Question

using hugynes construction draw a figure showing propogation of plane wave reflecting at the interface of a mirror verify law of reflection​

Answer 1

Answer:

Laws of reflection by Huygen's principle: Let PQ be reflecting surface. Let a plane wavefront AB moving through the medium (air) towards the surface PQ meet at the point B. Let c be the velocity of light and t be the time of A to reach A' then AA' = ct.

Explanation:

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Answer 2

Explanation:

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Now in right-angled triangles ABB' and AA'B'

∠ABB' = ∠AA'B' (both are equal to 90°)

side BB' = side AA' (both are equal to vt)

and side AB' is common.

i.e., both triangles are congruent.

∴ ∠ BAB' = ∠ AB'A'

i.e., incident wavefront AB and reflected wavefront A'B' make equal angles with the

reflecting surface XY. As the rays are always normal to the wavefront, therefore the

incident and the reflected rays make equal angles with the normal drawn on the

surface XY, i.e.,

Angle of incidence i = Angle of reflection r

This is the second law of reflection.

Since AB, A'B' and XY are all in the plane of paper, therefore the perpendiculars

dropped

on them will also be in the same plane. Therefore we conclude that the incident ray,

reflected ray and the normal at the point of incidence, all lie in the same plane. This is

the

first law of reflection.

Thus Huygen’s principle explains both the laws of reflection.