# Analytical treatment of Interference of Light

Consider ‘s’ be the sources of monochromatic light wave ‘S_{1}‘ and ‘S_{2}‘ are the two coherent sources emitted from single point ‘s’. Let λ be the wavelength of a monochromatic light, ‘a’ is the amplitude and ϕ is the phase difference between the two waves originating from the source ‘s’. Let the displacement of the wave originating from S_{1} be y_{1} = asinωT and the displacement of wave originating from S_{2} be y_{2} = asin(ωT + ϕ). After superposition, the displacement of the resultant wave is given by:

y = y1 + y2 = asinωT + asin(ωT + ϕ)

or, y = a(1 + cos ϕ)sinωt + asinϕ.cosωt……(i)

Let a(1 + cos ϕ) = Rcosθ

and asin ϕ = Rsinθ

Here, R is the amplitude of the resultant wave after superposition.

Now, equation (i) becomes,

y = Rsin θ.sinωT + Rsinθ .cosωT

or, y = Rsin(ωT + θ)……(ii)

Equation (ii) is the equation of simple harmonic motion.

Now, squaring and adding equation (i) and (ii),

R^{2} = 2a^{2}(1 + cos ϕ) = 4a^{2}cos^{2}ϕ/2)

The intensity at a point is given by:

I = R^{2} = 4a^{2}cos^{2}ϕ/2 = I_{o} cos^{2}ϕ/2

Here, I_{o} = 4a^{2 }.This is the expression for maximum intensity.

**Case I:**

When ϕ = 0, 2 π, 4 π…… I = 4a^{2}.

This is the case for bright fringe or constructive interference.

**Case II:**

When ϕ = π + 3π, 5 π…… then I = 0.

This is the condition for dark fringe or destructive interference.

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## One Response to “Analytical treatment of Interference of Light”

Thanks.Very Nice article.