Q:

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 $\theta$, $\theta \approx \sin\theta \approx \tan\theta$.

