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A parallel beam of light of wavelength ' $\lambda$ ' is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of incident beam. At the second minimum of the diffraction pattern, the phase difference between the ray coming from the two edges of slit is
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$4 \pi$
Condition for the $\mathrm{n}^{\text {th }}$ diffraction minimum is as follows:
$\Delta \mathrm{x}=\mathrm{n} \lambda$
For second minimum, $\mathrm{n}=2$ :
$\Delta \mathrm{x}=2 \lambda$
Corresponding to this minimum the phase difference is given by:
$\phi=2 \pi\left(\frac{\Delta x}{\lambda}\right)=2 \pi\left(\frac{2 \lambda}{\lambda}\right)=4 \pi$
$\Delta \mathrm{x}=\mathrm{n} \lambda$
For second minimum, $\mathrm{n}=2$ :
$\Delta \mathrm{x}=2 \lambda$
Corresponding to this minimum the phase difference is given by:
$\phi=2 \pi\left(\frac{\Delta x}{\lambda}\right)=2 \pi\left(\frac{2 \lambda}{\lambda}\right)=4 \pi$
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