jcc
Member
Let me preface this by saying that it's past 4 am (late to bed, not early riser), I can't quite think of the answer, and it's unimportant enough that I will forget it the next day. But, it's a curious thought...
(1) Inverse square law states that light will fall off exponentially (the equation doesn't matter for the purposes of this discussion--or does it?!).
(2) The camera records light that is reflected off of the subject.
(3) Light rays travel from source to subject, then bounce off the subject, then into the camera. The light travels the distance of D_total = D_source-to-subject + D_subject-to-camera.
(4) Assume the source and subject remain at a constant distance from each other.
(Q) Shouldn't changing the distance between subject and camera change light intensity (via inverse square law), since the light has to travel that much more/less? Let's say we have a properly exposured photo at X distance from subject. Why does changing subject-to-camera distances, by X/2 or 2X, require no stop-compensation?
(1) Inverse square law states that light will fall off exponentially (the equation doesn't matter for the purposes of this discussion--or does it?!).
(2) The camera records light that is reflected off of the subject.
(3) Light rays travel from source to subject, then bounce off the subject, then into the camera. The light travels the distance of D_total = D_source-to-subject + D_subject-to-camera.
(4) Assume the source and subject remain at a constant distance from each other.
(Q) Shouldn't changing the distance between subject and camera change light intensity (via inverse square law), since the light has to travel that much more/less? Let's say we have a properly exposured photo at X distance from subject. Why does changing subject-to-camera distances, by X/2 or 2X, require no stop-compensation?