Inverse Square Law - I have always wondered...

[xpatUSA]

I'm sort of inclined to say really to some of the comments but this is not an area that I have ever been really interested in other that when illuminating something - solid angles etc.
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Normal sources of illumination might be the sun or the moon. Their distance is so great in terrestrial terms that object distances on the planet are more or less irrelevant.

Not so flash guns and brolly lights etc.

I've never used a brolly light, John, but are they not used as sort of parabolic reflectors with a source of light on the handle that points away from the subject? If so, what kind of light source would that be, especially when the subject could be smaller than the umbrella?

[ajohnw]

I've never used a brolly light, John, but are they not used as sort of parabolic reflectors with a source of light on the handle that points away from the subject? If so, what kind of light source would that be, especially when the subject could be smaller than the umbrella?

Pass. LOL But in practice it's nearly impossible to produce a parallel beam of light. With a pure perfect parabola for instance an infinitely small light source is needed. - This effectively means that there is no simple way to produce a perfectly parallel beam of light of significant size. The source is reckoned to have to be under 1/2 the size of the diffraction spot size of the optics used to get even near it. For home interferometry laser diodes can help a lot as they have such a small source size but the beam sizes for this sort of use is tiny and still either diverges or converges.

The basis of my intuition for interest is that photography would be impossible unless what I have assumed is correct - we also would not be able to see very far as things would dim rapidly with distance. All we have to put up with is absorption. Solid angle geometry looks after the rest as the object what ever is it gets smaller as the distance to it is increased.

John
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[PhotomanJohn]

The basis of my intuition for interest is that photography would be impossible unless what I have assumed is correct - we also would not be able to see very far as things would dim rapidly with distance.

+1

John

[xpatUSA]

Pass. LOL But, in practice, it's nearly impossible to produce a parallel beam of light . . .

Sorry, I must have misled you by using the word 'parabolic', thereby implying a collimated beam of light. Didn't mean that, realizing that a) umbrellas are not truly parabolic and b) the lamp or flash can be placed anywhere on the handle. I was just thinking that, from the subject's point of view, an umbrella would look pretty big and would not itself act like a point source.

[george013]

I missed this thread.
I understand that when you divide an equal amount of light on a bigger surface, that the light per m2 or cm2 will be less. And that the surface of a plane is growing quadratic if you double the distance using a same angle.
But a convincing explantation of why a lens of 100mm and f4 is giving the same result as a 50mm f4 I didn't see. The diameter if the aperture at 100m is twice at 50mm.
George

[PhotomanJohn]

But a convincing explantation of why a lens of 100mm and f4 is giving the same result as a 50mm f4 I didn't see. The diameter if the aperture at 100m is twice at 50mm.
George

Wikipedia has this simple explanation:

"A 100 mm focal length f/4 lens has an entrance pupil diameter of 25 mm. A 200 mm focal length f/4 lens has an entrance pupil diameter of 50 mm. The 200 mm lens's entrance pupil has four times the area of the 100 mm lens's entrance pupil, and thus collects four times as much light from each object in the lens's field of view. But compared to the 100 mm lens, the 200 mm lens projects an image of each object twice as high and twice as wide, covering four times the area, and so both lenses produce the same illuminance at the focal plane when imaging a scene of a given luminance."

John

[george013]

Wikipedia has this simple explanation:

"A 100 mm focal length f/4 lens has an entrance pupil diameter of 25 mm. A 200 mm focal length f/4 lens has an entrance pupil diameter of 50 mm. The 200 mm lens's entrance pupil has four times the area of the 100 mm lens's entrance pupil, and thus collects four times as much light from each object in the lens's field of view. But compared to the 100 mm lens, the 200 mm lens projects an image of each object twice as high and twice as wide, covering four times the area, and so both lenses produce the same illuminance at the focal plane when imaging a scene of a given luminance."

John

Thanks. I was looking in that direction too but in another way.
George

[chauncey]

I've long since forgotten the math involved but, I equate it with heat, as from the sun...
the further away ya get, the less heat is transferred to the subject.

Same thing applies to light.

[xpatUSA]

I've long since forgotten the math involved but, I equate it with heat, as from the sun...
the further away ya get, the less heat is transferred to the subject.

Same thing applies to light.

A bit too simplistic for my taste, sorry to say, although the analogy is reasonable enough.

Think of a large white sheet, evenly back-illuminated. Think of a small subject, metered at various distances from the front of the sheet. Would the statement "the further away ya get, the less [light] is transferred to the subject" still apply?

[revi]

A bit too simplistic for my taste, sorry to say, although the analogy is reasonable enough.

Think of a large white sheet, evenly back-illuminated. Think of a small subject, metered at various distances from the front of the sheet. Would the statement "the further away ya get, the less [light] is transferred to the subject" still apply?

Yes, but not following the inverse square law which is exact only for a point source. I guess you could get an exact result by applying the inverse square law for each point of the sheet, and sum/integrate over the sheet surface, but that's going a bit far for my taste (for this application).

[george013]

Yes, but not following the inverse square law which is exact only for a point source. I guess you could get an exact result by applying the inverse square law for each point of the sheet, and sum/integrate over the sheet surface, but that's going a bit far for my taste (for this application).

Just thinking loud.

The inverse square law says that 1) a point source has a certain amount of light-energie, 2) the radiation is radial from that point and 3) the light intensity on a surface reduces quadratic with the distance. I don't know the units so I say it in my own words.

