Question
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.
Solution
We are given a plane mirror XY and let, O be a point object at a distane OP, in front of the plane mirror. A part RPQ of the wavefront touches the plane mirror at P and from this point spherical wavefronts start emanating.
Whereas disturbance from R and Q continue moving forward, along the normal rays OR and OQ, that reflects back v. When, disturbances from R, P and Q reach the mirror at A, B'and C respectively, reflected spehrical wavefront is formed.
Whereas disturbance from R and Q continue moving forward, along the normal rays OR and OQ, that reflects back v. When, disturbances from R, P and Q reach the mirror at A, B'and C respectively, reflected spehrical wavefront is formed.
The reflected wavefront AB'C, appears to start from I. Hence, I becomes virtual image for O as real point object.
Draw AN normal to XY, hence parallel to OP.
Now, OA is the incident ray (being normal to incident wavefront ABC) and AD is reflected ray (being normal to reflected wavefront AB'C).
Thus, ∠OAN = ∠DAN = ө [i = r]
But, ∠OAN = alternate ∠AOP
and ∠DAN = corresponding ∠AIP
∠AOP = ∠AIP
Now, in ∆AIP and ∆AOP
∠AIP = ∠AOP (each ө)
∠APl = ∠APO = 90° (each 90°)
AP is common to both
∆s become congruent
Hence, PI = PQ
i.e., normal distance of image from the mirror = normal distance of object from the mirror.
Thus, virtual image is formed as much behind the mirror as the object in front of it
