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.
The pinhole camera model is the simplest model of a perspective projection camera. In this model, the camera has a coordinate frame represented by a translation vector $d$ and a rotation matrix $R$ whose columns are the unit directions of the three axes. If $r(x,y,z)$ is a point expressed in world coordinates (relative to the absolute frame) and $p(x',y',z')$ is this same point but expressed in the camera coordinate frame then
$ r = d + R p $
So
$ p = R^T (r - d) $
In the pinhole camera model, a projection plane is created at the plane $ z' = f $ where $f $ is a positive constant. Given a point $r_1(x_1, y_1, z_1) $ then its image on the projection plane is generated as follows. Connect $r_1$ to the origin (which is $d$) and intersect this line with the projection plane.
Equation of this line of sight between $d$ and $r_1$ is
$r(t) = d + t (r_1 - d) $
Now the projection plane equation is $z' = f$, using this information, we can determine the value of $t$ that gives the intersection of the line of sight with the projection plane, simply by setting the $z'$ coordinate of $r(t)$ to $f$.
$p = R^T (r(t) - d) $, so $ p = t R^T (r_1 - d ) $. Setting the third coordinate to $f$ gives
$ t = \dfrac{f}{k^T R^T (r1 - d ) } $
where $k^T = [0, 0, 1] $
Thus the image is
$ p = \dfrac{f R^T (r_1 - d)} { k^T R^T (r_1 - d) } \hspace{15pt}(*) $
The coordinate of $p$ are the image coordinates on the projection plane. Since the projection plane in reality is not infinite but bounded, we can assume that its bounds are the region $[-x_0, x_0] \times [-y_0, y_0 ]$.
Then we can determine if a world point is visible through the image boundaries by calculating $p$ directly using equation $(*)$
Best Answer
The problem is that you are confusing two different things. What your images show is how much of the outside world you see in the camera. The impulse response is something different, like a resolution for the camera. Let's select a point outside, along the axis of the camera. You can see the image in the center of the camera as a tiny circle. If you make the pinhole larger, the image of that point becomes a larger circle. But nearby points, who would focus on slightly different position on the camera, would produce larger circles, which will overlap in the image, and everything becomes blurry.
The impulse response tells you how big is the image of a point. Let's call the point $A=(l,0)$, with $l<0$, the center of the pinhole $O=(0,0)$ and the center of the image $B=(l',0)$. Let's call $C=(0,R)$ the point on the pinhole right above $O$. A ray of light from $A$ will reach the image only if it will pass through the pinhole at a height less then $R$. At the limit, the last ray will just barely miss the pinhole's edge. It will reach the image at some point $D=(l', R')$.
Let's call $\theta=\angle OAC=\angle BAD$. The radius of the image is $$BD=AB\tan\theta$$ You have $AB=|l|+|l'|=-l+l'$, and for small angles $\tan\theta\approx\theta$, so $$BD\approx\theta(-l+l')$$ Now triangles $AOC$ and $ABD$ are similar. Then $$\frac{BD}{OC}=\frac{AB}{AO}$$ Plugging in the values, you get $$BD=\frac{(-l+l')R}{-l}=R\left(1-\frac{l'}l\right)$$