If a 90-W lightbulb emits 2.5 % of the input energy as visible light (average wa
ID: 1560711 • Letter: I
Question
If a 90-W lightbulb emits 2.5 % 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 2.2 m away?
Express your answer using two significant figures.
****The answer to Part A is: 1.3*10^12 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 1.1 km away?
Express your answer using two significant figures.
N2t = photons/s
N1t = 1.3*10^12 photons/sExplanation / Answer
Pin = P
Pout = 0.025*Pin = 0.025*P
intensity at r1 = 2.2 m
I = Pout/(4*pi*r1^2)
I = 0.025*p/(4*pi*r1^2)
energy absorbed by pupil
Eabsorbed = I*pi*r2^2*t
r2 = radius of pupil = 0.002 m
energy absorbed = N1*hc/L
N1*h*c/L = 0.025*p*pi*r2^2t/(4*pi*r1^2)
N1/t = 0.025*P*r2^2*L/(4*r1^2*h*c)
N1/t = 0.025*90*0.002)^2*550*10^-9/(4*2.2^2*6.626*10^-34*3*10^8)
N1/t = 1.3*10^12 photons/s
part B
for r1 = 1.1 km = 1100 m
N2/t = 0.025*90*0.002^2*550*10^-9/(4*1100^2*6.626*10^-34*3*10^8)
N2/t = 5.1*10^6 photons/s <<<<<----answer