I have looked at this, but it did not help with locations. Really this just comes down to mathematical manipulation, which for some reason I fail to see. Here is my paraphrased setup:

Consider two sources of light separated by some distance $a$, and suppose they are emitting uniformly in all directions, at the same angular frequency, and same wave number. Determine the locations on a plane, which is parallel to the line joining the two light sources, at which the intensity is a minimum.

This is what I have. Let $r_1$ denote the distance from the first source to a point on the screen, $r_2$ for the second source to the point, and $L$ be the distance from the midpoint of the charges to the screen. Consider some point $p = (y,z)$ on the screen. Letting the x-axis be orthogonal to the screen and its origin at the midpoint, and the y and z axes to be parallel to the screen, we can write $$r_1 = \sqrt{L^2+y^2+(z-a/2)^2} \space\space\space \text{&} \space\space\space r_2 = \sqrt{L^2+y^2+(z+a/2)^2}$$ From this, we know that we can write the total wave at a point $p$ as $$E(t) = e^{iwt}(E_1e^{ikr_1}+E_2e^{ikr_2})$$ I am not substituting in yet for clarity, and note that $E_1$ and $E_2$ are the amplitudes of the respective waves. Now we also know that the intensity at this point has the following relationship: $I \propto |E|^2$, and we know from above that

$$|E|^2 = E_1^2+E_2^2+2E_1E_2cos\biggl[k\Bigl(\sqrt{L^2+y^2+(z-\frac{a}{2})^2}-\sqrt{L^2+y^2+(z+\frac{a}{2})^2}\Bigr)\biggr]$$

If things are correct up to this point, then it is clear that the minimum intensity locations occur when this cosine is a minimum. Namely, when

$$\sqrt{L^2+y^2+(z-\frac{a}{2})^2}-\sqrt{L^2+y^2+(z+\frac{a}{2})^2} = \frac{(2n+1)\pi}{k} \space\space n \in \mathbb{Z}$$

However, I cannot seem to put this into a nice simplified form for z as a function of y. I suspect that it is a hyperbolic relationship, but is there a nice way to reduce this to a compact form?

  • 1
    $\begingroup$ I believe you're missing a + sign in your last two expressions, in the (z-a/2) part. Also to get large parenthesis around your big expressions there you can use \bigl( and \bigr) instead of ( and ). $\endgroup$ Apr 3, 2014 at 6:16
  • $\begingroup$ @Nuclear_Wizard Thanks for pointing that out, I forgot to change those after copy and pasting from the previous expression. $\endgroup$
    – dsm
    Apr 3, 2014 at 6:19
  • $\begingroup$ When you manage to answer your own question, you should post it as an answer to the question, not edit the question to include it. Could you please copy and paste your 'edit' into an answer? $\endgroup$ Apr 6, 2014 at 14:25
  • $\begingroup$ @Nuclear_Wizard Thanks for letting me know, I updated the thread accordingly. $\endgroup$
    – dsm
    Apr 7, 2014 at 6:30
  • $\begingroup$ @ChrisMueller The above comment was directed at you, not Nuclear_Wizard. Thanks for letting me know. $\endgroup$
    – dsm
    Apr 8, 2014 at 14:37

2 Answers 2


I asked the purely mathematical question over here, and received the most complete answer. While I thought there would be a simple trick to seeing the hyperbolic relationship, it looks like you just have to go through the tedious algebra to have it pop out. User JJacquelin found that it can be rearranged to the form $$\frac{z^2}{(\frac{(2n+1)\pi}{k})^2}-\frac{y^2}{L^2-(\frac{(2n+1)\pi}{k})^2}=\frac{1}{4}+\frac{a^2}{L^2-(\frac{(2n+1)\pi}{k})^2}$$ To show how this looks, I set $L = 20 \space \text{cm}$, $a= 2 \space \text{cm}$, and an infrared light source of $\lambda = 1000 \space \text{nm}$. Plotting a modest amount of these produced the following image: enter image description here


Mathematica gives $$ z=\gamma\frac{\sqrt{\gamma^2-a^2-4(y^2+L^2)}}{2\sqrt{\gamma^2-a^2}}, $$ where $$ \gamma=\frac{(2n+1)\pi}{k}. $$


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