In my a physics book i found this formula regarding minimum angular distance for resolution.
$$\theta_R \sim \sin \theta_R = 1.22~ \lambda/\mathrm{diameter}$$
I dont understand what the tilde is saying. Id guess $\theta_R$ is approximately $\sin\theta_R$ but that doesnt make sense.. does it?
On
The tilde is used for asymptotic equivalence.
If $f$ and $g$ are two functions then $f(x)\sim g(x)$ for $x$ in a neighborhood of $x_0$ means that $\lim\limits_{x\to x_0}\frac{f(x)}{g(x)}=1$
It works also for limits at infinity.
(provided of course that $g(x)\neq 0$).
If you know the $o$ notation it also means that $f(x)=g(x)+o(g(x))$
This means that $f(x)$ and $g(x)$ differ only by a quantity which is negligible compared to either $f(x)$ or $g(x)$.
For instance $f(x)=x+3x^2$ and $g(x)=x-5x^3$ when $x\to 0$. We have $f(x)\sim g(x)\sim x$, because they differ from $x$ by quantities $x^2,x^3$ that are even smaller.
For your example $\sin(\theta)\sim \theta$ when $\theta\ll 1$.
If you have a look at the Taylor expansion $\displaystyle \sin(\theta)=\theta-\frac{\theta^3}3+o(\theta^3)$
Remember the definition : $\displaystyle \frac{\sin(\theta)}{\theta}=1-\frac{\theta^2}3+o(\theta^2)\to 1\quad$ so $\sin(\theta)\sim \theta$.
By Taylor series expansion $sin(x)=x-\frac{x^3}{3!}+ \frac{x^5}{5!}...$ and this is approximately equal to $x$ for small $x$