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On the basis of Huygens Wave theory of light, show that angle of reflection is equal to the angle of incidence. You must draw a labelled diagram for this derivation

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#### Solution

Suppose that a plane wavefront of light is incident at a plane refracting surface MN. Let A_{1}B_{1} and AB be the successive positions of the incident wavefront. A_{1}A and B_{1}B the corresponding rays. When the wavefront reaches the point A, it becomes a secondary source and emits secondary waves in the same medium. Let c be the speed of light in the medIum If t is the time taken by the incident ray to cover the distance BC, then, BC = c t. During this time, the secondary waves originating at A cover same distance c t in the same medium.Therefore, the secondary spherical wavelet has a radius c t.

With A as the centre, draw a hemisphere of radius ct in the same medium. It represents the secondary wavelet.

According to Huygens’s principle locus of the tangent to all secondary wavelets represent new wavefront. Draw

a tangent CD to the secondary wavelet. As the points C and D are in the same phase of wave motion, CD

represents the corresponding reflected rays. Wavefront in the medium. It moves parallel to itself, tackling successive

positions C_{1}D_{1}, C_{2}D_{2} etc. AD_{2} and CC_{2} represent the corresponding reflected rays.

Proof: i be the angle of incidence and r be the angle of reflection.

From definition ∠BAC = i and ∠DCA = r

From Δ ABC and Δ BCA

BC = AD from construction

AC = AC common

`lfloorABC = lfloor ADC` right angles

Therefore Δ ABC and Δ DCA are congruent

∴ ∠BAC = ∠DCA

∴ i = r

Hence laws of reflection is proved.