Young's Double Slit Experiment

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?

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What's the question?

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

hello ... someone?

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.

So you mean this is fringe width ??

well this looks satisfying

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So you mean this is fringe width ??

well this looks satisfying
You have now marked the fringe width correctly.

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

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.