Metering Question

[Helen B]

This is exactly what I was thinking. But according to other responses, I'm wrong.

Oh dear. You are right and your buddy is wrong.

 

[Alex_B]

he is wrong in thinking he is wrong ...oh wait! ...

 

[dandaluzphotography]

lol

 

[480sparky]

I am probably completely wrong (I'm sure somebody will correct me very soon) but if he stands farther away, but zooms in on the same subject, isn't the reflective light going to be the same? In other words, isn't an object that fills the frame from 10' away at 25mm the same reflective light from 40' away at 75mm or 100' away at 170mm (numbers don't matter in my example--just making the point).

Wouldn't the inverse square law only apply if, for example, he was operating with a prime lens and the distance from the front of the lens to the reflective light had physically changed when he moved from 10' away to 40' away?

I guess another way of asking the same question would be "For metering purposes, is the "value" of the reflective light that falls on the the light sensor the same at different distances, provided that by zooming in on the subject, the same light fills the frame?"

Or this: By zooming in on the subject from farther away, will the "front" of the lens act as if it was the same distance from the subject as when the photographer is physically closer to the light source by the zoom of the lens has been shortened?

If the distances are fairly close, I can see how this makes more sense. When they get further away, I can see how my hypothetical quickly falls apart.

Comments?

Light doesn't know, or care, what lens you're using. Prime, zoom, parfocal, reversed macro, makes no difference. The light loss occurs before it gets to the lens. Inverse Square law still applies.

 

[Helen B]

Danny,

I'm not in a rush to get to B&H before it closes any more, so I have time to clear up any doubts you may have about being correct. The intensity of the light does fall off as you get further from the object but as you say, if the size of the image of the object remains the same, and the shutter speed and ISO also stay the same then the f-number stays the same - because the larger size of the aperture (actually the entrance pupil) exactly makes up for the fall-off in light intensity, so the total amount of light that makes up the image does stay the same. There's no need to re-meter.

Is there anything that needs clarification to persuade you that you are correct?

10' to 40' - light intensity falls off to 1/16 (I don't know why Sparky said that it was negligible - maybe he wasn't referring to the intensity);
- focal length must go up to 4x to maintain same image size;
- entrance pupil diameter must also go up 4x to keep same f-number;
- therefore entrance pupil area goes up 4^2 = 16x;
- overall amount of light that makes up the image stays the same (1/16 the intensity, 16 x the area of light collected)
- brightness of image is the same.

This all assumes that the subject illumination stays the same.

 

[fokker]

It all depends on what the light source is.

If we're talking about direct light from the sun then the inverse square law is irrelevant - 93 million miles is pretty much the same distance as 93 million miles and a few feet.

If we're talking about a light source that is at your reference point in the frame then it makes a big difference.

Of course, real life is never quite so black and white and most likely you have a mixture of light sources due to reflections etc, so the end result will be a combination.

 

[Helen B]

It all depends on what the light source is.

Not if the illumination of the subject is constant, which is what Danny asked about.

If we're talking about direct light from the sun then the inverse square law is irrelevant - 93 million miles is pretty much the same distance as 93 million miles and a few feet.

True for the illumination of the subject, but that is not what this question is about. This question is about how the subject illuminates the sensor or film, and the inverse square law does apply, no matter what the original source of the light is.

If we're talking about a light source that is at your reference point in the frame then it makes a big difference.

See my previous comments.

Of course, real life is never quite so black and white and most likely you have a mixture of light sources due to reflections etc, so the end result will be a combination.

Apart from being just plain wrong in this case, I think that you are confusing things badly. This has nothing to do with the original light source as long as it is constant.

 

[Tiberius47]

I think someone needs to actually DO this!

Put a flash in a softbox, set it on manual, and then go take photos of it from different distances.

 

[Helen B]

You do have to wonder 'Haven't they ever taken any pictures?' when someone says that exposure settings change as you get further away from the subject (apart from the case when the source is on the camera or similar), yet some experienced photographers get confused by this. Somehow the half-understood theory seems to blind people to the observed evidence and strange explanations flower. You can, however, often see how this sort of confusion germinated - in this case about the applicability of the inverse square law and whether or not it even occurs in some cases - and I guess that there are two ways out of the apparent paradox: either allow your curiosity to run free and increase your understanding of the theory until the theory matches the observations, or just abandon any attempt to understand the theory.

 

[dandaluzphotography]

You do have to wonder 'Haven't they ever taken any pictures?' when someone says that exposure settings change as you get further away from the subject (apart from the case when the source is on the camera or similar), yet some experienced photographers get confused by this. Somehow the half-understood theory seems to blind people to the observed evidence and strange explanations flower. You can, however, often see how this sort of confusion germinated - in this case about the applicability of the inverse square law and whether or not it even occurs in some cases - and I guess that there are two ways out of the apparent paradox: either allow your curiosity to run free and increase your understanding of the theory until the theory matches the observations, or just abandon any attempt to understand the theory.

I'm going to do this test tonight. One thing I was thinking about is that in very few situations would you actually be photographing a subject where the light source doesn't change or the subject itself doesn't move. I'm just curious about it. My nature I guess.

Thanks for all the responses on this. It's appreciated.

Danny

 

[austinpcherry]

It is about reflected light. Think in terms of cones. While you are closer to your subject you are getting more light from every surface. As you move back you are able to take a proportionally less amount of space from the "cone of light" being reflected from your subject. The zoom level is irrelevant as your actual lens is further away.

 

[Helen B]

...As you zoom in, with the same F-number, the larger physical hole lets in more light.

Just to clarify that: as you zoom a lens in at constant f-number, the physical size of the hole may not change, only the size of its image (ie the size of the entrance pupil) as the iris moves in relation to the front elements. The physical size may also change over part or all of the zoom range.

This is why a 70-200mm f/2.8 lens is so much bigger than a 70-300 f/5.6. The front element of the 70-300mm lens only needs to be 54mm in diameter for f/5.6. But the 70-200mm lens needs a front element of 71.5mm for f/2.8. That's why it's larger.

The F-number embodies this relationship between physical focal length and physical aperture. ...

Is there a danger of misleading beginners in referring to the diameter of the front element when talking about the f-number equation? The front element can't be smaller than the entrance pupil, but it may be larger - much larger in some cases - and it is the diameter of the entrance pupil (which is not a physical object) that is used in the calculation of the f-number.

I'm just wary about these things because they can be misunderstood, and lead to confusion.

 

[dandaluzphotography]

Thanks Guys! Good stuff!

 

[KmH]

But does the distinction further the understanding of exposure and the F-number?

It should.

 

[Fred Berg]

The amount of reflected light which reaches your lens decreases the further away you are from the subject. This can be easily demonstrated by fitting a manual zoom lens and setting your camera to A priority. Choose a subject, set your aperture and make a note of the calculated shutter speed. Now zoom in and out without changing the aperture and you'll notice that the shutter speeds will be recalculated by the camera's metering system. To keep the shutter speed the same as the original setting, you have to adjust the f-stop.