Interference with slits of different width

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SUMMARY

The discussion addresses the interference pattern produced by three equidistant slits where the central slit has a width 1.4 times greater than the other two, with slit separation d = 2.116 μm and wavelength λ = 650 nm. It is established that the amplitude contribution from the central slit is multiplied by the width factor w = 1.4 because the slit width directly scales the transmitted wave amplitude, resulting in a modified interference intensity formula. The problem is analyzed within the Fraunhofer diffraction regime, treating slits as point sources with amplitude proportional to slit width. The key conclusion is that the width factor w appears as a multiplicative coefficient in the interference term due to the proportionality of amplitude to slit width, which is consistent with classical wave optics principles.

PREREQUISITES

  • Fraunhofer diffraction and interference theory
  • Wave amplitude and intensity relationship in optics
  • Trigonometric identities for cosine expansions (e.g., cos(A±B))
  • Basic understanding of slit diffraction and superposition of waves

NEXT STEPS

  • Study detailed derivation of multi-slit interference intensity formulas including amplitude weighting
  • Explore Fraunhofer diffraction patterns for slits of varying widths using Kirchhoff’s diffraction integral
  • Learn to apply phasor diagrams to visualize amplitude and phase contributions from multiple slits
  • Investigate experimental methods to measure slit width effects on interference patterns

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Physics students preparing for olympiads, educators teaching wave optics, and anyone analyzing multi-slit interference with non-uniform slit widths in classical optics experiments.

weirdoguy
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Hello everyone!

So I'm working through problems from this years' polish physics olimpiad, and I have a problem with one problem (I google-translated it, hope it's not a problem):

A flat wall contains three equidistant, thin slits of identical length. The distance between adjacent slits is d = 2.116 μm. The width of the central slit is w = 1.4 times greater than the other slits. A monochromatic wave with wavelength λ = 650 nm is incident perpendicularly on the partition. Interference fringes can be observed on a screen located at a distance significantly greater than d. Determine the angle at which first interference minimum is observed.

Here is the drawing:

1780491663728.webp


So, authors state that the intensity of wave at the point we are interested in is proportional to:

1780491858605.webp


My question: is there a simple, high-school argument that there should be ##w## in there, multiplying second cosine? I found something on wikipedia, but complex exponents and all that are not a good look in high-school.

I know I should know such things as a physicists, but "waves and oscillations" were at the second year, exactly at the time when I started partying hard, so I'm not that good with waves 👨‍🦽 I'm still good at partying though.


PS. Interesting that google translator changed commas to periods in numbers (we use commas in Poland, so we write 2,116 instead of 2.116).
 
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I have learned expansion of
$$\cos(A+B),\cos(A-B),\sin(A+B),\sin(A-B)$$
in high school math.

Let w=0, d=0 in the formula we get 2 which corresponds to width of a synthesized single slit.
Let w=1, d=0, we get 3 which again corresponds to width of a single slit.
We deduce that width of a slit is coefficient for passing wave.
 
Last edited:
weirdoguy said:
My question: is there a simple, high-school argument that there should be w in there, multiplying second cosine?
Is it not basically treating the three slits as point sources (ok in the Fraunhofer regime), and making the amplitude from the central slit ##w## times higher because it lets through ##w## times more light?
 
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Ibix said:
making the amplitude from the central slit w times higher because it lets through w times more light?

Ok, I thought about it, but I felt that there is something more to that. Thanks! :oldbiggrin:
 
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