# 2 Questions on superpostion (Double slit experiment)

1. Nov 1, 2009

### Thevanquished

Question 1:
Why is it required for the slit separation to be at least one wavelength in the double slit experiment?

Question 2:
Why do we need to place a single slit before a double slit in the double slit experiment?

Thanks in advance

2. Nov 1, 2009

### mikelepore

(2) The single slit isn't always shown in all books, but is sometimes shown to ensure that the wave reaching the two slits are in phase. You want one slit to get a crest as the same time the other slit gets a crest.

3. Nov 1, 2009

### Thevanquished

how does the single slit ensure that the wave reaching the double slit are in phase? Won't light from the same source without the double slit already cause the wave reaching the double slit to be in phase?

4. Nov 1, 2009

### mikelepore

They're thinking that the light source isn't a perfect point source with zero dimensions. But if the light has to pass through a narrow slit, Huygen's principle will cause a new circular wave to begin from that new point.

On the other hand, an incandescent light bulb doesn't give us a single pure frequency either, which the double slit experiment also requires.

I've have aimed a laser (one frequency and one phase) directly at a diffraction grating before, without a single slit in front of it, and the expected pattern on the screen was obtained.

5. Nov 1, 2009

### Born2bwire

The slits can be sub-wavelength and the experiment will still work. It is probably rarely presented as such since most demos use lasers or some other radiation of such short wavelength that sub-wavelength slits are not feasible. In addition, there is an appreciable amount of attenuation when compared with larger slits.

The effect has to do, roughly, with the diffraction that occurs at the edge of the slits. A sub-wavelength slit will reproduce a point source, which is basically the limit when we require that only the diffracted edge contributions to pass through. So the sub-wavelength works rather nicely but since large slits still have the edge diffraction they will reproduce the effect too, they require a larger "problem space" before the interference pattern arises though. Again, for most practical demonstration this is irrelevant.

I agree with mike's reason for the addition of the single-slit. It should provide a point source regardless of the form of the incident wave. So it is a good way of setting up the source for the double slit experiment. Oddly enough, I never thought to do a double-slit with an offset point source. That should be an interesting exercise.

EDIT: Here we go:

A bit difficult to see here and the source is probably not in the far-field for the right-hand slit but.... You can see that the interference pattern does not arise. Primarily it's because of the difference in the relative amplitudes, the source's proximity to the right slit means that the wave from that slit overpowers most of the effects from the left slit. But we can see in the initial propagation of the waves from the slits that the phase difference (which we can observe by comparing the wavefronts of the source at the slits) means that the waves from the slits are not in the proper phase to setup the correct interference pattern. Neat.

Last edited by a moderator: Sep 25, 2014
6. Nov 3, 2009

### Thevanquished

Anyone has answers to

Question 1:
Why is it required for the slit separation to be at least one wavelength in the double slit experiment?

And thanks guys for answering question 2

7. Nov 3, 2009

### sweet springs

Hi.

We take a point A on the screen behind the slits. Let us think of
φ＝( distance between slit 1 and point A - distance between slit 2 and point A )/wave length
When you move point A on the screen the brightness changes as cosφ.

Brightness cahnges by other geometrical factors. One is the angle of diffraction or approximately the distance between A and the center of the screen
on which the vertical line from the middle of the slits falls. Smaller the distance, brighter the points because larger the angle of diffraction at the slits, weaker the wave amplitudes.

Another factor is distance from the slits. Longer the distance, weaker the wave amplitude in general. If you use laser beam it will be less sensitive.

So in these context you should set the experiment so that you can clearly see the interference pattern clearly enough, not too dense to investigate not so loose to find at a glance. I thought question one is one of these conditons.

Regards.

8. Nov 3, 2009

### Thevanquished

hmm.. sorry sweet springs but i still don't get why you need more than the length of a wavelength between the slits for the experiment

9. Nov 3, 2009

### Born2bwire

Like I said in my earlier posts, we do not need the slits to be larger than a wavelength.

10. Nov 4, 2009

### Staff: Mentor

It's not. Where did you read that it is?

11. Nov 4, 2009

### Thevanquished

It came out as one of the questions in my exams.

The exact question was:
Why would it be impossible to obtain interference fringes in a double slit experiment if the separation of the slits is less than the wavelength of the light used?

12. Nov 4, 2009

### Born2bwire

Ooooh, crap, sorry. I kept reading that as the slit width. We can certainly have slit widths that are sub-wavelength (although the transmitted wave will be severely attenuated since the fundamental frequency of a parallel plate waveguide is half a wavelength).

As for the slit separation, that is related to the fact that the difference in the paths that the light can go from the slits must differ by an integral factor of half a wavelength for interference to occur. So take a look at any derivation of the location of the min and maxima. http://www.matter.org.uk/schools/Content/interference/formula.html [Broken]

You will see that the derivation places a limit on the minimum slit separation because if the slits are too close, then the difference in their paths will never be an integer multiple of half a wavelength. You have done the Rayleigh limit, yes? In that theory, we look at the proximity of aperture sources and our ability to differentiate between what looks like one or two sources. This is the same idea, except instead of the merging of the amplitudes of the intensity, we are interested in the phase difference.

Last edited by a moderator: May 4, 2017
13. Nov 4, 2009

### Staff: Mentor

Sure, if the path difference never reaches half a wavelength, then the resultant intensity never reaches zero (assuming the intensities from the two slits are equal). Nevertheless, you still have partial cancellation if the path difference is e.g. a quarter of a wavelength, corresponding to a phase difference of $\delta = \pi / 2$. In this case, assuming the two sources have equal intensity, the resultant wave has an amplitude which is less than the maximum amplitude (equal path lengths) by a factor of $\cos^2 (\delta / 2) = \cos^2 (\pi / 4) = 0.5$.

So I would phrase the condition as: in order to see at least one minimum in the resultant intensity, the path difference has to be at least half a wavelength.

Now that I've looked at the exact wording of the question again:

I can see that it can be interpreted as requiring there to be at least one complete interference fringe. So the originally posted version of the question wasn't precise enough.

Last edited: Nov 4, 2009
14. Nov 4, 2009

### sweet springs

Hi.
I will correct my previous post.

d:separation of slits
λ:wavelength
θ:angle of diffraction
L:path difference
Then L=d sinθ. The phase difｆerence φ=2πL/λ=2πsinθ(d/λ）

Let us think of the first vanishing point φ=π
sinθ= (1/2) / (d/λ）. In the case d/λ < 1/2, any θ　cannot satisfy the equation, so no vanishing occur.
Similarly in the case d/λ < 1, the next brightest point does not appear.
I think it is an answer to your question 2.

Regards.

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