Young's Double Slit Experiment


by cupid.callin
Tags: double, experiment, slit, young
cupid.callin
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#1
May12-11, 11:24 AM
P: 1,135
1. The problem statement, all variables and given/known data




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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Gregg
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#2
May12-11, 12:02 PM
P: 463
What's the question?
cupid.callin
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#3
May12-11, 01:43 PM
P: 1,135
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

cupid.callin
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#4
May13-11, 12:03 AM
P: 1,135

Young's Double Slit Experiment


hello ... someone?
Dadface
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#5
May13-11, 02:56 AM
PF Gold
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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.
cupid.callin
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#6
May13-11, 04:28 AM
P: 1,135
So you mean this is fringe width ??

well this looks satisfying
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Dadface
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#7
May13-11, 05:05 AM
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Quote Quote by cupid.callin View Post
So you mean this is fringe width ??

well this looks satisfying
You have now marked the fringe width correctly.
cupid.callin
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#8
May14-11, 12:37 AM
P: 1,135
are you sure?
because fringe width means "width of 1 fringe"
but (as you said) i have marked width of 2 fringes
Dadface
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#9
May14-11, 02:36 AM
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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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