Double slit interference missing maximum problem

1. Jun 30, 2013

PsychonautQQ

Double slit interference "missing" maximum problem

1. The problem statement, all variables and given/known data
In the double slit interference experiment, d is the distance between the center of the slits and w is the width of each slit. For incident plane waves, an interference maximum is "missing" when what relationship between d and w is present?

2. Relevant equations
Really i don't even know where to begin... I know the condition for maxima is dsin∅=nλ and y=nλD/d, but i don't have much understanding of these equations and they don't include a width variable so i'm quite lost :-X

3. The attempt at a solution
Halp ;-(

Last edited: Jun 30, 2013
2. Jun 30, 2013

haruspex

3. Jun 30, 2013

PsychonautQQ

asin∅=nλ
where a is the slit width, n is there integer and λ is the wavelength. So in the double slit experiment slit width is not a factor? According to this answer sheet i'm suppose to get 2d = 3w.

4. Jun 30, 2013

haruspex

Slit width is still relevant in the double slit experiment, but usually it is arranged to be very small compared with the slit separation, so can be ignored. I'm not sure whether it gives the desired answer, but it seems to me you can take the diagram at that link and determine that as the slit width increases the dashed line envelope narrows until it falls inside the first minimum from the double slit interference.

Edit: I mean, falls inside the first non-central maximum.

5. Jul 1, 2013

PsychonautQQ

I don't think this is quite what I'm looking for. The relationship you are talking about between the angle and slit width is definitely on the right track but i don't know if there's a way i can get a numerical ratio from that diagram.

6. Jul 1, 2013

haruspex

You're right, that approach is too crude to get the desired answer. Looks like we need a general expression for the intensity at angle θ which takes into account both w and d. The equations at http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html#c1 and http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c1 should help. I see where the 3 would come from, but I haven't cracked it yet.

7. Jul 2, 2013

ehild

Try these
http://www.ualberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect35/lecture35.html
http://www.ualberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect36/lecture36.html
In case of a single slit, the diffracted intensity changes with θ as

$$I=I_0\left(\frac{\sin\left(\pi \frac{w}{\lambda} \sin\theta\right)}{\pi \frac{w}{\lambda} \sin\theta}\right)^2$$
The amplitude of the double-slit pattern is modulated by the single-slit one, which has its minima at sin(θ)=±mλ/w, m=1,2,3,4...while the maxima of the double-slit interference pattern are at sin(θ)=±kλ/d, k=0,1,2,3,... The maximum of the interference pattern is missing if it is at the minimum of the single-slit pattern.

ehild

8. Jul 2, 2013

haruspex

That was the basis of my initial responses, but it seems to me that gives w = d, not the target result, so I thought there must be a subtlety I was missing.
Having thought about it further, as w increases the first non-central maximum will move closer to the centre. The double-slit minima won't move. So there continues to be that 'second' maximum, merely in a different place.

Notice what happens when the first single slit minimum moves inside the second double slit minimum (w=d/4). At that point, we get an extra maximum.
I plotted up the intensity function from your link. Can't see the disappearance indicated.
Then again, the question says "an interference maximum", but I can't see that is going to give the desired answer either.

9. Jul 2, 2013

ehild

At the singe slit minima, the intensity is zero. If the double-slit interference maximum falls to that place, it would be missing.

I don't get you. The intensities do not add up: The interference pattern is modulated by the diffraction pattern. At the place of the diffraction minimum, the intensity is zero.

Now, the problem says that one interference maximum is missing. That means the second diffraction minimum does not appear (it would result in sinθ>1) So the position where the interference maximum is missing is at m=1. dsinθ=kλ and wsinθ=1*λ, so d/w=k. In case d=w as you suggested, it would not be a double slit but a single one. The slits would touch each other.

In the second picture, the small bright dots are the interference extrema. See where they are missing.

ehild

Last edited: Jul 2, 2013
10. Jul 2, 2013

haruspex

If you start with the theoretical case of infinitesimal slits, the maxima are at certain positions. As soon as the slit width is taken to be nonzero, the maxima (except for the central one) will not fall at the same locations; they all migrate a little towards the centre. I don't see that being reasonably described as their having gone missing. The number of maxima between the centre and the first single slit minimum does not, at first, change.
For a given width of slit, w, if you look at sufficiently distant maxima, one of the single slit minima will have passed through one of the original double slit minima. This creates additional maxima, but all the old maxima are still there - just displaced.
I attach a chart of the intensity curves as the width increases from 0.2d (top curve) to 0.8d between the first and second double slit minima. x-axis is d.sin(θ)π/λ from 1.1 to 5.95, so this shows the first two double slit minima. Note how the maximum in between migrates leftwards; in the bottom curve you can see a new maximum just starting to rise at the 65th datapoint.

