Q.You have learnt in the text how Huygens' principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Applying Huygens' construction to a spherical wavefront reflecting off a plane mirror shows that every point on the mirror is equidistant from the source and its mirror-image point, so the reflected wavefronts are spherical and appear to diverge from that image point — placed exactly as far behind the mirror as the source is in front.

Step 1: Set up the geometry

Let $S$ be a point source at perpendicular distance $a$ from a plane mirror $MM'$, with $O$ the foot of the perpendicular from $S$ onto the mirror. A spherical wavefront centred on $S$ expands outward and begins striking the mirror.

Step 2: Huygens' construction for the reflected wave

Each point on the mirror, as soon as the incident wavefront reaches it, becomes a source of secondary spherical wavelets travelling back into the region containing $S$ (this is Huygens' principle applied to reflection). Consider a general point $A$ on the mirror at distance $x$ from $O$. The wavefront from $S$ reaches $A$ after travelling a distance $SA = \sqrt{a^2+x^2}$, i.e. after a time $t_A = SA/c$. From that instant, $A$ emits its own secondary wavelet.

Step 3: Locate the candidate image point

Let $S'$ be the point on the far side of the mirror, on the normal through $O$, with $OS' = a$ (i.e. the geometric mirror image of $S$). Because $A$ lies in the mirror plane, which is the perpendicular bisector of segment $SS'$, every such point $A$ is automatically equidistant from $S$ and $S'$:

Step 4: Show the reflected wavefront is a sphere centred at $S'$

At some later total time $T$ (measured from when the wave left $S$), the secondary wavelet from $A$ has been expanding for a time $(T - SA/c)$, so it has radius

The point on this particular wavelet lying on the line from $S'$ through $A$, extended beyond $A$ (away from $S'$), is at a distance from $S'$ equal to:

This distance, $cT$, is the same for every point $A$ on the mirror, regardless of $x$ — it does not depend on which point of the mirror we picked.

Step 5: Conclusion

Since the envelope (tangent surface) of all the reflected secondary wavelets touches each one at a distance exactly $cT$ from $S'$, independent of position, the reflected wavefront is itself a sphere of radius $cT$ centred at $S'$. This is precisely a spherical wave diverging from the point $S'$ — so the mirror produces a virtual image at $S'$, located at perpendicular distance $OS' = a$ behind the mirror, exactly equal to the object distance $OS = a$ in front of it.