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:

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