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 08/17/2005 04:57:26 PM · #1
ISO 100 1/100 F16...

is that the same as...

ISO 200 1/200 F16

and is THAT the same as...

ISO 200 1/400 F8?

and while we're at it, would THAT be the same as...

ISO 400 1/200 f8?
 08/17/2005 05:02:35 PM · #2
Originally posted by deapee:
ISO 100 1/100 F16...

is that the same as...

ISO 200 1/200 F16

yes

and is THAT the same as...

ISO 200 1/400 F8?

no, that would be 1/400 @ f/11

and while we're at it, would THAT be the same as...

ISO 400 1/200 f8?

ISO 400 1/200 f/8 = ISO 200 1/200 f/11 = ISO 100 1/200 f/8


R.
 08/17/2005 05:07:57 PM · #3
hrmm why does one stop less than 16 equal 11 instead of 8?
 08/17/2005 05:11:15 PM · #4
hrmm does it have something to do with 1.4?

like 16 divided by 1.4 is about 11.5 -- so we just say 11? then 11/1.4 is about 8?

Where does the number 1.4 come from?
 08/17/2005 05:13:30 PM · #5
does that have something to do with the inversed squared law?

Something about the area of a circle being directly proportional to the change of the diameter squared...

but I dont see where we get the number 1.4 from or is it sort of like pi?
 08/17/2005 05:17:26 PM · #6
You mean like grandma's steaming apple pie?
 08/17/2005 05:18:33 PM · #7
Well, it has to do with a square root, but you seem to have a lot of questions, so I'll just point you at a couple of links and see if they are complete enough for you:

//www.paragon-press.com/lens/lenchart.htm

//www.uscoles.com/fstop.htm
08/17/2005 05:33:39 PM · #8
2.8 4 5.6 8 11 16 22 32

Those are the standard "stops"... Each is one stop from the next.

R.
 08/17/2005 05:38:44 PM · #9
OK cool, thanks...I was a bit confused..I'm all caught up now.
 08/17/2005 06:03:24 PM · #10
OK, a puzzle for you.

If light falls off in accordance with the inverse square law, why is the exposure for a subject the same no matter how far away the camera is from the subject, assuming that the lighting on the subject is constant?

In other words, if I have a subject that meters at 1/125 @ f11 with the camera 10 feet away, why doesn't the exposure change to 1/125 @ f8 when I move to 14 ft from the subject? The light reflected off of the subject that exposes the image on my sensor/film is now 1.4 times as far away.
 08/17/2005 06:35:30 PM · #11
Originally posted by Spazmo99:
OK, a puzzle for you.

If light falls off in accordance with the inverse square law, why is the exposure for a subject the same no matter how far away the camera is from the subject, assuming that the lighting on the subject is constant?

In other words, if I have a subject that meters at 1/125 @ f11 with the camera 10 feet away, why doesn't the exposure change to 1/125 @ f8 when I move to 14 ft from the subject? The light reflected off of the subject that exposes the image on my sensor/film is now 1.4 times as far away.


Because the inverse square law refers to the light SOURCE. It's how far the light is from the subject, not how far the camera is from the subject.

Robt.
 08/17/2005 07:05:37 PM · #12
oops


Message edited by author 2005-08-17 23:10:31.
 08/17/2005 07:08:26 PM · #13
Originally posted by bear_music:
Originally posted by Spazmo99:
OK, a puzzle for you.

If light falls off in accordance with the inverse square law, why is the exposure for a subject the same no matter how far away the camera is from the subject, assuming that the lighting on the subject is constant?

In other words, if I have a subject that meters at 1/125 @ f11 with the camera 10 feet away, why doesn't the exposure change to 1/125 @ f8 when I move to 14 ft from the subject? The light reflected off of the subject that exposes the image on my sensor/film is now 1.4 times as far away.


Because the inverse square law refers to the light SOURCE. It's how far the light is from the subject, not how far the camera is from the subject.

Robt.


But the same light is still travelling from the subject to my camera, and if my camera moves farther away, the light still falls off over that distance. The laws of physics apply to all light, reflected or incident.

My puzzle remains unanswered.

Message edited by author 2005-08-17 23:10:53.
 08/17/2005 07:49:50 PM · #14
Originally posted by Spazmo99:
Originally posted by bear_music:
Originally posted by Spazmo99:
OK, a puzzle for you.

If light falls off in accordance with the inverse square law, why is the exposure for a subject the same no matter how far away the camera is from the subject, assuming that the lighting on the subject is constant?

In other words, if I have a subject that meters at 1/125 @ f11 with the camera 10 feet away, why doesn't the exposure change to 1/125 @ f8 when I move to 14 ft from the subject? The light reflected off of the subject that exposes the image on my sensor/film is now 1.4 times as far away.


Because the inverse square law refers to the light SOURCE. It's how far the light is from the subject, not how far the camera is from the subject.

Robt.


But the same light is still travelling from the subject to my camera, and if my camera moves farther away, the light still falls off over that distance. The laws of physics apply to all light, reflected or incident.

My puzzle remains unanswered.


Look at it this way: assuming the sun is your light source, and assuming the subject has 100% relectivity (just for the sake of argument): then if the sun is a gazillion feet away, and you move your camera 10 feet further away, the light has to travel a gazillion-plus-10 feet, an insignicicant difference.

And in fact, the full moon is exposed with the same exposure as a daylit scene. The differnce between earth and moon is insignificant compared to earth and sun.

Now, if you are lighting in the studio, the moving of the camera is only significant as a percentage of the total distance the light has to travel. If the light's 15 feet from the subject and the camera's 5 feet fromn the subject, the total travel of light is like 20 feet. And if if you move the camera back 2 more feet, you might think 5 feet to 7 feet was significant, but the operative distance is 20 feet to 22 feet, and the difference is negligible.

Robt.


08/17/2005 08:00:10 PM · #15
The other confusion is because as you move the camera back, you are decreasing the area the reflected photons cover (the image gets smaller). If you were to zoom in so that the subject was the same apparent size, I think you'd have to open the aperture to get the same exposure.

There's a certain number of photons of a given energy reflected off any object. How large an area of sensor you expose them to should have an effect on the brightness; remember that the inverse-square law is describing the number of photons striking a given area. If you have a bigger area to cover, there's fewer photons/cm sq or whatever area unit yyou want to use.
 08/17/2005 09:01:36 PM · #16
Originally posted by deapee:

but I dont see where we get the number 1.4 from or is it sort of like pi?


1.4 is the square root of 2.
Basically, and increase of 1 stop lets in twice as much light. To get twice as much light, you need to increase the area of the aperture by a factor of 2.

So, for the sake of convenience, imagine a square aperture. To increase the area of the square by a factor of 2, you would need to increase the length of each side by the square root of 2, or approx. 1.4.

I hope that's both clear and concise, but it probably isn't. :)
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