First slide

Diffraction

Question

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

Easy

A

$2 π$
B

$3 π$
C

$4 π$
D

$π λ$

Solution

For the second minimum, Path difference $2 λ$
Therefore, corresponding value of phase difference is

$Δ ϕ = \frac{2 π}{λ} \times Path difference ∴ Δ ϕ = \frac{2 π}{λ} \times 2 λ = 4 π$

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