But in my question about the differences with the diameter of the aperture at different focal lengths it's backwards.
The sensor needs a certain amount of light. That's a fixed amount. The lightmeter measures the receiving amount and calculates an exposure so that it corrects the receiving amount of light to the quantity the sensor is build for. The receiving light is not send from a point-source but is reflecting light. Comparing a 50mm and a 100mm lens, the 100mm lens covers a smaller surface in the field or a smaller angel of view. So to let pass a same amount of light, the diameter of the aperture must be bigger.

I still have some black spots in my logics.

George

[JR1]

http://www.intl-lighttech.com/suppor...ght-calculator

or in my opinion, who cares

I have never cared,worried,bothered about it is 40 years

[ajohnw]

It might be easier for you to think in telescope terms George. Diameter sets the quantity of light captured and focal length sets the image scale. For a given diameter a larger image scale decreases the light levels in the image, conversely a smaller image scale increases it's intensity. As it's an area of one to another relation ship F4 etc is F4 what ever the diameter of the lens. Area calculations involve a ^2 in simplistic terms.

Resolution in normal photographic terms does involve an integral of fictitious point sources that form surfaces and objects. Departures from a pure circular aperture interfere with the diffraction pattern, The iris in a camera lens often isn't truly round.

In terms of the inverse square law the old way of describing it fits. Drop a round pebble in a pond. Ripples spread radially and the amplitude drops of with distance. In this case because energy is being absorbed. In light's case it's being spread out more. At an infinite distance they become a plain wave - a straight line rather than being radial. Drop a square pebble in and the shape of the ripples change. and there will be more interference between them.

Light is weird stuff. It's easy to see what a diffraction pattern should look like. Take 2 pieces of card and make a pin hole in them. Hold one in front of a clear bulb and view the pin hole through the other. With care the F ratio of the viewing card can be increased by enlarging the hole and the diffraction spot will shrink in size. Then try it against a frosted or opal bulb. The reason the clear bulb works is that the area of the filament being viewed is so small that the waves are effectively coherent.

A bloke called Abbe established a long time ago that on microscopes the F ratio of the illumination needs to be greater or equal to F ratio of the objective being used to view it to achieve it's theoretical resolution. In practice due to optical effects it's better to match it or even go below it. In this area there is a debated formulae that suggests that if the lighting is arranged in a certain way resolution can be halved. The resolution problem also disappears if the optic can be placed more or less in contact with the object being viewed - sub wavelength distances. A simple glass ball and a microscope objective can be used to do this but as always with this sort of thing the field of view is rather small.

John
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[ajohnw]

http://www.intl-lighttech.com/suppor...ght-calculator

or in my opinion, who cares

I have never cared,worried,bothered about it is 40 years

True Jeremy. Of academic interest only really. Best example in photography is flash guns - distance weaken the light levels they produce. Not enough light - buy a bigger one. Same with studio lighting.

John
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[JR1]

True Jeremy. Of academic interest only really. Best example in photography is flash guns - distance weaken the light levels they produce. Not enough light - buy a bigger one. Same with studio lighting.

John
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+1

[george013]

+1

It also means he cares.

George

[xpatUSA]

Yes, but not following the inverse square law which is exact only for a point source. I guess you could get an exact result by applying the inverse square law for each point of the sheet, and sum/integrate over the sheet surface, but that's going a bit far for my taste (for this application).

With the large sheet example, I was trying to refer to an Area source, not a point source. Having said that, applying the calculus is a little too complicated for my taste .

However, this graph tells us all we need to know:



Do observe above that, for an area source, "changes in distance hardly affect the irradiance".

Source: http://kronometric.org/phot/xfer/lig...20handbook.pdf

So, thinking again of a large white sheet, evenly back-illuminated, say 48" diameter, and of a small subject place at various distances from the front of the sheet. Would the statement "the further away ya get, the less [light] is transferred to the subject" still apply?

From the graph above, it seems that distances up to about 7" (24" x 0.3) make little difference to the irradiance on the subject.

On the other hand, for a CFL of say 2" diameter, the inverse square law would apply for a distance of 10" upward.

We should all (maybe even those who don't care) get some benefit from the concept of distance versus light-source radius and, at the very least, the "five times the diameter" rule described above.

[ajohnw]

Perhaps you might say an area source consists of many point sources which is probably how the maths was developed Ted. There are probably edge effects as well.

This video might amuse you and George. Trouble is some of it is about conjugates especially related to focusing onto the back focal plane of an objective - that will be where it's infinity focus is, not where the image focus is. On a microscope objective that may only be a few mms above the glass, or probably in some cases even in it. The image plane is usually much further away.



Some of the odd aspects have practical applications.

John
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[Downrigger]

Perhaps you might say an area source consists of many point sources which is probably how the maths was developed Ted.
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Ted's source is excellent and does provide clarity on ""extended area sources - which is what (I think) best addresses the original question of this thread.

Under "Radiance and Luminance", p.34, he gives us

"Radiance is a measure of the flux density per unit solid viewing angle, expressed in W/cm2/sr. Radiance is independent of distance for an extended area source, because the sampled area increases with distance, cancelling inverse square losses. (My underline)

I think subjects in photographs mostly reflect light as though "extended area sources", and what we get on the sensor or film at a given distance is the result of a "solid viewing angle". Thus, for instance, moving (or zooming) closer decreases the sampled area, cancelling the possibility of inverse square law gains because the decrease in sampling area.

[ajohnw]

Perhaps I didn't expand enough when I mentioned solid angles earlier Mark. They relate to viewing, imaging and to actual sources. On the later where it is a true source up pops the inverse square law.

The wiki sums it up well under justification. It doesn't directly apply to viewing or imaging other than it happens but the solid angles from that end cancel it out.

https://en.wikipedia.org/wiki/Inverse-square_law

John
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