A_{1}RB_{1}\cong\angle A_{1}PB_{1}\ \ (\text{both right angles})$$

$$A_{1}R=PB_{1}\ \ (\text{reflected waves travel equal distances in the same medium in equal times})$$

$$A_{1}B_{1}=A_{1}B_{1}\ \ (\text{common side})$$

$$\therefore\ \triangle A_{1}RB_{1}\cong\triangle A_{1}PB_{1}$$

$$\therefore\ \angle RA_{1}B_{1}=\angle PB_{1}A_{1}$$

$$\therefore\ 90-r=90-i$$

$$\therefore\ i=r$$

• This is the first law of reflection.
• It is also clear from the figure that the incident ray, the reflected ray and the normal all lie in the same plane — the plane of the paper. This is the second law of reflection.
• Hence reflection of light from a plane reflecting surface is explained on the basis of Huygens' principle.

Two consequences the textbook draws from the same figure

• Size of the image. If $AB$ is taken to be the size of the object, the distance between the corresponding reflected rays is the same as $AB$, because the two triangles are congruent. So the image in a plane mirror is the same size as the object.
• Lateral inversion. If $A$ is the right side of the object and $B$ the left side as it faces the mirror, then after reflection the right side at $A$ is seen at $R$ and the left side at $B$ is seen at $B_{1}$: right has become left and left has become right. This is called lateral reversal (lateral inversion). It does NOT occur during refraction at a plane surface.

Standard results

• $A_{1}R=PB_{1}=vt$ — equal distances, because both are travelled in the same medium.
• $\triangle A_{1}RB_{1}\cong\triangle A_{1}PB_{1}$ gives $i=r$.
• Incident ray, reflected ray and normal are coplanar.
• Image in a plane mirror: same size as the object, laterally inverted.