# Young's Double Slit Experiment

1. May 12, 2011

### cupid.callin

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 $$\frac{\lambda L}{2d}$$
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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Last edited: May 12, 2011
2. May 12, 2011

### Gregg

What's the question?

3. May 12, 2011

### 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

4. May 13, 2011

### cupid.callin

hello ... someone?

5. May 13, 2011

### 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.

6. May 13, 2011

### cupid.callin

So you mean this is fringe width ??

well this looks satisfying

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7. May 13, 2011

### Dadface

You have now marked the fringe width correctly.

8. May 14, 2011

### cupid.callin

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

9. May 14, 2011

### 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.

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