# Sources for the diffracted waves

For diffraction you can assume slits in screens contain sources for the diffracted waves.

But for a slit of finite width, how far apart are the sources? If the sources are continuously placed in the gap, then that means there are an infinite number of sources. An infinite number of sources with little spacing between them only produces a beam in one direction, and you won't get any diffraction.

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Hi, my opinions are as follows. I hope they can be helpful for you.
At the macroscopic level, the diffraction can be observed when the wavelength is comparable to the width of the slit.
At the microscopic level, according to your views, the diffraction can be thought of the superposition of infinite sources placed along the slit by a very small scale. Then, i think here is the key point. The very small scale means the difference of light path between two neighboring source are small. However, the superposition of infinite number of sources can arise an evident effect which can be seen on the screen. This is analogy to the calculation of area in calculus. When the delta x goes to infinity, the area of per rectangle approaches 0, however, the number of rectangle becomes infinity. The overall result of the sum of infinite rectangles is the real area of a certain curve.

jtbell
Mentor

If the sources are continuously placed in the gap, then that means there are an infinite number of sources. An infinite number of sources with little spacing between them only produces a beam in one direction,
No. When you integrate (add) the contributions from an infinite number of infinitesimally-wide sources, you get the usual intensity formula for single-slit diffraction. You can find the mathematical derivation using an integral, in any intermediate-level optics textbook. It's also possible to do the integration graphically using phasors (phase vectors), as on this page:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c1

You're all right. Most books only consider slits of infinitismal width, but I worked it out.

You can add n phasors using a geometric series:

$$1+e^{i\phi}+e^{2i\phi}+...+e^{(n-1)i\phi}= \frac{e^{in\phi}-1}{e^{i\phi}-1} =phase* \frac{sin(n \phi/2)}{sin(\phi/2)}$$

The intensity is $$I=\left(\frac{sin(n \phi/2)}{sin(\phi/2)}\right)^2$$

Now each phasor had amplitude 1, but I guess it would make more sense to say each amplitude is 1/n. That way no matter how many sources you put in the slit, the you'll get the same intensity at phi=0. So the new intensity is:

$$I=\frac{1}{n^2}\left(\frac{sin(n \phi/2)}{sin(\phi/2)}\right)^2$$

Allow the phase difference to be $$\phi=dsin\theta \frac{2\pi}{\lambda}$$ where theta is the angle to the screen and d is distance between sources, then:

$$I=\frac{1}{n^2}\left(\frac{sin(nd sin\theta \pi / \lambda )}{sin(\pi d sin\theta / \lambda)}\right)^2$$

But nd=L, or the length of the slit, and the denominator is small since d is small so this is:

$$I=\frac{1}{n^2} \frac{\lambda^2}{\pi^2d^2sin^2\theta} (sin(L sin\theta \pi / \lambda) )^2 = \frac{\lambda^2}{\pi^2L^2sin^2\theta} (sin(L sin\theta \pi / \lambda) )^2= \left(\frac{sin(x)}{x}\right)^2$$

So if x is small, you'll get peaks. So if the wavelength is a lot larger than the length, then interference comes from multiple slits, and not a single slit of finite width.