Young's Double Slit Experiment

[cupid.callin]

Homework Statement

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

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

but fringe width is $$\frac{\lambda L}{d}$$ , right ?

What mistake am i making?

 

[Gregg]

What's the question?

 

[cupid.callin]

i wrote it above

by using formula of distance of dark fringe ...

fringe width comes out to be ... $$\frac{\lambda L}{2d}$$

but fringe width is $$\frac{\lambda L}{d}$$

so why am i getting wrong answer

 

[cupid.callin]

hello ... someone?

 

[Dadface]

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]

So you mean this is fringe width ??

well this looks satisfying

 

[Dadface]

So you mean this is fringe width ??

well this looks satisfying

You have now marked the fringe width correctly.

 

[cupid.callin]

are you sure?
because fringe width means "width of 1 fringe"
but (as you said) i have marked width of 2 fringes

 

[Dadface]

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.