Far Field Approximation in Young's Double Slit Experiment 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.
 A: 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 $\theta$, $\theta \approx \sin\theta \approx \tan\theta$.
A: 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:
$2 \frac{e^{i(kr -\omega t)}}{r} \cos (\frac{k d}{2} \theta)$
