I like it too!
[long technical story alert]
regarding the final question from your first post that hasn't been answered yet (f-stops): in photography, a stop is a doubling or halving of the amount of light that reaches the film/sensor. So one stop brighter is twice as bright and one stop slower is half as bright (and 2 stops is 2x2=4 times as bright/dark, 3 stops 2x2x2=8 times, etcetera).
the aperture (the "bottleneck" in the lens that determines how much light is allowed to reach the sensor for any given amount of time) is commonly measured as "focal length (for instance 50mm) divided by "f-number" = physical aperture diameter". So if the physical aperture diameter is 25mm, the notation is 50/2, or just f/2. Note that the larger the physical aperture and the more light is transmitted, the lower the f-number (for a given focal length).
One f-stop is an increase or decrease of the aperture to allow twice as much or twice as little light to reach the sensor. Because the light is going through a round shaped aperture, it's the surface of that circle that needs to double in order to double the amount of light. In order to double the aperture circle's surface, the diameter needs to be multiplied by the square root of 2, which is ~1,41, usually rounded off to 1.4. So if you want to increase the aperture to allow one extra stop of light (in other words, double the amount of light) you need to multiply the aperture (which was 25mm) by roughly 1,4, which gives ~35mm. If you again calculate the f-number again, it's 50/35= ~1.4 (like I said, the larger the aperture and the brighter the image, the lower the f-number).
If you make the same calculation over and over again to get the lens one f-stop "brighter (aka faster)" or "darker (aka slower)" with each step, you'll see that the f-numbers are (from low=bright to high=dark) 1; 1,4; 2; 2,8; 4; 5,6; 8; 11,2; 16; 22,4. Each stop darker / slower makes the f-number 1,4 times higher, and each time you go 2 stops darker (so 2x2=4 times as little light) the f-number is doubled.