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.

Explanation

In the figure given below, let O be a point object that is placed in fromt of the plane mirror YZ, at a normal distance of OP from it.
From point O, the spherical wavefront originates. One part QPR of the wavefront touches the plane mirror at point P. The optical disturbances from points Q and R proceed forward along the rays OQ and OR, respectively, but from point P it reflects back . At the instant of time when the disturbances from points Q and R reach the mirror at points C and A, respectively , the disturbance from point P reaches B.. In this way, reflected spherical wavefront CB.A is formed. In the figure, CBA (shown as dashed) shows position of the wavefront if there were no mirror present.

It seems that the reflected wavefront CB.A starts from point O.. So O. becomes virtual point image for point object O.
Draw CM, normal to YZ, hence parallel to PO.
Refer to the figure, OC is the incident ray (as it is perpendicular to the incident wavefront CBA) and CD is reflected ray (as it is perpendicular to the incident wavefront CB.A).
Hence, `/_OCM = /_DCM = theta ( :. i = r)`
But `/_OCM = /_COP` (alternate angles)
and `/_DCM = /_CO.P` (corresponding angles)
`/_COP = /_CO.P`
Now in triangles `CO.P and COP`,
`/_CO.P = /_COP ("each" = theta)`
`/_CPO. = /_CPO ("each" = 90^@)`
CP is common to both the triangles. Thus, triangules CO.P and COP are congruent.
Hence, `PO. = PO`
This means that the normal distance of the object and the image from the mirror are equal.