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Young's Double Slit Experiment

  • #1
1,137
0

Homework Statement



attachment.php?attachmentid=35479&stc=1&d=1305217898.png



so here i have shown the distances of two dark fringes ...

from here ... fringe width is [tex]\frac{\lambda L}{2d}[/tex]
L is distance of screen ... d is distance b/w 2 fringes.

but fringe width is [tex]\frac{\lambda L}{d}[/tex] , right ???

What mistake am i making?
 

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Answers and Replies

  • #2
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What's the question?
 
  • #3
1,137
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i wrote it above

by using formula of distance of dark fringe ...

fringe width comes out to be ... [tex]
\frac{\lambda L}{2d}
[/tex]

but fringe width is [tex]
\frac{\lambda L}{d}
[/tex]

so why am i getting wrong answer
 
  • #4
1,137
0
hello ... someone?
 
  • #5
2,449
87
The fringe width is the distance between the centres of alternate dark(or bright) fringes.
Lamda D/d as marked on your diagram goes from the centre of a bright fringe to the centre of an adjacent dark fringe in other words you have marked in half the fringe width.
 
  • #6
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So you mean this is fringe width ??
attachment.php?attachmentid=35509&stc=1&d=1136054234.png

well this looks satisfying
 

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  • #7
2,449
87
So you mean this is fringe width ??
attachment.php?attachmentid=35509&stc=1&d=1136054234.png

well this looks satisfying
You have now marked the fringe width correctly.
 
  • #8
1,137
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are you sure?
because fringe width means "width of 1 fringe"
but (as you said) i have marked width of 2 fringes
 
  • #9
2,449
87
I did not say you marked in the width of two fringes I said you marked in the width of half a fringe.The fringe width as marked above in red is correct since it goes from the centre of a bright fringe to the centre of an adjacent bright fringe.
Look at a graph of how the intensity varies smoothly across an interference pattern and this should clarify things for you.
 

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