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Ferrera

Can anybody explain to me where this equation came from? It's for two point sources at the two listed points, and it's calculating the period of the wave on the substrate. It seems to be $\lambda/\sin(\theta)$, which seems contrary to what I would normally expect the period to be. i.e $\sin(\theta)\lambda$ where $\theta$ is the angle from the normal.

Brandon Enright
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1 Answers1

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Defining $$f(x,y,z)=\mbox{exp}\left(\frac{2\pi i\sqrt{x^2+y^2+z^2}}{\lambda}\right)$$

we find that $$|f(x-a,y,0-c)+f(x+a.y,0-c)|^2=4\mbox{cos}\left[\lambda^{-1}\pi\left(\sqrt{(a+x)^2+y^2+(c-z)^2}-\sqrt{(a-x)^2+y^2+(c-z)^2})\right)\right]^2.$$ To compute the spatial frequency of $\mbox{cos}(\phi(x,y))$ it suffices to compute $\frac{1}{2\pi}|\nabla \phi|$. Since $\mbox{cos}(x)^2$ oscillates twice as fast as $\mbox{cos}(x)$, we have the spatial frequency is $\frac{1}{4\pi}|\nabla \phi|$, where $\phi$ is the expression inside the $\mbox{cos}^2$ expression I gave above. Mathematica spat out the following result: $$\frac{1}{4\pi}|\nabla \phi|=\frac{\sqrt{y^2 \left(\frac{1}{\sqrt{(a-x)^2+c^2+y^2}}-\frac{1}{\sqrt{(a+x)^2+c^2+y^2}}\right)^2+\left(\frac{a-x}{\sqrt{(a-x)^2+c^2+y^ 2}}+\frac{a+x}{\sqrt{(a+x)^2+c^2+y^2}}\right)^2}}{\lambda }$$ and if you simply ignore the left term in the radical and invert the expression (to get period), this is the same result as in the book.

I am not sure why the book decides to omit the left expression in the radical, although maybe they're assuming $a\gg x,a\gg y$ (ie, that the light sources are far away from each other), in which case the approximation is valid.

Brandon Enright
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DumpsterDoofus
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  • Why do we define f to be that? –  Nov 21 '13 at 05:05
  • Are you confused as to why the spatial frequency is the gradient of the argument of a sinusoid divided by $2\pi$? – DumpsterDoofus Nov 21 '13 at 05:07
  • No, I don't know why you defined $f(x,y,z)$ like you did. –  Nov 21 '13 at 05:09
  • But also what you said. –  Nov 21 '13 at 05:10
  • Oh, sorry. I defined $f$ to be a radial plane wave centered at $(0,0,0)$. Then to model the system, I placed one of them at $(-a,0,c)$ and another at $(a,0,c)$ and added them and computed their squared norm. – DumpsterDoofus Nov 21 '13 at 05:12
  • Radial plane wave? Oh my, I've only done a little wave mechanics, and only an introductory E&M course, I'll read up on what a radical plane wave it. Why did you computer their squared norm? For amplitude? Also, what is wrong with just projecting the wavelength onto the xy plane, and saying that is the period? I know it's very simple, but I don't know why it's wrong. –  Nov 21 '13 at 05:14
  • I edited my answer because my original answer used the same symbol $f$ for the plane wave and for the argument of a spatially-modulated sinusoid, so hopefully this makes it more clear. Apologies, a better name for $f$ would be a spherical wave centered at the origin (I said plane wave, but I guess that's not quite right). – DumpsterDoofus Nov 21 '13 at 05:17
  • Wait, why is this not the same as the double slit experiment for points along the x-axis? If it were, the middle point should have period $\lambda/sin(\theta))$ right? But according to this, it picks up a two in the denominator. –  Nov 21 '13 at 16:26
  • And how did you get the intensity again? I just looked back at this and was confused. –  Jan 13 '14 at 07:33
  • I took the absolute value squared of the amplitude (recall that light intensity is proportional to the square of the $E$-field). The rest of the algebra was done by my computer. Just as a reminder, $y^2 \left(\frac{1}{\sqrt{(a-x)^2+c^2+y^2}}-\frac{1}{\sqrt{(a+x)^2+c^2+y^2}}\right)^2$ can be set approximately equal to zero in the last step to give the book's result, assuming the light sources are comparatively far away from each other. There might be (and probably is) a simpler way to show this, but this is the brute-force approach. – DumpsterDoofus Jan 13 '14 at 15:56
  • Also as a reminder, my equation for $f(x,y,z)$ is essentially just $e^{2\pi r/\lambda}$ where $r$ is radius, which gives an outgoing spherical wave emitted from a point source, like waves in a pond emanating from a rock that was dropped in. – DumpsterDoofus Jan 13 '14 at 16:01
  • Your step from the modulus squared to the $4\cos(...)$ confuses me... How did you do that? –  Jan 13 '14 at 16:34
  • Ah, yes. Well, my computer did it, but it can also be done by hand. You just need to remember that for any complex number $a$, you have $|a|^2=aa^$ where $a^$ is the complex conjugate of $a$. Also, you have $(e^{ix})^*=e^{-ix}$ for any real number $x$. Putting those together with the fact that $\mbox{cos}(x)=\frac{e^{-i x}}{2}+\frac{e^{i x}}{2}$ and doing a little fiddling around will get you the result. – DumpsterDoofus Jan 13 '14 at 18:39