Young's double slit experiment

AI Thread Summary
In Young's double slit experiment, when one slit is covered, the intensity at the center of the fringe pattern changes from I to 0.25I. The initial assumption that intensity should increase to 4I due to reduced area is incorrect, as intensity is defined as power per unit area. With one slit open, the interference pattern is lost, resulting in a decrease in intensity. The energy distribution is more concentrated with two slits, leading to greater density in specific areas. Thus, the correct intensity with one slit covered is indeed 0.25I.
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Homework Statement


In a Younf's double slit experiment , the intensity of light at the centre of fringe pattern is I. If one of the two identical slit is now covered , the intensity at the cnetre ANS is 0.25I.


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The Attempt at a Solution


In my opinion, it should be 4I. because the light ray is now concentrated at smaller area, but i know that intensity = power /area . As the area is decreased is decreased , the intensity should increase , assuming the power is constant.
 
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If you check against diffraction formulas, you'll see that the answer is \frac 1 4 I=0.25 I.
But if you want arguments using density considerations, I should say that when you have two slits open, there is interference which means you have dark and bright bands and so the energy coming from the source has less places to be at which means greater density in those places than when you have only one slit open where light has all the wall to be at.
 
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