# Double Slits Question

1. Feb 1, 2014

### cathy

1. The problem statement, all variables and given/known data

The figure shows the light intensity on a screen behind a double slit. The slit spacing is 0.20 mm and the wavelength of the light is 600 nm. What is the distance from the slits to the screen?

This is the figure
http://session.masteringphysics.com/problemAsset/1020631/6/jfk.Figure.17.P38.jpg

2. The attempt at a solution

I have the answer there (attached thumbnail) from what the teacher provided, but I do not understand where 1/4 came from. Will someone please explain this to me? I am very confused. I am also confused on how the intensity diagram shown is used.

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Last edited: Feb 1, 2014
2. Feb 2, 2014

### Simon Bridge

To understand a model answer - 1st ignore it and use your own understanding to complete the problem yourself.
Just work through the problem and you'll see why you need the intensity pic.

Last edited by a moderator: Feb 2, 2014
3. Feb 2, 2014

### cathy

http://imgur.com/XFvgXj7

Would you be able to explain to me where the 1/4 came from? I understand every other part except for that.

And I now understand that the intensity is just to show me the .20 mm

I know that Idouble= I 1(4)
Would it have something to with that? If so, how?

4. Feb 2, 2014

First let me make clear what does this intensity diagram depict:

1) The maxima i.e, the tip of the mountain/curve signifies that at that position the intensity is maximum or, the path difference between the light rays of two sources is n$\lambda$

2) The points where the graph touches the x-axis is where the intensity is zero, i.e, the paths difference is $(2n+1)\lambda/2$

3) Note that, the distance between the $1^{st}$ maxima/tip of the mountain and the $5^{th}$ tip is 2cm. And there are 4 gaps between the first and the 5th. So, if 4 gaps amount to a width of 2cm, then 1 gap is $(1/4)*2.0 cm$ wide.

I hope you got it now. Do notify if you need any further help.

5. Feb 2, 2014

### cathy

Thank you so much for explaining everything. I understand it a bit better. Just another question, when the intensity is 0, will $(2n+1)\lambda/2$ always be the path difference?

For some reason, my textbook is giving me the equation

(m+1)$\lambda/$L / d
and m$\lambda/$L /d
subtracted to give $\lambda/$L /d

Last edited: Feb 2, 2014
6. Feb 2, 2014

### Simon Bridge

For zero intensity, you need 100% destructive interference ... that means that the trough of one wave matches with the peak of the other. So - yeah.

7. Feb 2, 2014

YES.

Resultant intensity from two sources of intensities $I_{1}$ and $I_{2}$ is given by:

$I_{resultant} = I_{1}$ + $I_{2}$ + $2\sqrt{I_{1}*I_{2}}cosø$

where ø = Phase difference between the two waves

Since the slits are equidistant from the source, $I_{1} = I_{2}$

Now you have the formulae with you; think when and at what ø will $I_{resultant}$ be zero.

8. Feb 2, 2014

m$\lambda/$L represents the height from the central maximum where there is a bright fringe due to a path difference of m$\lambda$. Similarly m+1$\lambda/$L corresponds to the height from the central maximum where there is a bright fringe due to a path difference of m+1$\lambda$.

So the fringe width = distance between the above two fringes = $\lambda/$L

P.S : I think you need to revisit your class notes/ source for a clearer understanding of Young's double slit experiment. Or else, you can watch these videos:

9. Feb 2, 2014

### cathy

Okay, I understand.
Thank you everyone.

10. Feb 2, 2014

### cathy

Okay, I understand.
Thank you everyone.

11. Feb 18, 2014

### Tyrone Darryl

Thank you!

Thank you! Now I understand how to do this problem!