windowofhope
- 5
- 0
I'm faced with a problem here and would really appreciate anyone's help. The question is paraphrased as follows:
The eye can respond to single photons but there are certain requirements from the eye: it must receive at least 100 photons/s to process it. Assume the eye to have an area of 4E-6 m^2, from how far away can the eye pick up the radiated energy of a 400 nm wavelength photon from a 200W light bulb?
So I know that the photon flux is equal to the poynting vector (in W/m^2) divided by the energy of each photon i.e |S|/hf=S*Area*lambda/(hc).. I do (200W/m^2)*(4E-6 m^2)*(400E-9m)/(6.63E-34Js)(3E8m/s) = 1.6E15 photons/s which is clearly greater than the required 100 photons/s required for the eye to register the light. However, the formula doesn't say anything about distance away from the source.. All I know is that the surface of the eye picks up 1.6E15 photons/s but have no information as to what distance that is from and how much mroe distance until it gets reduced down to 100 photons/s. Any help would be really awesome!
thank you
The eye can respond to single photons but there are certain requirements from the eye: it must receive at least 100 photons/s to process it. Assume the eye to have an area of 4E-6 m^2, from how far away can the eye pick up the radiated energy of a 400 nm wavelength photon from a 200W light bulb?
So I know that the photon flux is equal to the poynting vector (in W/m^2) divided by the energy of each photon i.e |S|/hf=S*Area*lambda/(hc).. I do (200W/m^2)*(4E-6 m^2)*(400E-9m)/(6.63E-34Js)(3E8m/s) = 1.6E15 photons/s which is clearly greater than the required 100 photons/s required for the eye to register the light. However, the formula doesn't say anything about distance away from the source.. All I know is that the surface of the eye picks up 1.6E15 photons/s but have no information as to what distance that is from and how much mroe distance until it gets reduced down to 100 photons/s. Any help would be really awesome!
thank you