Find Photons per second on the eye from a source a distance away

[rocapp]

Homework Statement

A typical incandescent light bulb emits ~3x10^18 visible-light photons per second. Your eye, when it is fully dark adapted, can barely see the light from an incandescent light bulb 10 km away.

How many photons per second are incident at the image point on your retina? The diameter of a dark-adapted pupil is ~7 mm.

Homework Equations

N = n*(d/D)

The Attempt at a Solution

I found the above equation on the internet but it didn't work. I tried:

N = Number of photons*(Diameter/Distance)
=3x10^18 s^-1 * (7x10^-3 m / 1x10^4 m)
N= 2.1x10^12 photons/s

This is incorrect. Please help! Thanks!

 

[technician]

The light travels out in all directions so you need to look at the AREA through which the photons pass.
The surface through which the photons pass is the surface of a sphere.
Hope this gets you started

 

[rocapp]

Not sure of this either, but here is my thinking:

Number of Photons/s = (Source Photons/s)/(Surface Area of sphere formed by the distance) * (Surface Area of eye)

Photons/s = (3x10^18 photons/s)/(4*PI*1.0x10^8 m^2) * (4*PI*7x10^-6 m^2)

Photons/s = 2.1x10^5

 

[mfb]

The surface area of the eye is wrong (the pupil is like a circle and not a sphere, and 7mm is the diameter, not the radius), but the concept is right.

 

[rocapp]

So then:

((3x10^18)/(4*PI*1.0x10^8)) * (PI*3.5x10^-6)

= 26250 Photons/s

Correct?

 

[rocapp]

That wasn't correct. The correct answer was:

9.2×10^4 s^-1

 

[rocapp]

Anyone know why?

 

[mfb]

I don't know. The factor between those two answers is ~3.5. Not pi, not 4, or anything present in the problem statement. Very odd.