# Intensity, photons, human eye

1. Apr 4, 2009

### Yroyathon

hi there. there's an example in the book, but i'm having a little trouble here.

1. The problem statement, all variables and given/known data
Assume that the human eye can pick up as few as 9 photons/s in the visible range. Based on this, estimate the intensity of the dimmest star that can be detected by a night-adapted eye. What is the ratio of this intensity to the intensity of noon sunlight, some 1400 W/m2? This large intensity range means that the eye is indeed a very adaptable instrument.
Answer format = (intensity of 9 photons/s / intensity of noon sun)

use 3mm for the radius of the pupil.
use 550 nm for wavelength.

2. Relevant equations
I=Power/Area
I=N*E, N is the number of photons/(m^2*s)
E=h*f
f=c/lambda
A=Pi*r^2

3. The attempt at a solution

So I want the intensity, which is N * E, which is (N * h * c) / (Pi * r^2 * lambda). I get the intensity of the dimmest star on the human eye as 1.14592 * 10^(-13).

With the given intensity of noon sunlight (do I need to adjust this for the area of the pupil?...), I divide it. 1.14592*10^(-13) / 1440 = 7.95775*10^(-17).

but this is wrong, so... either I was supposed to adjust the noon intensity, or I've made one or several other mistakes.

I'd appreciate any insights or tips.

Thanks!

,Yroyathon

2. Apr 3, 2010

### grantjerry

check ur calculations^_^

3. Apr 3, 2010

### phyzguy

I got the same numbers. Why do you think it's wrong? What do you think the right answer is?

4. Apr 3, 2010

### grantjerry

I got 1.150429803e-13 for the intensity of the star

divide it by 1400, the final answer should be 8.217e-17