flash guide number f-stop calculation

[swchris]

Hi,

I understand that in order to calculate the required aperture one divides the guide number by the flash<->subject distance (Take care of film speed etc.).

What I don't understand is why just the flash<->subject distance is important, but not the camera<->subject distance.

For example, this scenario:

- Flash is put on a tripod at a given flash<->subject distance
- calculate f number for this distance
- take one picture with camera<->subject distance 1m
- take one picture with camera<->subject distance 100m (probably with a long tele lens)

Will the two pictures be exposed correctly?

regards,
chris

 

[spijker]

yes, the subject will be properly exposed in both cases. The brightness of the subject is not affected by the distance from which you look at it. Just think of replacing the flash with a regular lamp. Put the lamp on a stand and shine it on your subject. Now if you look at the subject, walk away from the subject and look again. Does the subject appear to get darker as you walk back or just as bright?

 

[swchris]

Does the subject appear to get darker as you walk back or just as bright?

Didn't try right now (it's day right now over here), but from past experience I'd say, no, it doesn't get darker.

But, and in fact that's really my question: There is a physical law that light intensity decreases with the square of the distance. That's why we have to take bellows extensions into account. And it's also buried in the guide number calculations (f stops, half or double amount of light, come in square root steps, so a double or half number means 4 times or 1/4 times the amount of light). But why is this law seemingly irrelevant for the watcher (camera), but not for the flash distance?

 

[cowanw]

As the distance gets larger so does the area on the sensor get smaller.

 

[Kawaiithulhu]

light intensity decreases with the square of the distance

This is for Light To Subject distance, has nothing to do with Camera To Subject. You're plugging in the wrong variables to the equation.

If Camera To Subject distance made any difference at all then pictures of the Moon would be impossible and mountains far away would all be black in everyone's family vacation pics. Size of image on sensor/film plane also makes no difference, otherwise DX and half-frame formats would all be super bright and require different exposures. Minox would be King of low light photography!

 

[swchris]

If Camera To Subject distance made any difference at all then pictures of the Moon would be impossible and mountains far away would all be black in everyone's family vacation pics. Size of image on sensor/film plane also makes no difference, otherwise DX and half-frame formats would all be super bright and require different exposures. Minox would be King of low light photography!

I didn't talk about sensor/film size.

But of course, when I'm on top of a mountain and look at the landscape, things far away aren't any darker.

This is for Light To Subject distance, has nothing to do with Camera To Subject. You're plugging in the wrong variables to the equation.!

Ok, probably this is the guts of my question: Why does the law (intensity decreases with square of distance) apply to "Light to Subject", but not to "Camera to Subject"? Isn't light light?

 

[cowanw]

The size of the sensor makes no difference, however the size of the image does; given a certain distance and focal length will be the same size on all sensors.
Let me try again.
When you are a foot from the candle, the light exposes the area that it covers on the sensor.
Twice as far away the light falls of by the square but the area that the candle covers on the sensor is smaller by the square, so the density of light is the same.

 

[swchris]

When you are a foot from the candle, the light exposes the area that it covers on the sensor.
Twice as far away the light falls of by the square but the area that the candle covers on the sensor is smaller by the square, so the density of light is the same.

Call me dense, and I'm feeling this way currently, but I still don't get it.

One thing, why does the size of the picture on the film/sensor account here?

Another thing, take my initial example, pictures 1m and 100m away (or make that 3ft and 300ft). When taking the close distance (1m/3ft) picture with an wide angle lens and the distant picture (100m/300ft) with an telephoto lens so that the size of the subject is the same in both cases, your explanation doesn't apply.

 

[Chan Tran]

Call me dense, and I'm feeling this way currently, but I still don't get it.

One thing, why does the size of the picture on the film/sensor account here?

Another thing, take my initial example, pictures 1m and 100m away (or make that 3ft and 300ft). When taking the close distance (1m/3ft) picture with an wide angle lens and the distant picture (100m/300ft) with an telephoto lens so that the size of the subject is the same in both cases, your explanation doesn't apply.

But the physical aperture of the telephoto is much larger than that of the wide angle lens although they are the same f/stop.

 

[MattKing]

This is always much harder to explain than to observe.
When light reflects off a subject and as a result of that reflection an image of the subject heads our way, that image doesn't get any dimmer as it travels. Whether our film or sensor stops it after it has traveled 1 meter, 2 meters or 10 meters, it will still be of the same intensity. It may, of course, end up appearing smaller on that film or sensor because of increasing distance.
This is different than a source of illumination - non image bearing light - gobs and gobs of it. As light spreads out from a source, its density/intensity decreases. The calculation is based on the square of the distance, but that just measures how dense/intense the light is at one particular distance.

 

[swchris]

Tthat image doesn't get any dimmer as it travels.

Why is it so? Doesn't the law apply here for some reason?

As light spreads out from a source, its density/intensity decreases..

But why doesn't it decrease when it doesn't spread out from a light source, but is reflected after hitting some subject?

 

[Maris]

The brightness of a illuminated subject is the amount of light coming off it divided by the visible area. If one doubles the distance to a subject then the amount of light arriving from it drops to a quarter: inverse square law. But the subject now looks only half as high and half as wide so its visible area is only a quarter of what it was before. A quarter of the light over a quarter of the area means the amount of light per unit area is the same - the brightness is the same. One effect exactly balances the other for all distances.

 

[MattKing]

Why is it so? Doesn't the law apply here for some reason?

Because the law applies to an emitting source, which by its very nature is spreading out, not fixed in size like an image.
If you project an image, it does spread, which does result in decreasing intensity.

Fibre optic cables transmit light without spread - and essentially they do so without any loss.

But why doesn't it decrease when it doesn't spread out from a light source, but is reflected after hitting some subject?

Because it is the spreading out that causes the decrease.

 

[cowanw]

What Maris said; and as usual in far better language than I.

 

[swchris]

Thanks guys, esp. Maris and Matt. I got it now.

 

[Dali]

I didn't talk about sensor/film size.

But of course, when I'm on top of a mountain and look at the landscape, things far away aren't any darker.



Ok, probably this is the guts of my question: Why does the law (intensity decreases with square of distance) apply to "Light to Subject", but not to "Camera to Subject"? Isn't light light?

Why would you want the light law to apply to the "camera to subject' case even if the camera does not emit any light? Don't you feel your question little absurd?

 

[swchris]

Why would you want the light law to apply to the "camera to subject' case even if the camera does not emit any light? Don't you feel your question little absurd?

I omitted the word "distance", but of course was just talking about the distances.