Period of Interference Pattern on a Substrate 
Can anybody explain to me where this equation came from? It's for two point sources at the two listed points, and it's calculating the period of the wave on the substrate. It seems to be $\lambda/\sin(\theta)$, which seems contrary to what I would normally expect the period to be. i.e $\sin(\theta)\lambda$ where $\theta$ is the angle from the normal.
 A: Defining $$f(x,y,z)=\mbox{exp}\left(\frac{2\pi i\sqrt{x^2+y^2+z^2}}{\lambda}\right)$$
we find that $$|f(x-a,y,0-c)+f(x+a.y,0-c)|^2=4\mbox{cos}\left[\lambda^{-1}\pi\left(\sqrt{(a+x)^2+y^2+(c-z)^2}-\sqrt{(a-x)^2+y^2+(c-z)^2})\right)\right]^2.$$
To compute the spatial frequency of $\mbox{cos}(\phi(x,y))$ it suffices to compute $\frac{1}{2\pi}|\nabla \phi|$. Since $\mbox{cos}(x)^2$ oscillates twice as fast as $\mbox{cos}(x)$, we have the spatial frequency is $\frac{1}{4\pi}|\nabla \phi|$, where $\phi$ is the expression inside the $\mbox{cos}^2$ expression I gave above. Mathematica spat out the following result:
$$\frac{1}{4\pi}|\nabla \phi|=\frac{\sqrt{y^2
   \left(\frac{1}{\sqrt{(a-x)^2+c^2+y^2}}-\frac{1}{\sqrt{(a+x)^2+c^2+y^2}}\right)^2+\left(\frac{a-x}{\sqrt{(a-x)^2+c^2+y^
   2}}+\frac{a+x}{\sqrt{(a+x)^2+c^2+y^2}}\right)^2}}{\lambda }$$
and if you simply ignore the left term in the radical and invert the expression (to get period), this is the same result as in the book.
I am not sure why the book decides to omit the left expression in the radical, although maybe they're assuming $a\gg x,a\gg y$ (ie, that the light sources are far away from each other), in which case the approximation is valid.
