photographic conundrum

[FrankS]

Following on the reciprocity failure thread, here's another technical issue:

Here's a question whose answer eluded me for many years. I wonder if anyone has thought about this. It has to do with the inverse square law: light intensity/strength is inversely proportional to the square of the distance from the source. No problem, I can dig that.

How come the light reflected from a subject doesn't act the same way? If I light a beautiful woman in a studio such that proper exposure at 2 meters is say, f8 (@1/60sec), why is the proper exposure at 4 meters (double the distance) still f8 (@1/60sec)? Shouldn't the light intensity/strength follow the inverse square law too?

 

[Xmas]

The size of the image shrinks, by *2, if you need the image the same size then the telephoto lens is bigger, (or zoom).

Noel

 

[uhligfd]

Xmas, your answer unfortunately does not make any sense; even with a tele lens of double the focal length from twice as far away, the f stop and exposure would not change.

And the moon at much greater distance still falls under the 1/16 sunshine rule, if there is no haze in the sky, no matter what focal length is used, 50mm or 1500 mm. So what now?

 

[Bryce]

Noel nailed it.
Consider for a moment what the world around us would look like if that weren't true- on a sunny, clear day, the person standing next to you would look as they do now, but the stop sign at the end of the black would be barely visible, and a mountain range 50 miles away would be pitch black.
I think I like things better as they are!

 

[Bryce]

uhligfd-
Noel's point about the tele lens is only that you would need a longer lens to make the same size image on film from a longer distance.
The F/ number system makes up for the difference in required aperture (entrance pupil). For a 50mm lens at F/8 this is about 6mm, and for a 1500mm about 180mm.

 

[ferider]

Think about it this way, Frank:

In one case you have a point light source, all the light leaves one point at
different angles in the directions determined by the exit angle (well defined
for a flash for instance). The square law is related to the cross-section of the
light cone leaving that point source.

In the other case you basically have a "wall of light", where all light rays
exit in parallel. Instead of a light cone you have a cylindrical shape, the
diameter/cross section of which does not change. Sun and Moon are great examples.
They are not point light sources, even though they seem like it.

Roland.

 

[Bryce]

And here is another example from the night sky to really confuse-
A true point light source for photographic purposes is one that is smaller than the circle of confusion produced by the lens. Makes sense, right?
Stars are true point sources by that definition. As a result, their brightness on film varies only by exposure time and the entrance pupil of the lens, not the F/ number of the lens.
So two images of a nebula with stars around it, both with the same length exposure, film, and F/ stop but taken through a long lens and a short one will look different. Not only will the image be larger in the frame taken with the longer lens, but the stars will appear brighter. The nebula will have the same brightness in both images, though, since it is not a point source.
Sounds wierd but it is true.

 

[FrankS]

Think about it this way, Frank:

In one case you have a point light source, all the light leaves one point at
different angles in the directions determined by the exit angle (well defined
for a flash for instance). The square law is related to the cross-section of the
light cone leaving that point source.

In the other case you basically have a "wall of light", where all light rays
exit in parallel. Instead of a light cone you have a cylindrical shape, the
diameter/cross section of which does not change. Sun and Moon are great examples.
They are not point light sources, even though they seem like it.

Roland.

Roland, but the same holds true for a softbox or umbrella that aren't point sources.

(Anyway, Noel got the answer right in the 2nd post to this thread. Damn Noel! 🙂)

 

[FrankS]

you're measuring reflected light off the surface of the subject, AT the subjects position or the amount of light reaching your subject with an incident meter, at the position of the subject. Either way the amount of light measured at the subjects position does not change just because you've decided to place your camera further away.

Don't understand this.

Yes, but why not?

 

[ferider]

Roland, but the same holds true for a softbox or umbrella that aren't point sources.

(Anyway, Noel got the answer right in the 2nd post to this thread. Damn Noel! 🙂)

Yes they are, all the softbox does is diffuse the light emission further.
A huge convex lens generating parallel light would be a different matter.

How about a wall of 20 x 20 soft boxes ?

You can always look at it from the point of view of the light source (my explanation)
or the observer's (Noel).

What happens if instead of at a beautiful woman lightened by several sources,
you shoot a single flash at a reflector and you take a photo of that reflector ?

Now make the reflector slowly less reflective. What happens ?

Roland.

 

[FrankS]

Yes they are, all the softbox does is diffuse the light emission further.
A huge convex lens generating parallel light would be a different matter.

You can always look at it from the point of view of the light source (my explanation)
or the observer's (Noel).

What happens if instead of at a beautiful woman you shoot a flash at a reflector ?

Roland.

This is quibbling about definitions and semantics on my part, but a softbox or umbrella is not a photographic point source, it is a broad light source. (wall of light)

 

[sepiareverb]

Geez Louise my head is spinning. This is why I need a camera with a meter.

 

[ferider]

This is quibbling about definitions and semantics on my part, but a softbox or umbrella is not a photographic point source, it is a broad light source. (wall of light)

Look Frank, forget the term point source. All that matters is the angle the rays leave
the source with, either parallel or as a cone. You put a softbox in a dark room on
the ground. Will the ceiling receive light ?

But you understand Noel's explanation so never mind.

 

[FrankS]

Here's another way of describing the situation. Light that travels from a light source, say a softbox, follows the inverse square law, falling off rapidly as the distance from the light to the subject increases. However, light that is reflected off the subject does not seem to follow the inverse square law, exposure doesn't change regardless of distance to the subject.

Also Roland, it doesn't matter how big the wall of light is, it's illumination of the subject will still follow the inverse square law. It doesn't matter if the light source is a point source or a braod source.

 

[ferider]

Also Roland, it doesn't matter how big the wall of light is, it's illumination of the subject will still follow the inverse square law. It doesn't matter if the light source is a point source or a braod source.

And this is incorrect. It does matter. Next time you watch soccer or football look at the
stadion lights and their beams. They do not decrease in intensity with the square of the
distance (partially due to amount of bulbs and partially due to the parabolic reflectors).

Roland.

 

[FrankS]

Roland, sorry if this is frustrating. I'm not picking an argument, I'm just posting a problem. I'm not intending a mean-spirited argument at all. 🙂 Please forgive me if I made you feet that way.

 

[FrankS]

Roland I get your point about projected parallel beams of light, the extreme being a laser. Yes you are right on that!

 

[ferider]

Roland, sorry if this is frustrating. I'm not picking an argument, I'm just posting a problem. I'm not intending a mean-spirited argument at all. 🙂 Please forgive me if I made you feet that way.

Not feeling bad at all, Frank. I'm content 🙂

Yes, the laser is a great example. A cop's flash light, too. And the Fresnel lens in a lighthouse or on
my Metz flash ...

Best,

Roland.

 

[bsdunek]

Boy, you guys made me think.
The light source has rays emitting in all directions. If you're close, you get a lot of them, as you get farther you get less rays.
When you take a photo (or a light reading), you are reading the rays that are parallel to the axis from the subject to the camera. Because they are parallel, they don't "fall off".
If I could draw a picture here, it would make more sense.😎

 

[sepiareverb]

So Frank, who is the girl anyway?

And if you light a girl in the dark and there is no camera to see it does she still reflect?