Which Interference Maxima Are Missing in the Slit Experiment?

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SUMMARY

The discussion focuses on the phenomenon of missing interference maxima in the double slit experiment when monochromatic light illuminates two thin parallel slits. It is established that every first, third, fifth, seventh, and ninth maxima will cancel due to the overlapping single slit diffraction patterns. The key equations discussed are \(\delta \sin \theta = n \lambda\) for maxima and \(\alpha \sin \theta = n \lambda\) for minima, where \(\alpha = \delta/7\). The conclusion is that the missing maxima are not absent but rather never existed due to the interaction of the single slit patterns with the double slit configuration.

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  • Understanding of wave optics principles, specifically interference and diffraction.
  • Familiarity with the double slit experiment and its implications in physics.
  • Knowledge of the equations governing maxima and minima in wave patterns.
  • Basic grasp of monochromatic light and its properties.
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  • Study the mathematical derivation of interference patterns in the double slit experiment.
  • Explore the impact of slit width on diffraction patterns in single and double slit experiments.
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Punkyc7
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Monochromatic light illuminates a pair of thin parallel slits at normal incidence, producing an interference pattern on a distant screen. The width of each slit is 1/7 the center-to-center distance between the slits.

Which interference maxima are missing in the pattern on the screen?

every first will cancel
every 3rd will cancel
every 5th will cancel
every 7th will cancel
every 9th will cancel

Im leaning towards the seventh but why does the maxima just disappear? Shoulnt the maxima still be a maxima?
 
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Remember, that the diffraction pattern for a single slit gives us a series of minimas at quantum points in the pattern. This pattern does not go away when a second slit is introduced into the equation. The minimas produces by the single slit will in fact occur at the same angle as every so many maximums produced by the double slit diffraction pattern.

for maxima, \deltasin\theta=n \lambda

and for minima \alphasin\theta=n\lambda

where \alpha=\delta/7

does that help?
 
Ok but how does a maxima disappear, shouldn't they be evely spaced apart? Are we considering the distance between the slit where a single slit would be open?
 
Punkyc7 said:
Ok but how does a maxima disappear, shouldn't they be evely spaced apart? Are we considering the distance between the slit where a single slit would be open?

The maxima does not disappear - it was never there to start with.
OR
The maxima are missing, they have not disappeared.

The double slit interference comes about from the "interaction" of the two, overlapping, single slit patterns.

Think like this:

For the single slit pattern, the central bright band may be 5 cm wide on the screen. on each side is a 5 mm band of nothing, this is flanked by couple of bands 4 cm wide, then 5 mm bands of nothing etc etc [I am not interested here in actual widths of band - just the concept that there are bands].

With the second slit present, the interference pattern means that there should now be alternating bright and dark bands "evenly spaced" as you said. Suppose they are are each 2mm wide, with a 1 mm gap between them.

There will be 12-15 of them spread across the previous central broad band. the next one or two won't be there because they "should" have been in a position where the single slit diffraction patterns didn't produce any light anyway. Then comes another bunch of bands [across the secondary maxima] then one or two more missing, then some more in the tertiary maxima etc, etc.

This explanation was intended as a conceptual/pictorial description, not some analytic numerical specification.

Peter
 
Consider the single slit experiment. You get a series of minima as a result, right? When you add a second slit, the single slit result doesn't go away. Rather it rules over the double slit phenomenon and makes it its slave. If you do the math, you'll find that the value of theta will coincide on a regular basis for both the minima produced by the size of the single slit and the maxima produced by the size of the double slit. Solve each equation for theta and set them equal. You will see why your intuition is correct.
 
So what this question is asking is when will the minimum for the single slit the same as the maximum for the double slit. So that would mean that the it doesn't matter what wavelength.

If that is correct than I think I understand it
 
Punkyc7 said:
So what this question is asking is when will the minimum for the single slit the same as the maximum for the double slit. So that would mean that the it doesn't matter what wavelength.

If that is correct than I think I understand it

If the wavelength had been numerically important, you would have been given it. The fact that there was only one wavelength [monochromatic light] was vital.Peter
 
why is that?
 
Punkyc7 said:
why is that?

I re-read, and found it less unusual and a legitimate way of seeing the problem - thus my edit.

Peter
 
  • #10
What Petero is trying to say is that the wavelength does not matter in this particular solution. " It is not numerically important"
You are correct in thinking that this problem is true for all lambda.
 
  • #11
So if it is not depenedent on the wave length if you change the wave length would the missing maximum occur in the same spot. I am thinking it wouldn't but i want to be sure
 
  • #12
Punkyc7 said:
So if it is not depenedent on the wave length if you change the wave length would the missing maximum occur in the same spot. I am thinking it wouldn't but i want to be sure

Correct, it should still be the nth maximum that is missing, it is just that the nth maximum would have been expected in a different place.

Peter
 

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