A sunbeam with intensity 1.8 watts / m^2 shines through a square window .6 meters on a side. The beam makes an angle of 64 degrees with normal to the plane of the window. What is the power flux through this window?
The area of the window is ( .6 m) ^ 2 = .36 m ^ 2, so the flux of a perpendicular 1.8 watt/m ^ 2 beam would be
The 64 degree angle means that less light passes through the window. The proportion of the change is cos( 64 deg) = .4392, so the flux is
The power flux of an energy intensity `intensity (typically measured in Watts/square meter) intercepted at a perpendicular by an area A is simply
If the beam makes angle `theta with the perpendicular, then flux is reduced according to the cosine of the angle: