Distance from subject to camera - How does it affect lighting?

Just wondering how you compensate lighting when the distance from

your camera changes. I know that the ammount of light decreases by

the Umm... Inverse square isn't it? so I think it must come into

play.

 

I guess what makes me wonder is this... say I do a head shot.. I

don't want to be right up on the model using a 50mm lens, I'd rather

be back using a longer lens... so won't the strobes need to be more

powerful to give me the same amount of light hitting my film when I

take the shot? if this is correct, how do I know how to compensate

for distance when metering?

Camera to subject distance doesn't affect exposure. It'll be the same. Inverse square law only applies to light to subject distance.

If the flash is *on* the camera, then the inverse square law applies. It's flash-to-subject distance that matters.

At short distances, subject to camera(flash)doesnt matter.At longer distances, it certainly matters!

 

Example:What if the flash is right next to the subject, and the camera is 100 feet away?

Josh,

 

It makes no difference. The light intensity decreases with distance but is counterbalanced by the fact that it occupies correspondingly less space, i.e the same amount is concentrated into a smaller physical area, so it all ends up square.

 

*I know that the ammount of light decreases by the Umm... Inverse square isn't it?* Not it doesn't, unless the light is a point source.

'Example: What if the flash is right next to the subject, and the camera is 100 feet

away?'

 

It makes no difference. The exposure is the same.

when you are talking about light source to subject distance, the inverse square law applies regardless of what type of light source it is.

"<i>Example:What if the flash is right next to the subject, and the camera is 100 feet away?</i>... use a telephoto lens, but don't change the exposure... t

re: The Inverse Square Law theory

 

The inverse square law does not apply equally to all light sources, or at least not in practical terms. It applies only to point sources of light. If a larger light source is used (and all practical light sources are larger) then although the intensity of the light reaching the subjectdoes fall off as the distance increases, it does not do so in a linear way and in accordance with the inverse square law.

 

The larger the light source in relation to the subject, the less the fall-off of light over a given distance.

 

If you feel otherwise, please test it for yourselves!

'Example: What if the flash is right next to the subject, and the camera is 100 feet away?'

It makes no difference. The exposure is the same.

 

 

No way! Try shooting out doors at night with the camera 100 feet away from the flash,and let me know how much density your negs get.I shoot night time exteriors, and flash to camera distance has to be compensated for.What if the flash is pointed at a subject outdoors at night, and the camera is 100 yards away?Light reflected from the subject headed to the lens, falls off.This is high school physics, not complicated stuff.

'This is high school physics, not complicated stuff.'

 

I'm interested in how you factor in your distance to the subject when making your

exposure calculations.

At night, during a full moon, you can take a photo of the moon's surface and use the Sunny 16 rule and get a perfectly exposed image of the moon's surface. A full moon has as much light hitting it's surface as the earth does in the daytime.

 

The distance from the earth to the moon doesn't seem to make much difference.

Of course things like reflectors and soft boxes will affect the inverse square law--closer to a half rather than a fourth change in light, but for the purposes of photography, and trying not to add confusion to the asker of the original question, it is essentially the same. To say only point sources follow the inverse square law may lead beginners to believe that changing non-point source light positions to alter fall off would be futile.

Steve... just try it. Set up 50 feet away from your subject with a light 4 ft from the subject. Make a photo with a 24mm lens that fills the frame wth that subject. Then put your longest lens on the camera, fill the frame with the same subject and without changing anything else (except that <i>you and the camera</i> will be much further away) and make the same exposure. Use slide film or check your histogram... t

On a pitch dark night.Place a 100 guide number strobe atop a small stand.Place it 5 feet away from a model's face.Next make exposures while moving only the camera.Shoot a frame at 10 feet,100 feet and 1000 feet.The density of the face on the film will be zero, after the camera moves far enough away.The reflected light from the model's face falls off, sorry to disagree.I have been there, done this.

Look,

 

I really don't want to sidetrack this thread away from the subject of the question but we really should be sticking to facts here.

 

Brooks Short said that *The distance from the earth to the moon doesn't seem to make much difference.*

 

Of course it doesn't, for the reason I gave in my first post. The only reason that the (very considerable) distance makes ANY DIFFERENCE is not related to the actual distance but to particles of matter (pollution) that block some of the light.

 

Ken says that

*To say only point sources follow the inverse square law may lead beginners to believe that changing non-point source light positions to alter fall off would be futile.* I take your point, we don't want to mislead people into believing that there is no fall-off with non point sources of light, but that's not what I said anyway. And the fact remains that only point sources DO follow the inverse square law. If a photographer uses a candle as a light source (which distributes its light in all directions) and places it far enough away from the subject (so that it becomes a true point source) then the Inverse Square Law (light) will apply - but for normal photographic lighting it doesn't, because the light source itself doesn't qualify.

