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Single slit diffraction and minimum intensity

In the single slit diffraction the condition for minimum intensity is that
Del S= n*lamda n=1,2,3,.......natural number
the extreme path difference is equal to integral numbers of wavelength ( in the double slit experiment it is the condition for maximum intensity).

It is understandable in the case of double slit experiment when the path difference is equal to integral numbers of wavelength, then the two wave meets at the same phase, then we have maximum intensity at that particular point.
But when it comes with single slit the scenario is reversed. Isn't it the only factor that determine the intensity of light wave at any particular point -with which EXTENT OF PHASE DIFFERENCE two wave meet there?

So why Del S= n*lamda for minimum intensity? If the extreme path difference Del S is the only and net path difference then the waves having extreme path difference Del S= n*lamda will meet with same phase, giving rise to a maximum.
when the light wave pass through the slit it is diffracted to certain amount and fall on the lens, these are parallel rays and the lens converge them on the focal plane. But these rays are not parallel to the principle axis of the lens. So the converging path of the two rays are not equal. I justifies (i may be wrong , cos i have no clues; my textbooks explained nothing about this)that this path difference should be odd multiples of half wavelength (i.e m*lamda/2 where m=1,3,5,7.....) . SO that when two rays have the path difference Del S= n*lamda on to the lens ( extreme path difference) they would still have to follow a converging path that will force them to meet out of phase and thus giving rise to minumum intensity.
These are how i can think so far. Please make correction of my analysis.
Waiting for you sensible response. :shy:
 
Last edited:
http://hyperphysics.phy-astr.gsu.edu...t/sinslit.html [Broken] says the condition for minimum intensity is asinθ=mλ
Isn't asinθ=extreme path difference?
 
Last edited by a moderator:

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