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Question:

The intensity at the central maximum (O) in a Young's double slit experimental set-up is I_O. If the distance OP equals one-third of the fringe width of the pattern, show that the intensity at point P would equal I_O/4.

Solution:

In YDSE the expression for intensity is given by: I = I₁ + I₂ + 2√(I₁I₂)cos φ. (1)
Where,
I₁ is the intensity of slit S₁
I₂ is the intensity of slit S₂
φ = phase difference
Take I₁ = I₂ = I (Both slits have same intensity).
φ = (2π/λ)Δx
Also, Δx = (x d)/D ⇒ φ = (2πλ × x d)/D
At O:
Δx = 0 (No path difference at O)
φ = 0
Imax = I₀ = I + I + 2I²cos 0 = 2I + 2I = 4I
I₀ = 4I ∴ I = I₀/4 (2)
Fringe width = β = (λD)/d
Also, x = (1/3)β = (1/3)(λD)/d
Intensity at D = Ip = I₁ + I₂ + 2√(I₁I₂)cos[(2πλ × x d)/D]
Ip = I + I + 2I²cos[(2πλ × d/D × (1/3)(λD)/d)]
Ip = 2I + 2Icos(2π/3) = 2I + 2Icos(π - π/3)
Ip = 2I + 2I × (-cos π/3) = 2I + 2I × (-1/2) = 2I - I = I
Ip = I ∴ Ip = I₀/4
Thus intensity at point P will be I₀/4
Hence proved