Interference Analytical Treatment

Consider a monochromatic source of light S emitting waves of wavelength λ and two narrow pinholes A and B. A and B are equidistant from S and act as two virtual coherent sources. Let a be the amplitude of the waves. The phase difference between the two waves reaching the point P, at any instant is δ.

If y1 and y₂ are the displacements

y₁ = a sin ωt

y₂ = a sin (ωt + δ)

y = y₁ + y₂ = a sin ωt +  a sin (ωt + δ)

y = a sin ωt +  a sin ωt cos δ + a cos ωt sin δ

= a sin ωt (1 + cos δ) a cos ωt sin δ.

Taking a(1 + cos δ) = R cos θ                            (i)

and , a sin δ = R sin θ                                          (ii)

Y = R sin ωt cos θ + R cos ωt sin θ

Y = R sin (ωt + θ)                                                  (iii)

which represents the equation of simple harmonic vibration of amplitude R.

Squaring (i) and (ii) and adding.

R² sin²θ = R² cos²θ = a² sin²δ + a² (1 + cos δ)²

R² = a² sin²δ + a² (1 + cos²δ + 2cosδ)

R² = a² sin²δ + a² + a² cos²δ + 2a² cos δ

= 2a² (1 + cos δ)

R² = 2a² + 2 cos² δ/2 = 4a² cos² δ/2

The intensity at a point is given by the square of the amplitude

∴ I = R²

Or, I = 4a² cos² δ/2                                                (iv)

Special cases: (i) When the phase difference δ = 0, 2π, 2(2π), …… n(2π), or the path difference x = 0, λ, 2λ, nλ.

Intensity is maximum when the phase difference is a whole number multiple of 2π or the path difference is a whole number multiple of wavelength.

(ii) When the phase difference, δ = π, 3π, …. (2n + 1)π, or the path difference x = λ/2, 3λ/2, 5λ/2, …. (2n + 1) λ/2.

Intensity is minimum when the path difference is an odd number multiple of half wavelength.

(iii) Energy distribution: From equation (iv), it is found that the intensity at bright points is 4a² and at dark points it is zero. According to the law of conservation of energy, the energy cannot be destroyed. Here also the energy is not destroyed but only transferred from the points of minimum intensity. For, at bright points, the intensity due to the two waves should be 2a² but actually it is 4a². As shown in fig. the intensity varies from 0 to 4a², and the average is still 2a². It is equal to the uniform intensity 2a² which will be present in the absence of the interference phenomenon due to the two waves. Therefore the formation of interference fringes is in accordance with the law of conservation of energy.

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