A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is

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2π

3π

4π

πλ

answer is C.

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Detailed Solution

For the second minimum, Path difference 2λ Therefore, corresponding value of phase difference is Δϕ=2πλ× Path difference ∴ Δϕ=2πλ×2λ=4π

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