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

\frac{π}{2}

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π

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2 π

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4 π

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Solution

The correct option is C $2 π$
For minima in diffraction $Δ x = n λ$

For first minima $Δ x = λ (∵ n = 1)$
and, $Δ ϕ = \frac{2 π}{λ} \times Δ x$

So, putting the value of $Δ x$ in above equation we get
$Δ ϕ = \frac{2 π}{λ} \times λ = 2 π$

Hence, $(C)$ is the correct answer.

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

The correct option is C 2πFor minima in diffraction Δx=nλ For first minima Δx=λ (∵n=1) and, Δϕ=2πλ×Δx So, putting the value of Δx in above equation we get Δϕ=2πλ×λ=2π Hence, (C) is the correct answer.

The correct option is C $2 π$
For minima in diffraction $Δ x = n λ$

For first minima $Δ x = λ (∵ n = 1)$
and, $Δ ϕ = \frac{2 π}{λ} \times Δ x$

So, putting the value of $Δ x$ in above equation we get
$Δ ϕ = \frac{2 π}{λ} \times λ = 2 π$

Hence, $(C)$ is the correct answer.