If a 90-W lightbulb emits 3.5 % of the input energy as visible light (average wa
ID: 1499870 • Letter: I
Question
If a 90-W lightbulb emits 3.5 % 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 2.5 m away?
Express your answer using two significant figures.
N_1/t = photons/s
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?
Express your answer using two significant figures.
N_2/t = photons/s
Explanation / Answer
a) Total power radiated by light bulb P is = 90* 3.5/100 = 3.15 watts
Intensity of light at 2.5 m away I = P / A
I = P / (4pi R^2 )
= (3.15 W) / (4pi (2)^2* 10^-6) = 0.6270 W /m^2
No. of photons striking to that surface
N = IA / Ephoton
where A = area of surface at which photon striking
and E = hc / lambda
N = (I A lambda) / hc
N = (0.6270 x pi x 0.004^2 x 550nm) / (1240 nm-eV x 1.60 x 10^-19 J/eV) = 1.59 x 10^12 photons
b)
Intensity of light at 1.5 km away I = P / A
I = P / (4pi R^2 )
= (3.15 W) / (4pi (1500)^2) = 1.672 x 10^-7 W /m^2
No. of photons striking to that surface
N = IA / Ephoton
where A = area of surface at which photon striking
and E = hc / lambda
N = (I A lambda) / hc
N = ( 1.672 x 10^-7 x pi x 0.004^2 x 550nm) / (1240 nm-eV x 1.60 x 10^-19 J/eV) = 4233935.48 photons