Luminous intensity in different directions

# The intensity of direct sunlight on a surface normal to the rays is Io. What is the intensity of direct sunlight on a surface whose normal makes an angle of 60 degrees with the rays of the sun?
I think the answer is Io itself because for an isotropic point source of light the luminous intensity is the same in all directions. Is it correct?
I may be barking up the wrong tree, so correct me someone if I'm wrong but....

By your reasoning, if the angle of normal to the surface tended towards zero, then if the intensity at all points were Io then the total energy across the surface would tend to infinity (for an infinite surface).

The intensity must decrease with distance.



Gold Member
Why wouldn't it be cos(60)?
I think lamberts inverse square law is true only for illuminance and not for luminous intensity.
Illuminance = [(luminous intensity) x (cos(theta))]/R^2
Luminous intensity = (Luminous flux)/(Solid angle)
Luminous intensity is equal to the Luminous flux for unit solid angle. By definition of Luminous Flux, it is the energy of the visible part of the radiation emitted,transmitted or received per second. As the solid angle is increased, the total flux goes on increasing and reaches a maximum value at 4(pi) steradians. So I think in a particular direction the luminous intensity is constant irrespective of the distance. I may be wrong.

The question as you typed it asks for the intensity falling on a surface rather than for a given solid angle. The intensity - a measure of the power received per unit area must change with angle.


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