Question

A light detector (your eye) has an area of 1.95×10-6 m2 and absorbs 75% of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.40 m from the source. If the detector absorbs photons at the rate of exactly 3.950 s-1, at what power does the emitter emit light?

Answer 1

Answer:

Power=9.74*10^8ev/s

Explanation:

A light detector (your eye) has an area of 1.95×10-6 m2 and absorbs 75% of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.40 m from the source. If the detector absorbs photons at the rate of exactly 3.950 s-1, at what power does the emitter emit light?

the rate at which proton are absorbed by the detector is related to the rate at which proton is emitted by light source

Rabs=0.75(Aabs  *Remit)/4$\pi$r^2

Aabs=1.95×10-6 m2

Rabs=3.950 s-1

r=3.40 m

Remit=4*3.142*(3.4)^2*3.95

Remit=573.8/(.75*1.95×10-6 )

3.928*10^8 photon/s

enery of poton=

E=hc/lambda

E=1240ev/500nm

E=2.48ev

power of source

Remit*E=3.928*10^8*2.48

Power=9.74*10^8ev/s