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• slits.pdf
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11. Jul 2, 2013

ehild

I understand that the real maxima shift a bit, as the intensity is the product of two function, both of them having maxima and minima. But the problem says that one of the interference maxima is missing. First you have an interference pattern with infinite narrow slits, and when the slits are wider, the intensity of the original pattern changes, as in the pictures. A simple person sees in interference maximum missing. If I understood you correctly, you see that the original maximum is shifted, the intensity is zero at the place of the original maximum and a new maximum appeared at the other side of the minimum. The problem is written for students, who might look at a real diffraction pattern, as the one in my previous post. There are dark places where one expects a clear interference maximum.
See my figure. It was calculated for w/lambda=2 and d/lambda=10. The black line is the single-slit, the red line is the interference pattern and the blue one is the diffraction pattern. I see one interference peak missing at each arrow. You see two peaks instead of one. Anyway, their intensity is very small.
By the way, double-slit experiments and diffraction gratings use well separated slits. I mean, w=0.8d is a bit awkward set-up.

ehild

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• doubleslit210.jpg
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12. Jul 2, 2013

haruspex

Maybe, but if you're right I think it's a pretty shoddy question. If missing is to be interpreted as there no longer being a max at that point then any w > 0 will do it.
Anyway, what solution does that give you? It doesn't give me 3w=2d. What I see at that ratio is that the second minimum of the double slit factor coincides with the first minimum of the single slit factor. That reduces the total number of minima, and 2d/3 is the smallest w for which this happens, so maybe there's a typo in the question.

EDIT: Doh! If a minimum goes missing then a maximum must too!
OK, so here's how it works:
There are minima because of zeroes in the double-slit factor and those in the single-slit factor. Between any two such minima there is a maximum. As w increases, the single-slit minima migrate centrewards, as do the maxima, while the double-slit minima stay put. When a single-slit minimum reaches a double-slit minimum the maximum that had been between them disappears. So the question becomes, for what w do two minima coincide?
But I see nothing special about the 2d/3 solution (I was wrong about that too). There are both smaller w and larger w solutions. It is the w (<d) for which the most central maximum disappears.

Last edited: Jul 2, 2013
13. Jul 2, 2013

ehild

An interference maximum disappears if it is at the same position as the zero of the diffraction pattern. That happens if d/w is a rational number. Best seen when an interference maximum is at the first minimum of the diffraction pattern, so d/w is integer.
If you see the resultant two-slit pattern for d/w=3/2, (attachment) the first maxima are quite shifted with respect to the interference maxima. It is not clear what the problem means on missing interference maximum, and why d/w=3/2 is the suggested solution. Interference maximum is missing at the second minimum of the diffraction pattern, where it looks rather two little maxima with a minimum in between.
I have seen quite many related problems and every time missing maximum meant the interference maximum at the position of the zero of the single-slit diffraction pattern.

ehild

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14. Jul 3, 2013

haruspex

But what does that mean exactly? If we don't worry too much about exactly where the maxima are, or what happens in the centre, we can simplify the function to f(x)=sin(Ax)cos(x), where A = w/d < 1 and x > 0, and we refer to peak amplitudes rather than maxima. (More accurate would be sin(Ax)cos(x)/x, which shifts the peak amplitudes inward a little, but leaves the zeroes unchanged.) The sin(Ax) describes the single slit factor and cos(x) the double slit factor. I think you mean when a zero of the sin(Ax) lands on a peak of the cos(x); that does indeed happen for every rational A. For A=2/3, it happens at x = 3π, but that is the third cos(x) peak from the centre - there's nothing at all special about that. Why not pick A=1/2? Remember, the OP doesn't ask one to prove it happens at 2/3. That answer was not initially given.
I'm now convinced that it means repeated roots of f(x). The 2/3 then at least makes some sense. As A increases, peaks of |f(x)| are getting squeezed out by adjacent zeroes all over the place, specifically at every solution of A = 2m/(2n+1). (The peaks reappear the other side of the zero.) What's special about 2/3 is that the disappearance occurs at x = 3π/2, and that's the closest to x=0 that any disappearance occurs (until w=d and the slits join up). That said, this should be made clearer in the question.

15. Jul 3, 2013

ehild

What do you mean on f(x)? After the double slit, the intensity in the diffracted beam depends on the diffraction angle theta,

a is the slit width, (w here) d is the distance between the centres of the slits, lambda is the wavelength.

ehild

16. Jul 3, 2013

haruspex

Put x = πd sin(θ)/λ, A = w/d. Peaks and zeroes of I will be the same as the peaks and zeroes of |f(x)|, where f(x)=sin(Ax)cos(x)/x. If we drop the /x, the zeroes will be the same (except an extra one at x=0) and the peaks will shift a little away from the origin.