Why Isn't Reflected Light Subject To Inverse Square Law?

Why isn't reflected light subject to Inverse Square Law?

For example, if you take a photo of an object at a given aperture, shutter
speed and ISO based on the light reading. Why is it also possible to
correctly expose the same object from much further away with the same
settings?

On 03/02/2011 09:44 PM, GA wrote:
Why isn't reflected light subject to Inverse Square Law?

For example, if you take a photo of an object at a given aperture,
shutter speed and ISO based on the light reading. Why is it also
possible to correctly expose the same object from much further away with
the same settings?

Because the aperture is taken in account by the settings. If you double
the distance, to keep the same frame you must double the focal length,
and to keep the same aperture number with a doubled focal length you
should also double the diameter of the input pupil, so you multiply by
four the amount of light you receive, and this exactly compensates the
reduction due to the inverse square law.

--
Bertrand

On Wed, 2 Mar 2011 20:44:05 -0000, "GA" wrote:

Why isn't reflected light subject to Inverse Square Law?

For example, if you take a photo of an object at a given aperture, shutter
speed and ISO based on the light reading. Why is it also possible to
correctly expose the same object from much further away with the same
settings?

Because while less light reflected/scattered from the object is
hitting the sensor, it is also concentrated in a smaller area of the
sensor.

I haven't done or looked up the math, but off the top of my head I
believe the two effects cancel each other out.

On 2011-03-02 13:03:32 -0800, DaveS said:

On 3/2/2011 2:44 PM, GA wrote:
Why isn't reflected light subject to Inverse Square Law?

For example, if you take a photo of an object at a given aperture,
shutter speed and ISO based on the light reading. Why is it also
possible to correctly expose the same object from much further away with
the same settings?



Inverse Square law applies to force of gravity or magnetism.
Light intensity is not a force.

Dave S.

Wrong.
The Inverse Square Law also applies to light and sound, and several
other types of wave form energy.

--
Regards,

Savageduck

On Wed, 02 Mar 2011 16:31:19 -0500, John A.
wrote:

On Wed, 2 Mar 2011 20:44:05 -0000, "GA" wrote:

Why isn't reflected light subject to Inverse Square Law?

For example, if you take a photo of an object at a given aperture, shutter
speed and ISO based on the light reading. Why is it also possible to
correctly expose the same object from much further away with the same
settings?

Because while less light reflected/scattered from the object is
hitting the sensor, it is also concentrated in a smaller area of the
sensor.

I haven't done or looked up the math, but off the top of my head I
believe the two effects cancel each other out.


John's right. That idea bothered me when it first occurred to me.
Move twice as far away, light is four times less. (one fourth the
amount of energy.) The image is one half linear size on the sensor,
the area one fourth.
Same energy per unit area.

"charles" wrote in message
...
Why isn't reflected light subject to Inverse Square Law?

For example, if you take a photo of an object at a given aperture,
shutter
speed and ISO based on the light reading. Why is it also possible to
correctly expose the same object from much further away with the same
settings?

Because while less light reflected/scattered from the object is
hitting the sensor, it is also concentrated in a smaller area of the
sensor.

I haven't done or looked up the math, but off the top of my head I
believe the two effects cancel each other out.


John's right. That idea bothered me when it first occurred to me.
Move twice as far away, light is four times less. (one fourth the
amount of energy.) The image is one half linear size on the sensor,
the area one fourth.
Same energy per unit area.



But it still doesn't explain why you can take an incident reading from a
subject and still use the same aperture, shutter speed and ISO whether you
are 1 meter away, or 10 meters away.

So, what causes inverse square law? After all, we can see stars on a clear
night.

On Wed, 2 Mar 2011 22:52:34 -0000, "GA" wrote:

"charles" wrote in message
.. .
Why isn't reflected light subject to Inverse Square Law?

For example, if you take a photo of an object at a given aperture,
shutter
speed and ISO based on the light reading. Why is it also possible to
correctly expose the same object from much further away with the same
settings?

Because while less light reflected/scattered from the object is
hitting the sensor, it is also concentrated in a smaller area of the
sensor.

I haven't done or looked up the math, but off the top of my head I
believe the two effects cancel each other out.


