Metering Question

[dandaluzphotography]

Hi Guys,

I just had a conversation with my buddy and we have a disagreement hopefully someone here can clear up for us.

Assume that I am metering an area with a constant light source that doesn't change and to fill the frame with only that area I need to stand 10 feet away and be at a 70mm zoom. Also, assume a shutter speed of 1/200th, aperture of F/11 and ISO 100. Assume the camera tells me the metering for this is perfect and my exposure is perfect.

My contention is that if I now stand 40 feet away and have to zoom in to 175mm to fill the frame with the same area as when I was 10 feet away and assuming that the same light is falling upon that area and that my shutter/aperture/ISO have not changed, the metering will be exactly the same.

My friend says that because I am physically further away from the subject, that the light changes and I will have to change one of my three settings to get the proper exposure. I say that because the area filling the frame is not changing, including the amount of light falling on the area, and I am getting to the same area via the zoom, I will not have to re-meter.

The only thing I am not sure of is that because I am zoomed in, or out for that matter, that this is somehow causing a change that would require me to change shutter/aperture/ISO to get the same perfect exposure.

Any help with this would be greatly appreciated!

Thanks,
Danny

 

[480sparky]

Inverse Square Law dictates that if you're further away, there's less light reaching the lens. So technically, your buddy is correct.

BUT: The light loss between 10' and 40' is negligible in this case.

 

[Helen B]

The light loss is definitely significant, but it is compensated for by either the physical aperture being bigger (longer focal length lens, same f-number = larger physical aperture) or by the smaller image (if the lens f/l was the same, which it isn't in this case).

 

[dandaluzphotography]

Inverse Square Law dictates that if you're further away, there's less light reaching the lens. So technically, your buddy is correct.

BUT: The light loss between 10' and 40' is negligible in this case.

Hmm. Ok, but my thinking was that if the lens is zoomed in to get to the same area as when I was 10 feet away, the camera would have no idea and it would provide the same result. Of course, if I stand 40 feet away and I don't zoom in, now the area filling the frame is different and that would cause a change that may require me to re-meter, but this isn't the case.

 

[dandaluzphotography]

The light loss is definitely significant, but it is compensated for by either the physical aperture being bigger (longer focal length lens, same f-number = larger physical aperture) or by the smaller image (if the lens f/l was the same, which it isn't in this case).

Hi Helen,

Can you explain this a bit more? I didn't quite get it.

Thanks,
Danny

 

[Helen B]

A 280 mm lens would cover the same field at 40' that a 70 mm would cover at 10'

Therefore at the same f-number, the aperture (entrance pupil) diameter of the 280 mm lens would be 4x that of the 70 mm. 4x linear = 4^2 x area = 16x area.

Light intensity at 40' = 1/4^2 than at 10' = 1/16 (inverse square law).

It works out. It is not a coincidence.

It also works out at any other lens length, because of the difference in image size.

Suppose you had the same lens. There is still 1/16 the intensity of light coming from any finite area of the object, but the entrance pupil is the same size this time (same f/l, same f-number). However, because you have moved to 4x the distance, the area of the image of that finite object area is 1/16 of what it was at 10', therefore there is 1/16 the light falling on 1/16 the area. Same image brightness. This is not a coincidence either.

 

[Alex_B]

The amount of light falling on the film or sensor does get reduced, since you catch light from a smaller solid angle - However, if you zoom in, at the same f-stop, your aperture changes accordingly to the change in focal length / solid angle.

In other words: Your f-stop remains at 11, but your aperture diameter scales with f/11 (f being 70mm in one case and 175mm in the other case).

 

[Alex_B]

bummer, I was to slow for the techies ... should not have gone to the fridge while typing

 

[dandaluzphotography]

Thanks Helen!

 

[dandaluzphotography]

The amount of light falling on the film or sensor does get reduced, since you catch light from a smaller solid angle - However, if you zoom in, at the same f-stop, your aperture changes accordingly to the change in focal length / solid angle.

In other words: Your f-stop remains at 11, but your aperture diameter scales with f/11 (f being 70mm in one case and 175mm in the other case).

So then I would have to remeter...

Damn it! I hate being wrong but at least i have learned something!

Thanks!

Danny

 

[Alex_B]

The amount of light falling on the film or sensor does get reduced, since you catch light from a smaller solid angle - However, if you zoom in, at the same f-stop, your aperture changes accordingly to the change in focal length / solid angle.

In other words: Your f-stop remains at 11, but your aperture diameter scales with f/11 (f being 70mm in one case and 175mm in the other case).

So then I would have to remeter...

Damn it! I hate being wrong but at least i have learned something!

Thanks!

Danny

No, you can use the same settings ... as you do not set the aperture diameter directly, but you set the f-stop to 11. And the f-stop is the same for both situations.

 

[jwbryson1]

Inverse Square Law dictates that if you're further away, there's less light reaching the lens. So technically, your buddy is correct.

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?

 

[Alex_B]

I think I mucked it up with my handwaving explanation we need someone who is better at explaining this tonight ....

 

[Helen B]

Inverse Square Law dictates that if you're further away, there's less light reaching the lens. So technically, your buddy is correct.

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?

I'm not quite sure what you are saying, but it has nothing to do with a prime or zoom. If you move further away, the inverse square law always applies in this case. I hope that my answer explains it, but you need more of an explanation I'm sure someone will provide it.

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?

No, that's not how zoom lenses work. The important part of the lens is the 'entrance pupil' - the image of the iris as seen from the front of the lens. That is where the lens sees the world from, and what the light from the object passes through to reach the image. That is where the inverse square law needs to be measured to. It's inside the lens or right at the front, even on zooms.

 

[dandaluzphotography]

Inverse Square Law dictates that if you're further away, there's less light reaching the lens. So technically, your buddy is correct.

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?

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