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Young's Double Slit Experiment 
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#1
May1211, 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? 


#2
May1211, 12:02 PM

P: 463

What's the question?



#3
May1211, 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 


#4
May1311, 12:03 AM

P: 1,135

Young's Double Slit Experiment
hello ... someone?



#5
May1311, 02:56 AM

PF Gold
P: 2,031

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
May1311, 04:28 AM

P: 1,135

So you mean this is fringe width ??
well this looks satisfying 


#8
May1411, 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 


#9
May1411, 02:36 AM

PF Gold
P: 2,031

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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