How many Photons per second are entering the pupil?

[clamatoman]

Homework Statement

(a)Estimate the number of photons per second emitted by a 100-W lightbulb, assuming a photon wavelength of 550nm.(b) A person can just see this bulb for a distance of 800m, with the pupil dilated to 7.5mm. How many photons per second are entering the pupil?

Homework Equations

E=hc/λ
h=6.63*10^-34 J s
c=3.0*10^8 m/s
1eV=1.602*10^-19 J
1 watt second = 100 joules

The Attempt at a Solution

For part (a)
E=hc/λ=(6.63*10^-34 J s * 3.0*10^8 m/s)/(550nm) =3.616*10^-19 J

convert to eV: (3.616*10^-19 J * (1eV/1.602*10^-19J)=2.257 eV per photon

100 watt sec = 100 joules

(100 joules * (1eV/1.602*10^-19J))/2.257eV) = 2.77*10^20 photons

For part (b):[/B]
Not excatly sure how to proceed. I made a circle and determined the percent of the circumference was the 7.5mm pupil then multiplied that by the total number of photons, but the answer I got did not mach the book.
C=pi*d=pi*1600m=5026.548m
(7.5mm/5026.548m)*2.77*10^20 photons = 4.13*10^14 photons.
The answer in the book is 6.1*10^9 photons.
How can I get to this answer?

 

[gneill]

I made a circle and determined the percent of the circumference was the 7.5mm pupil

Does a lightbulb send out all its photons in a plane?

 

[clamatoman]

Ah, I need to consider a sphere, and a conical section of that sphere with the circular base being the 7.5mm pupil. I shall attempt.

 

[clamatoman]

Volume of the sphere = 4/3 pi r^3 = 2.14×10^9 m^3
Volume of the cone = pi r^2 h/3 = 0.047 m^3
ratio = 2.196 * 10^-11
2.196 * 10^-11 * 2.77*10^20 =6.1*10^9 protons!
Thanks!

 

[gneill]

Actually, I believe that the given answer is incorrect. You want to look at the flux of photons passing through the surface area of the pupil per second, so the ratio should be between surface areas, not volumes.

 

[willem2]

Actually, I believe that the given answer is incorrect. You want to look at the flux of photons passing through the surface area of the pupil per second, so the ratio should be between surface areas, not volumes.

Wouldn't that come out the same? The volume of a cone is proportional to the area of the base.

 

[Delta2]

I think there is a conspiracy in this so that they come out the same.
The volume of sphere with radius r, $$V_s=\frac{4\pi r^3}{3}=S_s\frac{r}{3}$$
The volume of the cone with height r $$V_c=S_c\frac{r}{3}$$
From the above we conclude that the $$\frac{V_s}{V_c}=\frac{S_s}{S_c}$$ that is that the ratio of volumes equal the ratio of surfaces. The "conspiracy" is that both volumes can be written as the product of the respective surfaces times the same factor $\frac{r}{3}$.