Young's double slit experiment

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SUMMARY

In Young's double slit experiment, when one of the two identical slits is covered, the intensity at the center of the fringe pattern reduces to 0.25I from I. This reduction is due to the interference effects present when both slits are open, which create a distribution of light intensity across multiple bands. The argument presented highlights that while covering one slit decreases the area for light distribution, the intensity does not increase as initially assumed, but rather decreases due to the lack of interference, confirming that intensity is calculated as power divided by area.

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  • Understanding of Young's double slit experiment
  • Knowledge of light intensity and its relationship to power and area
  • Familiarity with interference patterns in wave physics
  • Basic grasp of diffraction principles
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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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