Help with single slit diffraction (fraunhoffer)

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In single slit diffraction, a minimum occurs at a*sin(θ) = m*λ because when light from opposite ends of the slit has a full wavelength phase difference, it results in destructive interference. This contrasts with double slit interference, where path differences lead to constructive interference at certain points. The key concept is that the light waves from different parts of the slit can be out of phase, causing cancellation. Understanding this phase relationship is crucial for analyzing diffraction patterns. Thus, the condition for minima is based on the principle of destructive interference.
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urgent Help with single slit diffraction (fraunhoffer)

why is it a minimum at a*sin (\theta) = m*\lambda,

a = slit separation I would have thought if the path difference is equal to a whole wavelength then there would be constructive interference??

See http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
 
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Don't confuse the double slit interference condition (http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html#c1").

In the single slit condition, when the light from opposite ends of the slit have a full wavelength phase difference, that means that half the light from the slit is exactly one half wavelength out of phase with the other half. Which mean destructive interference.
 
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