• So we have a floodlamp in the midst of the night, in the Grand Canyon, there is no moon, and it is bloody dark. The floodlight illuminates a piano, that we placed near the floodlight. The camera is 300 m away, it has got a 1000 mm lens, and we suppose for the sake of the argument that it frames the piano quite exactly.

The floodlamp light falling on the piano follows the inverse square law. We measure the light coming from the lamp with an incident light meter, without dome, pointing at the light source. We calculate the exposure based on that reading.

Then we walk 300 m to our camera, and set that exposure on it. Will it work?

Or, even, we don't have a 5000 mm but an ordinary 50 mm and the piano looks quite tiny in our frame. Nonetheless, is the tiny spot that represents the piano going to be correctly exposed?

"Experience" seems to suggest that the piano is going to be well exposed.
My (shaky) understanding of physics seems to suggest to me that the light reflected by the piano should indeed obey to the inverse square law, and the reading I take with the incident light meter should produce underexposure. If we had placed the lamp at twice the distance, the light we would have measured would have been 2 stops less. Now we walk 300 m away from the piano reflecting our light in all directions, we should observe a fall of light reflected from the piano with distance.

Fabrizio

PS In this example, the piano fills the image. We observe the light coming under that smaller angle that is given by our tele lens. We are not gathering on film more light than is necessary to describe the piano. A reflected light meter should give indicate a less bright object.