 

And to get back to Josh's question, no compensation is necessary.

 

And Steve says that his experience proves him right. Fair enough, but please test again and see whether you still feel the same way.

 

The physics are simple enough. http://hyperphysics.phy-astr.gsu.edu/hbase/vision/isql.html

My actual test, involved shooting a train passing through a small town at night.This was shot as a time exposure,with flash to freeze the train.The camera was located 100 yards down the street from the rail crossing.I hid several flash units behind utility poles.These were placed about 20 feet from the train in 2 spots about 50 feet apart from each other.The test film showed that the flashes were'nt bright enough to reach back to the lens.Even though,on the rail road tracks, the flashes read F11 !

 

My point here is that the test polaroid taken at 10 feet ,was exposed fine with flash.At 100 yards, the area illuminated by the flash was no brighter than the ambient lit areas on the film.I assumed that the light falls off over the distance to the lens.

 

As for the moon,its 150,000 miles away! If you went half way closer to,or twice as far away from, Id also assume it would get brighter and darker respectively.

Steve, I don't know why your train shot was underexposed. Maybe the flash was out of synch. Things happen.

 

What I do know is that your surmise cannot be right. As I said in my first answer, if the camera moves further from the subject the light reflected from the subject loses power over the greater distance, but the light reflected from the subject occupies a correspondingly smaller area of the film. The net results is that the exposure is exactly the same regardless of the distance between camera and subject.

 

The distance between the light source and the subject is of course another matter, if the distance is increased then the subject receives less light, although not in accordance with the Inverse Square Law (light) which only applies, in linear terms, to point sources of light.

 

Hope this helps

Garry,

 

I know that's not what you said, but some may think that is what you said. True it doesn't follow the inverse square law but only in the respect that the light fall off is not quartered, it is more like a half--so not really a big difference in the photographic world.

The inverse square law applies between the subject and the camera, but it applies twice and it balances. Once from the distance, and inversely because the area on the film is cut proportionately to the distance moved.

 

Take a reflective light source, say a 4x8 sheet of plywood lit be a hairlight. Shoot it from a convenient distance so you have an image on your film of 12 x 24 mm. Now move the camera (not the light or the plywood) back twice as far. You will have 1/4 the amount of light reaching the camera, but now you are exposing only 6 x 12 mm of film. The exact same light intensity will fall in the reduced area, the film density will be exactly the same.

 

This happens regardless of the distance traveled or any other factor, as long as no factor other than distance is changed.

 

Van

was it a black train?... t :^)

ok.. what if you move back 15 feet and zoom in to get roughly the same composition you had before... but now the light takes up the same ammount of space on the film and you are further away...

 

any adjustments need to be made?

 

sorry to raise such a stink about the inverse square thing..

Guys, guys... just think about real life! Do things get brighter when you get closer to them? If I'm in a movie theater, and I can't see my watch well enough to read it, can I bring it up closer to my eyes and make it brighter?

 

Can I read a book outside at night if I just move it close enough?

 

When you're in a plane, does the ground get darker and darker below you as you ascend? Cause you start out maybe 10 feet up, then you're 100, then 1,000, then 10,000... so it should be really dark from way up there, right?

 

No of course not. You don't even need to think about cameras to understand this one. Your eyes and retinas are similar to camera lenses and film anyways, at least in that they work on the same principles of physics and all.

Josh: about zooming. That is a little more complex. If you are at the maximum aperture of your zoom, and your lens varies in "speed" over it's range, then it may, in fact, affect exposure.

 

For example if your lens is F/3.5-4.5, and you are right near the subject, zoomed out and metering 1/60th @ f/3.5 then step back, zoom all the way in to get the same framing, you will now be at F/4.5 and exposure will have to change to 1/30th, or flash power will have to be doubled.

 

However in most cases, and if your lens is stopped down at all, the zoom will retain the same aperture, say f/8, when you zoom in.

 

The way this works is that the light *is* farther away, and projecting onto the same area on the film, but as you zoom in your lens gets "bigger", in diameter, to absorb the same amount of light. Look into the lens as you zoom and you can actually see this, the area of the lens that is absorbing light and projecting it to the film is increasing in size. No peices of glass are physically changing size of course, but you'll see what I mean.

Wow.. cool... thanks for all the answers I feel...*enlightened*

 

 

Thanks

-Josh