brightness of the surface of the earth,when the sun's rays fall at an angle of 45 is 8 lx.find brightness when light falls perpendicular tothe earth's surface
The brightness on a given surface is variable with the angle between the incoming light rays and normal to surface.When falling at a certain angle `alpha` the total surface that is illuminated by the same amount of available light is larger. Thus the total surface that is illuminated depends on the angle as `S =S_0/sin(alpha)` where `alpha` is the angle between the incoming light and horizontal (for very small angles of illumination the surface tends to very high values), and `S_0` is the total surface iluminated at normal incidence. Since the total available flux is the same in both cases
`Phi= I_1*S= I_2*S_0`
where `I_1` is the brightness at `45` degree and `I_2` is the brightness at `90` degree incidence angle.
It follows that
`I_1*S_0/sin(45) = I_2*S_0`
`I_2 =I_1/sin(45) =8/sin(45) =11.31 lx`
Answer: The brightness when the same light flux is falling at normal incidence is 11.31 lx.