Why intensity of light reaching the sensor or film with a particular lens directly proportional to $\frac{D^2}{f^2}$? The following is quoted from my book:
"The intensity of light reaching the sensor or film is proportional to the area viewed by the camera lens and to the effective area of the lens. The size of the area that the lens "sees" is proportional to the square of the angle of view of the lens, and so is roughly proportional to $\frac{1}{f^2}$. The effective area of the lens is controlled by means of an adjustable lens aperture, or diaphragm, a nearly circular hole with variable diameter $D$; hence the effective area is propor tional to $D^2$. Putting these factors together, we see that the intensity of light reaching the sensor or film with a particular lens is proportional to $\frac{D^2}{f^2}$."
My question is how did they conclude that the area that the lens "sees" is roughly proportional to the square of angle of view of the lens and $\frac{1}{f^2}$ and how is the effective area proportional to $D^2$? Ultimately, my question is how is the intensity of light proportional to $\frac{D^2}{f^2}$? Can someone please explain? I did not understand what the paragraph explained. Please help.
 A: The first term, proportional to $D^2$, is rather simple: The larger your lens aperture, i.e. the larger the area that collects light, the more photons you get.
The second term is a bit misleading. Actually, the smaller the focal length $f$, the larger the angle of view $\alpha$ that the camera sees, both in the horizontal and vertical:
$$
\alpha = 2 \cdot \arctan \left(\frac{d}{2 f} \right)
$$
(see https://en.wikipedia.org/wiki/Angle_of_view). $d$ is either the width or height of your camera chip, $\alpha$ the corresponding angle (so there is $\alpha_w$ and $\alpha_h$).
The width of the observed scene is proportional to $\tan(\alpha_w)$, the height is proportional to $\tan(\alpha_h)$. Thus according to the equation above you see that both width and height are proportional to $\frac{1}{f}$. The total amount of photons your camera collects is of course proportional to the image area, therefore to the product of width and height and thus proportional to $\frac{1}{f^2}$.
A: Since we are talking about a camera, it's more appropriate to talk about the illumination of an image which translates to the amount of light per unit area of the image. The dependence on D$^2$ is straightforward as that is proportional to the area of the lens aperture.
The magnification due to the camera lens is given approximately by the image distance divided by the object distance. For a far away object, the image distance is approximately equal to f. So the linear extent of the image depends on f and the area of the image depends on f$^2$. The smaller the area, the more concentrated is the light on the image.
So the image brightness depends, for a distant object, on D$^2$/f$^2$. This is just the inverse of the f/number and as we know the smaller the f/number (hence the larger D$^2$/f$^2$), the brighter the image.
