Calculating half-angle of cone of rays entering pinhole camera I have the following pinhole camera:

The cone of rays that would enter the pinhole from the object would resemble the following:
 
(Image from https://www.optilayer.com/products-and-services/tools/cone-angle )
I want to calculate the half-angle $\alpha$ of the cone of rays that enters the pinhole camera:

(Image from this (Bubba's) answer.)
I am told that the half-angle is $\alpha = \dfrac{R}{-l}$, where $R$ is the radius of the pinhole. However, I am unsure of what trigonometry/geometry was used to find this. I would greatly appreciate it if people could please take the time to carefully explain this to me.
 A: 
The expression for the half-angle $\alpha$ makes sense if we draw the incident light cone differently--with a point on the object as apex and the circular pinhole as base. For consider that the object is emitting light in all directions from each point on its surface, but just a cone of the hemisphere of light rays from each point passes through the pinhole. The pinhole is the base of all these cones, but each has for apex a different point on the luminous object.
In the figure, let $GH$ be the object, $CD$ the diameter of the pinhole, and $EF$ the projection screen. If $J$ is a point on the object, $K$ the center of the pinhole, and $JK\perp AB$, then$$JK=-l$$and $$CK=R$$ $\triangle CJD$ is a cross-section of the cone of light emanating from $J$, and half-angle $$\alpha=\angle CJK$$
Thus$$\tan \alpha=\frac{CK}{JK}=\frac{R}{-l}$$Of course this supposes that the cone of light passing through the pinhole is a right cone. For a point like $G$ the cone is slightly oblique, and the expression for $\tan \alpha$ is only approximately correct. Thus it holds best if the object is small in relation to its distance from the pinhole. Note too that if $R$ is too great, the image of a point on the object, such as $J$, will be blurred from $J'$ to $J''$ on the camera back. Thus the smaller $\alpha$ is the better, and then, as @Matti P. notes, in radians$$\tan\alpha\approx\alpha$$
There appear to be various conditions and limitations on the workings of a pinhole camera to be taken into account, but this may be enough to resolve the difficulty in question.
