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Find the intensity at a point on a screen in YDSE, where the interfering waves of equal intensity have a path difference of (i) `lambda//4` (ii) `lambda//3`.

Question

Find the intensity at a point on a screen in YDSE, where the interfering waves of equal intensity have a path difference of (i) lambda//4 (ii) lambda//3.

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Solution

In Young's Double Slit Experiment (YDSE), the intensity at a point on the screen is given by the formula:

I = 4I₀ cos²(Φ/2)

where I₀ is the intensity of each of the two interfering waves and Φ is the phase difference between the two waves.

(i) When the path difference is λ/4, the phase difference Φ is (2π/λ) * (λ/4) = π/2. Substituting this into the intensity formula gives:

I = 4I₀ cos²(π/4) = 4I₀ * (1/√2)² = 2I₀

(ii) When the path difference is λ/3, the phase difference Φ is (2π/λ) * (λ/3) = 2π/3. Substituting this into the intensity formula gives:

I = 4I₀ cos²(π/3) = 4I₀ * (1/2)² = I₀

So, the intensity at a point on the screen is 2I₀ when the path difference is λ/4 and I₀ when the path difference is λ/3.

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