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I am studying some things surrounding the Young's double slit experiment and am trying to understand the derivations. The part that is not clear to me is the far-field approximation. That is, I understand what it means, but am failing to obtain the same equation as the tutorial.

We start with a wave of wavelength $\lambda = \frac{2 \pi}{k}$ incident on a plate with two pinholes. Each pinhole or slit acts like a source of wavelength $\lambda$.

The resultant wave at a point with distances $r_1, r_2$ from the slits is $\frac{e^{i(kr_1-\omega t)}}{r_1} + \frac{e^{i(kr_2-\omega t)}}{r_2}$

The far-field approximation we make is $r_1,r_2 \gg d$, where $d$ is the distance between the slits.

The expression for the resultant wave should be $2 \frac{e^{i(kr-\omega t)}}{r} \cos(\frac{k d}{2}\theta)$, where $r = \frac{r_1 + r_2}{2}$ and $\theta$ - small angle of deviation from the normal to the screen on which the slits are located.

It is the latter expression that I would like to obtain. Any advice or hint (preferred) is appreciated.

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You asked for a hint... express your equations as $r_1 = r+\delta$ and $r_2 = r-\delta$; then note that the intensity term ($1/r_1$ and $1/r_2$) will basically be the same for both (replace as above, and the $\delta$ term will vanish), and things will fall into place. You might need to be reminded that $e^{i\phi} = \cos\phi + i\sin\phi$

I will leave it as an exercise to see how $\delta$ relates to $d$, $\lambda$ and $\theta$... as Emilio Pisanty points out in the comment, you may need to remember that for small $\theta$, $\theta \approx \sin\theta \approx \tan\theta$.

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  • $\begingroup$ It might be worth remarking that the strict geometrical relationship between $d$, $\lambda$ and $\theta$ will not yield the result as quoted in the question. In addition to the geometry, one needs to consider the relationship between $\sin(\theta)$, $\tan(\theta)$, and $\theta$ when the latter is small. $\endgroup$ – Emilio Pisanty Sep 19 '17 at 13:32
  • $\begingroup$ Thank you. What is the $\delta$ term? Is it some arbitrarily small value? $\endgroup$ – MadPhysicist Sep 19 '17 at 13:53
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    $\begingroup$ @MadPhysicist - it is not arbitrarily small - if you draw a diagram of the rays from the two pinholes reaching the screen, you will see that as the distance to the screen increases, the two rays will become "more and more parallel"; the difference in distance to the screen then follows from simple geometry. $\endgroup$ – Floris Sep 19 '17 at 13:55
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    $\begingroup$ Alright. I understand. It is the difference in the length of the paths traveled. $\endgroup$ – MadPhysicist Sep 19 '17 at 13:57
  • $\begingroup$ I appreciate both of yours help! I have posted my own answer based on the suggestions you gave. As the last bit, does the wavelength figure in derivation at all? Are there any limiting cases or intuition lurking there? $\endgroup$ – MadPhysicist Sep 23 '17 at 21:10
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I am answering my own question with the help of @Emilio Pisanty and @Floris. Much appreciated!

Here it goes.

Consider the difference between the paths traveled by the wave emitted from slit 1 and the wave emitted from slit 2. Call them $r_1$ and $r_2$. The difference is $2\delta = r_1 - r_2$. Then, $r_1 = r + \delta$ and $r_2 = r - \delta$. That is, $r$ - average between $r_1$ and $r_2$.

Furthermore, consider the intensity terms $\frac{1}{r_1} = \frac{1}{r+\delta}$ and $\frac{1}{r_2} = \frac{1}{r-\delta}$. As $r_1,r_2 >> d$, the two rays become more and more parallel. That is, the difference between them becomes smaller and smaller. Since $\delta = d\sin\theta$, where $\theta \rightarrow 0 $, we have $\delta \rightarrow 0$. The intensities are the same for all practical purposes in far-field approximation. This makes sense intuitively.

Let's consider the original expression:

$\frac{e^{i(kr_1 -\omega t)}}{r_1} + \frac{e^{i(kr_2 -\omega t)}}{r_2} = \frac{e^{i(kr +k\delta -\omega t)}}{r} + \frac{e^{i(kr -k\delta -\omega t)}}{r} = \frac{e^{i(kr-\omega t)}}{r} \left( e^{ik\delta} -e^{-ik\delta} \right) = 2 \frac{e^{i(kr-\omega t)}}{r} \cos{k\delta}$

Since $\delta = \frac{\left( \sin{\theta} \right)d}{2}$ and $\sin \theta \rightarrow \theta$ as $\theta \rightarrow 0$, we obtain the final expressions:

$2 \frac{e^{i(kr -\omega t)}}{r} \cos (\frac{k d}{2} \theta)$

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    $\begingroup$ I am glad you were able to get the derivation with "a little help from your friends". Well done. $\endgroup$ – Floris Sep 23 '17 at 23:36

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