I had a confusion in the diffraction experiment about which I was taught in school.

According to the formula we Derived The width of the central maxima decreases on decreasing width of slit and vice versa.

But what i wonder is what happens to the max intensity of central maxima in these cases? I asked two physics teachers and they gave me opposite answers-one said it increases on decreasing slit width which i feel is right but the other said it would be the same as light is of constant wavelength(monochromatic).

What do you think? Please explain:)

  • 1
    $\begingroup$ I think that the maximum intensity of the central peak should decrease in intensity as decrease the slit width for the simple reason that less photons can get through the slit at strike the screen directly in front of it at zero-angle. So now you have three different opinions! BTW, all this should be very easy to confirm experimentally. I remember playing around with single-slit diffraction and lasers as an undergrad physics lab, and I certainly don't recall the central peak getting brighter as I closed down the slits. I recall all the features getting dimmer. $\endgroup$
    – user93237
    Commented Nov 5, 2015 at 15:45

2 Answers 2


I did a quick calculation of the single-slit diffraction patterns for three different slit widths of 100, 60, and 40 microns. Wavelength was that of green light (about 5000 Angstroms). Intensity plot is shown below for a progression of decreasing slit widths from 100 microns width (blue curve), to 60 microns width (green curve), and then to 40 microns width (red curve):

Single-Slit Diffraction

Note that, as expected, the central peak gets broader as the slit width decreases. Also, again as expected, the intensity of the central peak decreases as the slit width decreases.


Short answer:

On axis or central lobe field increases with the area of the slit (and thus the intensity $I\propto area^2$).

Long answer:

If you have a monochromatic plane wave traveling along the $z$ axis incident on a diffraction slit of width $w$ (as shown in the image below), then the initial field is $E(r,t) = E_0 e^{ik_0 z- \omega t}$. Now immediately after the slit (assumed to be at $z=0$) the field is (ignoring the temporal component since this is a monochromatic wave) $$E(r) = E_0\operatorname{rect}(x/w),$$ where $\operatorname{rect}$ is the rectangle function (equal to one if $|x/w|<1/2$ and zero outside this range).


Now we can still write the field as a sum of plane waves in what is known as the angular spectrum representation, i.e. $$E(r) = \int \tilde E(k_x) e^{ik_x x} dk_x/2\pi,$$ which we can Fourier transform to find the amplitude in the plane wave $\tilde E(k_x)$, i.e. $$\tilde E(k_x) = \int E(x) e^{ik_x x} dx.$$ The image above shows a sample of three of these plane wave components (at $\theta = 0$ and at $\pm 45$ degrees).

Now if $E(x)$ has a finite width (e.g. by a slit of width $w$), then we can change variables to $dx = wdx'$ and we see that this equation is proportional to the width $w$. In particular, if the slit is a hard slit given by the rectangle function, then we know that the Fourier transform is simply given by $$\tilde E(k_x) = wE_0\operatorname{sinc}(wk_x/2\pi),$$ where $\operatorname{sinc}$ is the sinc function For the on axis (central lobe) intensity, this gives $$|\tilde E(k_x=0)|^2 = w^2|E_0|^2.$$

Note for a 2D "slit" or hole, the field for $k_x = k_y = 0$ would be proportional to $\int dxdy \to area$, and thus the intensity would be $|E_0|^2\times area^2$.


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