Question about light getting through a aperture (full stops, f/#)

[mr magoo]

The term aperture is used in reference to the definition given to it by photography.

Here is the two f stops: f/1.0, f/22.
These two f stops represent the absolute opening of the aperture and the absolute closing of the aperture.

I want to theoretically, put a box over a light source on all four sides of the light source, and on top of the box there is a aperture of f/11. I want to know if the light getting through the aperture will equal f/1.4?

Or another way of putting the question is: do I always get the equal amount of light through a aperture the amount a aperture is open; If the aperture is only open 20%, does only 20% of the light from the light source get through the aperture even if the aperture is on top of a box covering the light source?

I am asking because I'm going to draw this box in autocad and want to know how to measure the light getting through the aperture.

 

[NobodySpecial]

f# and apertures only make sense in the case of an optical system - and only then when they are the limiting stop in the lens.

If you had a light source that emitted uniformly in all directions then a circular aperture in a box at a certain distance would emit a proportion of the light which depended on the solid angle of the hole

 

[mr magoo]

It's a rectangular aperture, and the light source is a pixel from a lcd display. See the picture:

Once the light gets through the box's rectangular aperture that's on the far side of the box, the light hits square mirrors and gets reflected onto a second aperture that's in the shape of a square.
These rectangular and square apertures do not close, but the pixel flashes and this acts as the opening and closing of light.

I want to know if I had the aperture open on the far side, how would I get 80 percent of the light through the first aperture to the mirrors: how big would the aperture on the box have to be?

 

[NobodySpecial]

If the beam is collimated and if the beam is uniformally illuminated then the brightness will depend on the area of the aperture that the beam goes through.

 

[mr magoo]

Yes, the beam from the pixel is collimated, and the pixel area is uniformally illuminated.
So, this means that if the aperture I described takes 20% or the total area on top of the box covering this light source, only 20% brightness will show.

Thank you for showing me how to calculate the brightness, NobodySpecial! :)

 

[Andy Resnick]

Here is the two f stops: f/1.0, f/22.
These two f stops represent the absolute opening of the aperture and the absolute closing of the aperture.

I want to theoretically, put a box over a light source on all four sides of the light source, and on top of the box there is a aperture of f/11. I want to know if the light getting through the aperture will equal f/1.4?

Or another way of putting the question is: do I always get the equal amount of light through a aperture the amount a aperture is open; If the aperture is only open 20%, does only 20% of the light from the light source get through the aperture even if the aperture is on top of a box covering the light source?

Something about this question doesn't make sense- you seem to be using imaging concepts where there is no imaging in your system. For example, if I have a lens set at f/1, it's *always* at f/1, regardless of how much light goes through the entrance pupil. Perhaps you are thinking about vignetting- the lens aperture is not the limiting surface in that case.

Or am I missing something?

 

[mr magoo]

When NobodySpecial first answered I realized I used full stop terminology wrong and I should have just used aperture shape and size and light source description.

I was told that if the light source my box is covering, sends collimated light uniformally over the entire covered area. Then the area of the aperture equals the brightness percent that gets through the aperture.

After I send this through the first aperture, the aperture on top of the box covering the light source. I will reflect this light that gets through the aperture by square Plane mirrors that are smaller than the area of the first aperture. These plane mirrors reflect the light down and at a angle.

This reflected light from the Plane mirrors does not redirect all the light from the first aperture because they are square and since there are four plane mirrors they have space inbetween them and so the brightness they reflect is not equal to the brightness that got through the first aperture.

This reflected light is then put through glass plates. The glass plates serve to collimate the light but act as a third aperture, and this third aperture does not receive all light from the second aperture because it's too small and the light from the mirror was not perfectly sent through.

Then when the light leaves the glass plates it reflects off of another plane mirror, and this plane mirror is in a set of 4 individual mirrors and they steer the light to the left or right eye of four people. This light beams the light up and at different angles set by the designer.

I was going to try and put the entire box on top of one pixel, but after reading NobodySpecial's answer I figured the aperture on top of the box would not be covered at all but expose all the light from the pixel to the mirrors.
Then the mirrors would reflect to a space beside the pixel and this space beside the pixel would hold the part that reflects the light back up to the eye(s).

I have a youtube channel with three videos about my experiment: http://www.youtube.com/user/foxhoundfan"

I'm learning autocad so I can try and see how much total light brightness is reflected by the second set of plane mirrors to the eyes. So I'll know how bright the pixel has to be so the eyes can see enough light the picture looks special.