Sorry if the previous attempt didn't hit the mark. Permit me to try
First: When we're talking about general illumination of something (e.g.
a grey card), an extended light source, when you're close, does *not*
behave like a point source. This is because as you pull away from the
surface you start "seeing" more of the source at a higher angle, so as
you pull away the spread of the light is compensated for by bigger
contributions from more of the source. For an infinite plane, if you do
the integral then as with an electric field you should see *no falloff
at all*. This will be true whether the plane is emitting or reflecting
light (assuming it reflects equally at all angles).
But: People on stage are too far away for such effects to be
significant - essentially we already see the whole surface. And anyway
we're not interested in how their reflected light is illuminating
something - we're going to focus that light into an image.
The reason a lens is relevant is that it forms an image; it ideally
takes all the light that falls on its front element from a point on the
subject and directs it to a single point in the image, and does so for
all points that are in the field of view and in focus (let's assume
everything is in focus for the moment). That means a pixel (or
whatever) is not illuminated by the whole extended source, as it would
be without the lens - it's illuminated by all the light that was
reflected from a very small area of the subject and hit the front
element (I wish I could draw here).
Now: You're absolutely right that the light from the object is
spreading out and therefore falls off with the inverse square law. What
happens if you double your distance to the subject, for example? The
amount of light hitting the front element drops fourfold. But the image
has shrunk to half dimensions, and the pixel is now getting light from
an area on the subject that is twice as high and twice as wide, thus
four times the area. That increases the amount of light fourfold,
exactly compensating for the falloff.
How'd I do that time?
Finally, you're right about the experiment. Perhaps I misread - I
wasn't sure you'd accepted the conclusion, so I was trying to establish
that camera-subject distance does not affect exposure.
On Jan 5, 2009, at 12:55, Chuck Norcutt wrote:
> I'm afraid I don't understand you and Ken bringing the action of the
> lens into this discussion as I don't see the relevance. This started
> trying to understand why an extended light source didn't behave as a
> point source and lose light according to the inverse square law. I
> thought that the link I had posted adequately explained the physics of
> that without reference to cameras or lenses. After all, the lens can
> only work with what's impinging on its surface and has no part in how
> in what intensity the light arrives. Then, what happens between front
> element and film/sensor is immaterial since it's the same regardless of
> the light source or light intensity.
> Finally, I don't see the relevance of the experiment you propose. All
> it can do is show what we all know to be true but does nothing to
> elucidate why that is so. But maybe if the dimensions of that card are
> only 5% or less of the distance between card and camera we'd see that
> begins to behave as a point source. But that info came from my own
> What am I missing?
> Chuck Norcutt
> Andrew Gullen wrote:
>> Sorry, I should have addressed that too.
>> You are correct that line and plane sources have different falloff of
>> illumination, like electric fields - but only when you are close
>> that this makes a difference. See page 61 of this reference, where it
>> However, as a practical matter, whenever the longest dimension of
>> the surface
>> of an emitting source is less than 1/20 of the distance from which
>> the light is
>> being measured, it is usually acceptable to treat it as a point
>> But anyway, this is relevant only when considering gross illumination
>> as when you light a reflector to illuminate a subject, or use a
>> softbox, and you're only concerned with *how much light in total* is
>> falling on an area. It's not relevant when you focus an image of
>> something, because in that case the contributions from each little
>> are not summed but fall on different parts of the film/sensor. As Ken
>> just said. Extended light sources are a red herring in this
>> But words are cheap - try an experiment!
>> - Use a camera where you can lock ISO, focal length, aperture,
>> speed and white balance.
>> (An OM-1 with film and a fixed lens would be good. :-) )
>> - Set up a small lit object in an otherwise dark space, e.g. a card
>> lit with a flashlight (torch)
>> - Determine a correct exposure by incident metering, spot metering,
>> trial and error.
>> - Take a sequence of shots ranging from close to far.
>> - In all shots, though the object's size will vary it will be
>> (I'm assuming you'll actually use a digital camera. Don't use
>> print film as your photofinisher
>> will adjust and invalidate everything. Slide would be OK.)
>> You can also see this in everyday shooting, though. We don't change
>> exposure when varying distance to the subject (except for macro, which
>> is another topic). Sunny 16, for example, holds no matter how far you
>> Manual exposure would be excruciating if this were not so - you'd have
>> to adjust every time you changed distance.
>> It does take some time to get one's head around this - I remember.
>> On Jan 5, 2009, at 9:30, Chuck Norcutt wrote:
>>> The memory is weak but not wrong. I knew it had something to do with
>>> point vs. extended light sources. Read pages 60 and 62 of:
>>> Perception of the Visual Environment By Ronald G. Boothe and note the
>>> distinction between "intensity" (point source) and "luminance"
>>> source) Page 63 goes on to discuss luminance from reflection.
>>> Chuck Norcutt
>>> Andrew Gullen wrote:
>>>> Hi -
>>>> Ian has the right answer here.
>>>> There is no difference between "source" light and reflected light.
>>>> reflected light from a person on stage that falls on a given area
>>>> the front element of your lens, or your cornea) does indeed fall off
>>>> with the square of the distance. But the area of the formed image
>>>> goes down with the square, so everything balances out.
>>>> Note that if you double your distance (and cut the light fourfold),
>>>> go for a lens with twice the focal length to keep the image size the
>>>> same, you need to double the diameter of the front element (I'm
>>>> approximating a bit here) and thus quadruple the area of the front
>>>> element, in order to gather enough light to maintain the
>>>> of the film/sensor. But that's just keeping the same f-stop (focal
>>>> length divided by diameter). It's lovely that the physics and math
>>>> optics make photography so simple, except when we stop to think
>>>> it. :-)
>>>> On Jan 4, 2009, at 13:53, Ian Nichols wrote:
>>>>> Right answer, but I think your maths is a bit out - moving from 4
>>>>> to 8 feet, the image fills 25% of the viewfinder (it's an area,
>>>>> not a
>>>>> length) and the light from the subject has decreased by a factor of
>>>>> So 1/4 of the light gets focused onto 1/4 of the area, hence same
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