In Fraunhoffer diffraction, at the angular position of first diffraction minimum, phase difference (in radian) between wavelets from opposite edges of the slit is

Question

In Fraunhoffer diffraction, at the angular position of first diffraction minimum, phase difference (in radian) between wavelets from opposite edges of the slit is (A) (π/2) (B) π (C) 2 π (D) 4 π

Tardigrade Admin · Accepted Answer

For minima in diffraction Δ x=m λ For first minima Δ x=λ Δ φ=2 π

Q. In Fraunhoffer diffraction, at the angular position of first diffraction minimum, phase difference (in radian) between wavelets from opposite edges of the slit is

$\frac{π}{2}$

$π$

$2 π$

$4 π$

For minima in diffraction $Δ x = mλ$
For first minima $Δ x = λ$
$Δ ϕ = 2 π$

Q. In Fraunhoffer diffraction, at the angular position of first diffraction minimum, phase difference (in radian) between wavelets from opposite edges of the slit is

2π​

π

2π

4π

For minima in diffraction Δx=mλ For first minima Δx=λ Δϕ=2π

$\frac{π}{2}$

$π$

$2 π$

$4 π$

For minima in diffraction $Δ x = mλ$
For first minima $Δ x = λ$
$Δ ϕ = 2 π$