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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!?

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up vote 4 down vote accepted

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: $$ \Delta \theta = \frac{2\pi}{\lambda} \cdot L \cdot (1 + \frac{1}{2} x + o(x)- 1) \approx \frac{2\pi L}{\lambda} \cdot \frac{D^2}{8L^2} = \frac{\pi D^2}{4 \lambda L} $$

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it appears to me there is an extra factor of 1/2 gained in the last line after the approximation? – user2646276 Jan 11 '14 at 16:59
Actually I misprinted $x$. Will fix now. – Valeriy Sokolov Jan 11 '14 at 17:13

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 $\Delta\theta=\frac{\pi D^2}{4\lambda L}$

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