Exercise 0: Certain Basics in Photography

There are a few questions in the "Get to know your camera" section that caught my eyes:


What the f-stop numbers stand for?
How the lens aperture controls the light reaching the film?
How the shutter speed controls the light?


Ok, do I know the answer? I know the new few exercises will be providing "some understanding". But before that, why don't I go to the drawing board and try to run this through theoretically and see if I roughly get the concepts?



Consider for a moment that the camera is a box with a hole, and light enters from the hole in front. Simple logic: the bigger the hole is, the more light we can possibly get. Entrance area is approximately circle, the area is approximate Pi*r^2. If we speak of radius, everybody can easily understand why increase the radius by sqrt(2), the area of the circle increases by 2.


Let's introduce f-stop = focal length / diameter = focal length / (2* radius)


This f-stop is also known as relative aperture, or just "aperture".


So we want to double the light entrance, we either make the radius sqrt(2) time larger, or make f-stop swinks by sqrt(2). Let's start with f-stop of 1 and calculate sqrt(2) multiple of it and go down the row.

Actual	Rounded (usual f-stop)
1	1.0
1.4142135623731	1.4
2	2.0
2.82842712474619	2.8
4	4.0
5.65685424949238	5.7
8	8.0
11.3137084989848	11.3
16	16.0
22.6274169979695	22.6

Take my Nikon 50mm f/1.8D as an example:

f/#	Diameter (mm)	Area (mm^2)
1.8	27.78	606
2.8	17.86	250
4	12.50	123
5.6	8.93	63
8	6.25	31
11	4.55	16
16	3.13	8
22	2.27	4

The lens take 52mm filter so its diameter is about 52mm. In theory if we don't need the element that control focus, apeature size, ignore distrotion around the edges of the lens and assume we can hold the lens in the tube, it is possible to make the 50mm an f/1 without increasing the diameter.


Let's try one more example with a Sigma 70-200mm f/2.8, say I set the lens at 200mm.

f/#	Diameter (mm)	Area (mm^2)
2.8	71.43	4,007
4	50.00	1,963
5.6	35.71	1,002
8	25.00	491
11	18.18	260
16	12.50	123
22	9.09	65

The lens takes 77mm filter, so roughly the diameter of the lens is about 77mm. At it widest aperture, the lens really "wide open". With the same aperature, the diameter is different if the focal length is different. This tells me that if a lens is a telephoto and also a fast (low f-stop) lens, it probably will be bigger. So if there is a lens with focal length 200mm with f/1, I am expecting the lens has diameter at least 20 cm, which is going to be huge and heavy.


One strange thing I noticed. If you keep both lens at f/2.8, the calculation indicates that the lens with 50mm focal length will have aperture size of 250 mm^2, while the 200mm one will have aperture size over 4000 m^2. Well, the general photography knowledge says that the exposure by increasing the aperture size. If this is true, then isn't 200mm f/2.8 can handle low light photography better than a 50mm f/2.8? After all, it has 16 times the area. But they are both f/2.8!


I found it mystifying when I first come to this conclusion. After some research, I found an American Electrical Engineer, Mr Doug Karr has written an article about this. The misconception is when we said "exposure", it is not just light intensity, but illuminance.The lens in front of the camera transforms the light from the object to the final image hitting the sensor. During this transformation, one multiples the incoming light flux per area (luminance) by the solid angle (in unit steradian). If a camera is a box with a hole, we care about the real size of the aperture (the hole), but because of the lens, we care about the solid angle span by some parallel beams hitting the lens and focus at the focal point. In this case although the aperture area is large, the focal length is also longer. Thus the solid angle of the light cone merges at the focal point is unchanged. In that calculation of the solid angle (area/ radius^2), we get the ratio of (lens diameter)/(focal length). What is (lens diameter)/(focal length)? The inverse of aperture (f-stop)!


For the details, please read Mr. Kerr's article here:


http://dougkerr.net/pumpkin/articles/Photographic_Optics.pdf


Try this: grab a zoom lens and set the f-stop to the largest (easier to see), now move the focal length to short to long, watch the area of the aperture.


It is interesting to note that I usually think of light as photons. So if I have a larger hole for light beam to enter, in theory I should larger number of photon and more energy. However, I suspect that photon is really a concept in quantum theory while we are mainly looking at the classical wave theory here. If we really want to go pure quantum theory, it is hard to describe what exactly happen to the wavefunction when it interacts with the lens. We can probably find answer in a Quantum Optics text, which I doubt I can manage to read one.


Honestly, I am so glad that this course is called "ART of photography" but not "SCIENCE of photograph". 

Let me get my camera out and take some shots. This is probably easier.