Question

If a 80-W lightbulb emits 3.0 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions.

Part A

How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.3 m away?

Part B

How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.5 km away?

Answer 1

PARTA)

Power = work/time.

80J/s *3/100% => 2.4 J/s.

Now, we want to find how many photons are in 2.4J.

E = hf = hc/lambda.

E = 3.6142*10^(-19)J/photon.

So, we divide and find that there are going to be 6.64*10^(18)photons/s to be emitted.

Now we want to find the fractional part of the surface area that the eye occupies.
pi*(0.002m)^2 = area of eye.

4pi*(1.3m)^2 = total area

Number of photons/s = [(0.002)^2/(4*1.3^2)] * 6.64*10^18 = 3.929*10^12 photons/s

part b)

Number of photons/s = [(0.002)^2/(4*1500^2)] * 6.64*10^18 = 2.95*10^6 photons/s