Why does this assumption change the formula this way I am working through some notes and I cannot understand why the following assumption changes the formula as such.
The formula is basically referring to a right angled triangle of base $ L $ and height $ \frac{D}{2} $.  The difference between the hypotenuse and the base being $ \Delta L $.
The formula is as follows
$$ \Delta\theta = \frac{2\pi}{\lambda}[\Delta L] $$
$$ \Delta\theta = \frac{2\pi}{\lambda}[\sqrt{L^2+\frac{D^2}{4}}-L] $$
It then states that assuming $ L >> \frac{D}{2} $
$$ \Delta\theta \approx \frac{\pi D^2}{4\lambda L} $$
But, why!?
 A: Let us rewrite the second formula this way:
$$
\Delta\theta = \frac{2\pi}{\lambda} \cdot L \cdot \left[ \left(1 + \frac{D^2}{4L^2}\right)^{1/2} - 1\right] = \frac{2\pi}{\lambda} \cdot L \cdot \left[ \left(1 + x \right)^{1/2} - 1\right],
$$
where $x = \frac{D^2}{4L^2}$.
Then, since $D << L$, we can approximate $\left(1 + x \right)^{1/2}$ around $x \approx 0$ by Taylor series:
$$
\left(1 + x \right)^{1/2} = 1 + \frac{1}{2} x +o(x).
$$
So the expression of $\Delta\theta$ can be rewritten like this:
$$
\Delta \theta = \frac{2\pi}{\lambda} \cdot L \cdot (1 + \frac{1}{2} x + o(x)- 1) \approx \frac{2\pi L}{\lambda} \cdot \frac{D^2}{8L^2} = \frac{\pi D^2}{4 \lambda L}
$$
A: Just do this modification $\Delta\theta=\frac{\frac{2\pi}{\lambda}[\sqrt{L^2+\frac{D^2}{4}}-L]\times [\sqrt{L^2+\frac{D^2}{4}}+L]}{[\sqrt{L^2+\frac{D^2}{4}}+L]}$
hence you get: 
$\Delta\theta=\frac{\frac{2\pi}{\lambda}[(L^2+\frac{D^2}{4})-L^2]}{[\sqrt{L^2+\frac{D^2}{4}}+L]}=\frac{\frac{2\pi}{\lambda}(\frac{D^2}{4})}{[\sqrt{L^2+\frac{D^2}{4}}+L]}=\frac{\frac{\pi D^2}{2\lambda}}{[\sqrt{L^2+\frac{D^2}{4}}+L]}$
Now the denominator is $[\sqrt{L^2+\frac{D^2}{4}}+L]=[L\sqrt{1+(\frac{D}{2L})^2}+L]$
Using your assumption: $\frac{D}{2L} << 1$ then $\sqrt{1+(\frac{D}{2L})^2}$ is almost equal $1$. Therefore:
$\Delta\theta=\frac{\frac{\pi D^2}{2\lambda}}{2L}$ or $\Delta\theta=\frac{\pi D^2}{4\lambda L}$
