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In a Young’s double-slit setup, the intensity at a point where the path difference is λ/6 (λ being the wavelength of the light used) is I. If I₀ represents the maximum possible intensity, then the ratio I/I₀ is

1. 3/4
2. 1/√2
3. √3/2
4. 1/2

Correct answer: 3/4

Solution

The intensity at a point in a double-slit experiment can be calculated using the formula I = I₀ cos²(φ/2), where φ is the phase difference. A path difference of λ/6 corresponds to a phase difference of π/3, leading to cos²(π/6) = (√3/2)² = 3/4, thus the ratio I/I₀ is 3/4.

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