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I've found a few results on the internet for this question, all with different explanations or different answers. Some say it'll become 4I and some that it'll not change. I see that there are a few things going on here. I know that the intensity depends directly on the slit width, so that's one factor. Increasing the intensity also concentrates the beam. This is what I have figured out. Can someone please explain this to me in layman's terms? I'm just beginning to solve problems like this and a good guide on how the thought process and approach should be will help a lot.

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The intensity profile of the single slit diffraction can be found using Huygens' principle. It involves summing the contributions of all the rays from the slit to the detector (each with length $r+x\sin\theta$).

$$I(\theta) \propto \left|\int_{-a/2}^{+a/2} E_0\ e^{i\frac{2\pi}{\lambda}(r+x\sin\theta)}dx \right|^2$$ where $\theta$ is the angle of the diffracted light, $a$ is the slit width, $\lambda$ is the wavelength, and $r$ is the distance between the detector and the center of the slit.
In the formula above some over-all constants (like $\epsilon_0$ and may be more) are deliberately omitted. That's why there is $\propto$ and not $=$.

Doing the math the final result is $$I(\theta) \propto E_0^2a^2\ \text{sinc}^2\left(\frac{\pi a}{\lambda}\sin\theta\right)$$ where $\text{sinc}(y)=\frac{\sin y}{y}$ is the so-called sinc function.

So the intensity profile looks like this:

enter image description here

From this you see:
When you double the slit width $a$, then the peak intensity $I(0)=E_0^2a^2$ is multiplied by $4$, and the peak width $\Delta\theta\approx \frac{2\lambda}{a}$ is divided by $2$. That means, the total energy through-put (which is proportional to $I(0)\ \Delta\theta$) is doubled, as you would expect it.

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  • $\begingroup$ how did you get the first relation? Isn't it $I(\theta)=\frac{c\epsilon_0}{2}E_0^2 sinc ^2(\frac{\pi a \sin(\theta)}{\lambda})$? $\endgroup$ Jan 22, 2023 at 6:53
  • $\begingroup$ @SolitarySolus see my updated answer $\endgroup$ Jan 22, 2023 at 8:44
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With a coherently illuminated screen whose slit width is much smaller than its distance to the point of observation the amplitudes add up in phase so if the width doubles then the intensity quadruples. If the distance is not much larger than the width, then the complex amplitudes (phasors) are not aligned and they do not add up in phase because of the substantially varying propagation distance to the point of observation, and there the intensity will be less than 4x.

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  • $\begingroup$ In the case of Young's double slit experiment, when the sources are coherent and the slit width of any of the sources is, say, doubled, then the intensity of that source also doubles. So why here in the case of diffraction there is quadratic relation instead of a linear one? $\endgroup$ Jan 22, 2023 at 6:03
  • $\begingroup$ As I said above, in coherent illumination at all observation points the complex amplitudes (phasors) add up and the detector measures the resulting intensity, not the sum of intensities; $(3+3)^2 > 3^2+3^2$. This actually is true always that the for EM waves the complex amplitudes add up but when the components are not coherent the detector (your eyes, here) cannot follow the fluctuation and averages it. If the slit is wide then from the edges the travel time is much longer to the middle than from the center line, this is the reason for the curve above shown by @ThomasFritsch. $\endgroup$
    – hyportnex
    Jan 22, 2023 at 13:33
  • $\begingroup$ got it! and @John Doty's answer also clears things up. $\endgroup$ Jan 23, 2023 at 3:21
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There are three regimes. Given slit width $a$, distance to the screen $d$ and wavelength $\lambda$, if $d<<a^2/\lambda$, the projected slit image on the screen just has the slit width, and the intensity of that image is simply the intensity of the illumination of the slit, independent of the slit width.

Otherwise, in the case where $a<<\lambda$, there's no "beam" of radiation from the slit. There's no destructive interference between the waves across the slit to produce a narrow beam. The intensity is determined by the power through the slit, so it's proportional to $a$.

In the intermediate case, widening the slit narrows the beam, so the beamed intensity is the product of the beam effect and the power through the slit, thus proportional to $a^2$.

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  • $\begingroup$ Oh, okay! I didn't understand that the proportionality depended on their magnitudes at first. How did you arrive at those expressions, can you give me a bit more mathematical context behind these? Thanks! $\endgroup$ Jan 23, 2023 at 3:28

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