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Just what the title states.

Assuming identical conditions, excellent visibility - If a 1W monochromatic light source, and a 1W non-monochromatic light source were viewed at a location in deep space, which of the two would be visible from a greater distance?

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Are you asking about sensitivity or scattering? – Ron Maimon Aug 31 '12 at 3:23
Not sure I understand what is asked. Could you put it in a layman's terms please? – Everyone Aug 31 '12 at 17:52

Actually, that all depends on the wavelength frequency of your monochromatic light source.

Obviously, the $1/r^2$ falloff of the intensity is the same for both cases. What is more important here is the sensitivity of the human eye to specific colors. As you can read here, the human eye is most sensitive for light around 550nm, which is yellowish (makes sense, since the Sun is also yellow).

Somewhat simplified: your 1W panchromatic light source will have less energy in the neighbourhood of the 550nm wavelength, if the 1W monochromatic light source will be emitting closely to or on this specific wavelength.


  • if the monochromatic light source is yellow, you will be able to see it further out than a panchromatic source.
  • If the monochromatic one is deep red, the panchromatic source will win.
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Monochromatic simply in electromagnetic radiation means, staying at a single frequency...

Simple example: If your non-monochromatic light has some colors (say red) which have longer wavelength, and your monochromatic light is violet, then the non-monochromatic one would reach you at greater distances... But, you'd need highly sensitive detectors to search that tiny 1W lights...

Ultimately, we could conclude that distance traveled by these EM waves depends only on their wavelength (their energy) and not their state (mono or non-mono)

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To me it is not quite clear about what physical effects you are talking here, do you mean refraction ? In addition, the propagation of light rays depends on how the are scattered, absorbed, and reflected by the medium. This is described by the radiative transfer equation (RTE). I was not the downvoter here ... ;-) – Dilaton Sep 5 '12 at 8:47
@Dilation: What you said was right. But, the OP has mentioned identical conditions to be assumed in deep space.. But with this availability (no obstacles, no medium, excellent visibility), we could consider only the energy... to reach the observer. And Thank you... – Waffle's Crazy Peanut Sep 5 '12 at 15:21

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