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# Far-Field Approximation in Young's Double Slit Experiment

## In this post we derive a far-field approximation for intensity of a wave (interference pattern) at a distance far removed from the slits.

Posted by Yakov Boyko on September 23, 2017

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:

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