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When double slit experiment is carried with white light , the resulting pattern on the screen is predicted by assuming white light as combination of light of different colors. Then it is assumed that the situation is equivalent to light of different frequencies performing double slit diffraction individualy and each will form their maxima and minima and the resulting pattern on the screen is a rainbow pattern . But how are the situations equivalent . When different frequencies are producing interference pattern , different frequencies interface with each other . But if we consider that many frequencies are producing their respective interference pattern we are just ignoring interference of different frequencies we are considering interference of like frequencies only . Then what is the correct explanation for that ignoring?

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Interference between different frequencies between the source and the screen doesn't matter because light propagation is linear. You can compute the fields at the screen for each frequency separately and add the results.

Adding the observed intensities on the screen from different frequencies isn't strictly correct, but it's a very good approximation for visible light, because the frequency of visible light is in the hundreds of terahertz, so the beats between different frequencies/colors are also very high. E.g., two frequencies differing by 0.0001% would beat in the hundreds of megahertz, so you would need a camera with a nanosecond shutter speed to see a deviation from the approximation.

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It should not be ignored. Presence of multiple frequencies will definitely lead to blurring of the pattern. You can see that if you do the experiment with incandescent light instead of LED or laser light. (Perhaps candle light was the source in the original experiment).

Treating this in a mathematically correct way can be done just with classical EM. But it is complicated because, unless the slits are made extremely narrow, even for a single frequency we already have single-slit diffraction and 2-slit diffraction combined in the pattern. This can be easily described by [Fraunhofer diffraction], where the Fourier transform of a function like this:

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will give the result (ignoring some constants) for the $E$-field pattern as: $$E(y) \sim \cos\left(\frac aL\,k\,y\right)\ \sin\left(\frac bL\,k\,y\right)$$ where $k$ is the wave number, $L$ the distance to the screen and $a$ and $b$ are proportional to the slit separation (center-to-center) and the slit width, respectively. The extra $\sin$ factor is due to the finite slit width.

After we square this $E$-field to get the illumination we can integrate over a certain bandwidth of frequencies to get the result for a mix of frequencies. The expression becomes messy, with lots of $\cos$, $\sin$, and $\operatorname{Si}$ functions. For a bandwidth of $\pm 10$% the result is shown in the following figure (in orange), together with the monochromatic result (blue), each for $a=8b$:

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There we can clearly see the three things that matter, the strong 2-slit fringes, the undulating envelope of the single-slit diffraction, and the smoothing by the finite frequency range.

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When talking about the Intensity of monochromatic light one usually argues with waves of the form $\exp(ik\rho)$, $\rho$ beeing the distance to the corresponding slit. Letting them interfere on a far away screen one assumes that the length difference of the paths only depends on the angle and thus produces the known interference pattern.

Now the attentive reader might be wondering about the expression “waves of the form $\exp(ik\rho)$”. Why would one call those waves, shouldn’t waves show some kind of time development? Indeed the correct expression would be $\exp(i(k\rho + \omega t))$ with in the case of light $\omega = k \cdot c$. However, when talking about monochromatic light $\omega$ will be the same for each slit and thus one will just have a factor $\exp(i \omega t)$ in front of the sums of the spacial parts of the waves, which will vanish, when we take the absolute squared to get the time averaged Intensity.

For non monochromatic light this can no longer be neglected. If we look at the interference of waves with different $k$s, we will also have different $\omega$s and therefore at any angle one could get any phase difference depending on the point of time of the measurement. If you look at the time averages, you just get the sum of the Intensities of the different wavelength contributions.

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When different frequencies are producing interference pattern , different frequencies interfere with each other.

If the frequencies are different the interference pattern is not stationary with the result of no observable interference pattern.

You may have seen a demonstration with two loudspeakers producing sound of slightly different frequencies?
The demonstration is usually used to show beats however you can also consider the arrangement as producing a moving two source interference pattern and standing in one position you hear the maxima and minima of the interference pattern passing you.

Now replace the loudspeakers with two light sources of wavelengths $500 \,\rm nm$ (frequency $6\times 10^{14}\rm Hz$) and $501 \,\rm nm$ (frequency $5.988 \times 10^{14}\rm Hz)$ and a frequency difference of approximately $10^{12}\,\rm Hz$.
How would you see such beats noting the fact that the two sources would also not be emitting "infinite" wave trains?

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  • $\begingroup$ According to you there should be no interference pattern at all . But there is a real rainbow kind of interference pattern . I remember the comments in my previous question and it said particles can interference with itself only . Thus a particular photon having a definite frequency cannot interfare with another photon with another frequency . $\endgroup$
    – Users
    Commented May 6 at 10:56
  • $\begingroup$ Here's the link physics.stackexchange.com/questions/813043/… where I finally found my answer $\endgroup$
    – Users
    Commented May 6 at 10:57
  • $\begingroup$ According to you there should be no interference pattern at all . I disagree what I have tried to explain is the necessity for coherent sources to produce a stationary interference pattern. What is achieved in two slit interference by "division of wavefront" so the waves emerging from the two slits are of the same frequency and of constant phase difference. If the frequencies are the same, the "beat" frequency is zero and a stationary interference pattern results. $\endgroup$
    – Farcher
    Commented May 6 at 11:18

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