need clarity...

[jeetujee]

will somebody clarify this confusion about the fall of light at a distance----

 

FACT:

1) we generally take incident readings. typically at the point at which the subject will be.

2) as per laws of optics light falling on an object reflects.

3) as per laws of photography light reflected from an object travells through the lens, reaches the film plane and reacts propotionately to the amount of light falling and creates a certain amount of density on the film plane.

 

LIGHT also follows the inverse square law

 

now taking a case where the incident reading on the subject is A and the distance between the object and the image plane is D If the density on the film plane is X (negative) then if the distance is increased to 2D and the incident reading at the subject is still A then the density on the film should be 4D (following the inverse square law) (CONSIDERING THAT THE REST OF THE ASPECTS OF EXPOSURE REMAIN CONSTANT like f-stop)

 

ie., WE ARE UNDEREXPOSING BY 2 stops!!!!!!!)

 

Why is it that we still expose to the incident reading of the subject without considering the amount of distance between the camera and the subject which would invaribly determine the amount of light falling on the image plane.

 

AND how is it that we still get rightly exposed images

 

the same is the case with the reflected meter. If i read the a reflected reading of an area from a certain distance and then go futher away the reading still does not change?

[Rob Belics]

The amount of light, the number of photons, is constant in the same direction. Increasing the distance to the camera does not change that.

 

Changing the distance from the object to the light changes the number of photons spread over the object area. This increases (or decreases) the number of photons concentrated in an area. This concentration is the number of photons per square foot, thus the inverse square law.