John's right. That idea bothered me when it first occurred to me.
Move twice as far away, light is four times less. (one fourth the
amount of energy.) The image is one half linear size on the sensor,
the area one fourth.
Same energy per unit area.



But it still doesn't explain why you can take an incident reading from a
subject and still use the same aperture, shutter speed and ISO whether you
are 1 meter away, or 10 meters away.

So, what causes inverse square law? After all, we can see stars on a clear
night.



It's a funny thing, once you get your head wrapped around the idea it
is clear, until then it makes no sense.

Square law is just the light spreading out as it gets further from the
source.

Make a hole 1 inch square in a sheet of cardboard, hold it one foot
from the source of light, and measure the light there. Then draw
lines extending from the source out to two feet from the source, the
lines will have spread to two inches, the area now is 4 square inches.
Same amount of light, twice the distance, one fourth the light per
unit area. The total amount of light is the same, just spread over
four square inches rather than one square inch.

The photographic image doesn't matter to the total energy hitting the
entire sensor, film, or whatever, it is concerned with the energy per
pixel, or per square millimeter, or whatever size unit you want to
select. When you move ten times as far away, the illuminated area on
the sensor shrinks by the square of the distance, so the intensity per
pixel is the same, just fewer pixels.

To some extent this is why teleconverters eat light, they spread the
incoming amount of light over a larger area on the sensor, diminishing
the intensity per unit area, thus you need to let in more light, or
expose for a longer period.

The simple explanation neglects diffraction and quantum effects, but
they are small compared to what it does cover. This is for point
sources of light, or small sources, extended sources have a different
effect. Light loss from the light spreading out is offset by light
spreading in from other areas.

Stars are pretty bright up close.

GA wrote:

Why isn't reflected light subject to Inverse Square Law?

Because the inverse square law applies to point sources, and I assume you're
thinking of a uniformly radiant source filling the frame.

For example, if you take a photo of an object at a given aperture, shutter
speed and ISO based on the light reading. Why is it also possible to
correctly expose the same object from much further away with the same
settings?

Others have explained how a uniformly radiant source filling the frame
behaves, but if you back off far enough that the object appears as a point
source, the inverse square law will apply. Results in the region between the
two extremes may be unpredictable. :- )

On 2011.03.02 15:44 , GA wrote:
Why isn't reflected light subject to Inverse Square Law?

It is subject to 1/x^2 law.


For example, if you take a photo of an object at a given aperture,
shutter speed and ISO based on the light reading. Why is it also
possible to correctly expose the same object from much further away with
the same settings?

See Ofnuts reply.

It is easily demonstrate with a constant aperture zoom (such as an
80-200 f/2.8). Leave the aperture wide open and zoom in and out and you
can see that as the internal components change position, there is a
circle (pupil) that moves forward and back effectively controlling the
max amount of light coming in (before the aperture blades).

As you zoom "out" (wide) it gets closer to the front of the lens
(restricting more light) and as you zoom in it gets closer to the
aperture restricting less light. (This is on my 80-200 f/2.8, on other
lens designs it may be different).

As to a meter reading it generally covers a constant angle, so
regardless of distance it gets the same unit area amount of light.

So if you're measuring the light off of a 1m^2 area of fence you get L
amount of light.

Step back so it's a 4m^2 area and you receive 1/4 L of the light from
the first 1m^2 area - but you get 4 of those and you're back to square
1. So to speak.

--
gmail originated posts filtered due to spam.

On 03/03/2011 08:40 PM, Alan Browne wrote:
On 2011.03.02 15:44 , GA wrote:
Why isn't reflected light subject to Inverse Square Law?

It is subject to 1/x^2 law.


For example, if you take a photo of an object at a given aperture,
shutter speed and ISO based on the light reading. Why is it also
possible to correctly expose the same object from much further away with
the same settings?

See Ofnuts reply.

I was starting to wonder if my post was anywhere but on my ISP's servers :-)


So if you're measuring the light off of a 1m^2 area of fence you get L
amount of light.

Step back so it's a 4m^2 area and you receive 1/4 L of the light from
the first 1m^2 area - but you get 4 of those and you're back to square
1. So to speak.

Indeed. The fixed focal length case.

--
Bertrand