Optics: Electric fields, beams, irradiance

[blorpinbloo]

Homework Statement

An electromagnetic wave in a vacuum /w magnetic field of 265 mG.

If the wave is spatially confined to a beam with a diameter of .75mm, how much energy is measured by a photodetector with an active area with diameter 5 cm in one second.

The Attempt at a Solution

This is one part of the question, but its the only one that has me stuck. I found the max electric field to be 7950 N/c, and the irradiance I found to be 83941.14 W/m^2. I know that irradiance is power/area, so to get the beam power, I'd take 83941.14*(pi*(.00075/2)^2). Then power is energy per unit time, so I'd just divide by one second to get the beam energy. The confusion comes when the photodetector comes into play. I'm not sure how the active area of the photodetector affects the energy that it reads. This is where I'm stumped. Any help is appreciated

 

[Redbelly98]

Since the detector is larger than the beam, it receives the full energy of the beam.

If the detector were smaller than the beam, you would have to figure out what fraction of the beam energy actually hits the detector. (But for this problem you do not need to do this.)