Preface: if there is a better-suited forum for this question, please direct me. I figured it'd be better to post into Photography and not Physics as some of you might have already examined this particular issue. Thank you.
I am wondering about the physics side of things. Let's say I position a camera 25cm away from an ideal 4W light bulb (all 4 Watts go into electromagnetic radiation, and we won't differentiate between visible and IR/UV spectrums). Let's assume that the distance is measured from the bulb's filament to the camera's sensor surface. To have somesthing specific, let's say it's Canon 5D Mk4, with a Zeiss Milvus 2/35 lens attached. The flange focal distance is 4.4cm, and the lens' length until the front glass element is about 8.5cm.
My question is: as the light travels from the bulb to the glass, it loses denstiry (pardon my lack of proper terminology). Once it enters the lens system, however, does the process continue in some form? Or is all the light transferred to the imaging circle? I do not consider factors such as vignetting and glass transmission in this case. In the end, I would like to be able to estimate the amount of energy that reaches my sensor by knowing the distance to the light source and its initial brightness, etc.
Below is my initial understanding of the issue; however, it might be 100% wrong in every aspect. You can safely ignore it.
If I wanted to calculate the amount of light that hits the sensor, consulting the inverse square law to a reasonable precision, would I use the distance to the lens' surface, or to the sensor's surface? Or something else, like the focal center of the lens? The difference of ~12cm at this distance is quite significant. I assume that the relevant measure is the lens' entrance pupil. Perhaps it should be the whole frontal glass element?
I cannot measure the entrance pupil, but let us assume its diameter is 1cm. The area would be ~3.14cm^2 (or 0.000314m^2), that should be close enough. The end of the lens is approximately 13cm (0.13m) away from the light source. From that it is easy to calculate that the lens' entrance pupil receives about 0.15% of the total light emitted by the bulb, or 0.006 Watts. Meanwhile, if the lens' tip was positioned full 25cm away from the bulb, it would receive only 0.04%, or 0.0016 Watts. Nearly 4 times less.
But, the tip is positioned 13cm away from the light. My question is, does the light continue to dissipate within the lens? Is the "distance" within the lens the same as its physical length, or is it expanded or compressed somehow? Or is all the energy that enters the Entrance Pupil transferred to the imaging circle? Or, maybe I am wrong about everything from the get-go?