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    What affects the intensity of photons in EM radiation, and also do photons lose their energy, such as when photons from the sun hit Venus in comparison to the Earth?

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    Intensity is subject to the 'inverse square law'. I=k/(x^2)

    Basically, if you are twice as far away, (distance becomes '2x', but it's "x^2" so it's actually 2x^2.... Which is 2^2, which is 4!) then the intensity will become a quarter of what it was at that known point.

    So double distance = quarter the intensity. Treble the distance = one ninth the intensity, etc.

    The other equation you could use is "L1/L2 = (x2/x1)^2". Which means that the ratio between the intensities at two points is equal to the ratio of their distances, squared. Thus, you can find intensities at a distance, or distances from an intensity.

    Through a vacuum (EG, space) the photons will maintain their energy so long as they don't come into contact with anything. The only difference between the photons at Earth and the photons at Venus are that the ones at Venus aren't as spread out. They're closer to the source, so they have a greater intensity (things seem brighter). When they get to Earth, the photons have spready out more, so things are darker... Less energy arrives.


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Updated: June 28, 2012
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