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I have a question related to the interference (thought)experiment with water waves given in the book Feynman Lectures on Physics Vol.3. When only one hole (hole 1) is open the measured wave intensity at the second wall varies with the distance from the center. It is shown in the figure as $I_1$ which has a peak right at the point exactly opposite to the hole 1. Now my question is : Why is the wave intensity has the variation as given by $I_1$. Shouldn't it be constant. The hole 1 acts as a source with wave propagating in all directions from it.

Let this wave be given as real part of $e^{j(kx-\omega t)}$ where $x$ is the distance travelled by the wave along the direction of propagation. Then by this equation the intensity of the wave at any point on the second wall should be the same equal to 1.

Feynman assumes the wave be given as $h_1 e^{j\omega t}$ where $h_1$ is some complex number but does not mention what it is. I wonder how he got the intensity as $I_1$ when only hole 1 is open. Kindly request you to explain me in detail (possibly with a reference) as evidently I seem to miss something in wave propagation.

enter image description here

Image Source : The Feynman Lectures on Physics Vol. 3.

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up vote 4 down vote accepted

You have a 2D point source, so the energy of the wave is spread out along a circle of circumference $2\pi r$ where r is the distance from the point source. This means the intensity of the wave varies as 1/r. It's a 2D version of the inverse square law.

Obviously the amplitude can't be constant everywhere on the second wall or the energy of the wave would be infinite.

You get the curve drawn for the amplitude on the second wall because the distance from the point source to the wall increases as you move along the away from the point immediately opposite the slit.

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The damping factor, I missed it. thanks for the answer. – Rajesh Dachiraju Jul 15 '12 at 13:19

It's not in fact a damping factor but rather a geometrical effect due to the fact that the waves are two-dimensional. The form $e^{i(kx-\omega t)}$ you quote for the amplitude is wrong - it only applies for plane waves. In this 2D experiment, travelling waves are described by Bessel functions, for which a good asymptotic form (ignoring constants) is $$H_0^{(1)}(kr)e^{-i\omega t}\sim \frac{e^{i(kr-\omega t)}}{\sqrt{2\pi r}} \textrm{ (where of course }r=\sqrt{x^2+y^2} \textrm{).}$$ Thus the intensity profile at a distance $L$ from the slits goes like $\frac{1}{\sqrt{L^2+x^2}}$, which is the Lorentzian-like shape in the illustration.

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My $x$ is equal to your $r$. Thats what I had in my mind. There apparently one more $x$ in the diagram which I am noticing just now. May be that's what has caused confusion. The only thing you have added is the damping factor $\frac{1}{r}$ which is what is explained by John Rennie in his answer. – Rajesh Dachiraju Jul 15 '12 at 15:34
Using $x$ for a radial coordinate is certainly a recipe for confusion. Keep in mind that the factor of $1/r$ in the intensity is NOT a damping factor: it is a geometrical effect due to the fact that an equal amount of energy (or particle flow, in this instance) is being spread over half-cylinders of increasing area. There is no damping going on in this situation. – Emilio Pisanty Jul 15 '12 at 18:24
Thank you for the answer and comment. Could you give any good reference on the net for introduction to 2D waves? – Rajesh Dachiraju Jul 16 '12 at 1:53

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