# Homework Help: Interference-Diffraction Pattern

1. Mar 25, 2013

### jaumzaum

I'm having trouble to understand some concepts about interference and diffraction in waves. I've already seen thousands of exercise like the following: "Consider a 2-slit experiment in which we have both interference and diffraction in the slits. The distance between the slits is d and the slit length is a, the distance between the slits and the screen is L. We then put a material of length l and refraction index n behind the first slit. If the maximum of interference is displaced by N fringes... " And then we have to calculate the wave length or any other stuff.

The question is: If the problem says the maximum of interference is displaced by some distance, it's intuitive to say that the maximum of interference is something that can be easily identified. Is the maximum of interference the brightest maximum in both cases? Why?

I can't even understand why it is the brightest in the first case (when it is right between the fringes). For example: If d = 2Lλ/a, We would have a destructive interference right between the slits and the central "maximum" would be no bright at all !

Please someone help me, I'm pretty lost in this

Thanks, John

2. Mar 25, 2013

### Emilyjoint

One fact that may be useful: if white light shines through 2slits then the central max is white. All of the other max show as white light spectra.
Also, as you mention, the central fringe is brightest.

3. Mar 25, 2013

### jaumzaum

Thanks Emilyjoint. I had not thought of that.

For the middle fringe, we have destructive diffraction when the distance x in screen is kLλ/a, where k = 1,2,3...
If d/2 = Lλ/a the central maximum would have intensity zero, so the central fringe would not the the brightest, but the least bright at all. Right?
The question is, can d = 2Lλ/a in the real world?

4. Mar 25, 2013

### Sunil Simha

For the central fringe, think of the path difference between the light waves from the two slits. Isn't it a fixed value regardless of the double slit apparatus?

5. Mar 25, 2013

### jaumzaum

Yes, there is no path difference in the first experiment (without the refraction object). But you have to consider the refraction in each slit too right? What if the refraction is destructive (as the example I gave)?

6. Mar 25, 2013

### Sunil Simha

I'm guessing the following

Say we are given a point on the screen that is x units away from the center, can we find the intensities of light from the individual slits on that point due to diffraction? If yes we can proceed by using these intensities as initial intensities of the individual slits and then applying the standard procedure to find the resultant intensity at the required point due to interference.

In a nutshell I'm asking whether we can first find the effects due to diffraction and then find the effects due to interference as I'm not too sure.

As for the doubt regarding the presence of the central maximum, I guess it must exist as during a real interference experiment, diffraction does take place (because its the real world) but the central bright fringe is still obtained.

7. Mar 25, 2013

### jaumzaum

Sorry by the last post. For refraction I mean diffraction

I would say that for small diffractions (x very small), there is no variation of the angle of the resultant phasor due to each slit, that way we can just use the interference rule to obtain the pattern in the screen. But what if x becomes larger, the phasor angle due to diffraction differs from each slit and we have then a phase difference. That way I think we can't just use the standard procedure, it becomes very more complicated.

But I'm not sure. Besides, this is one of the questions of my first thread.
Does anyone know if it's right?