# Help me understand this

1. Apr 26, 2005

### Reshma

I seem to have a problem understanding this problem let alone solving it

A star emits 2x1045 photons per sec, 80% of which are in the visible range. The star is 1200 light years away, and the night adapted eye, whose pupil makes an opening 4mm in diameter, can barely see the star. How many photons per sec are necessary to trigger the eye as a detector?

I figured out the number of photons in the visible range is = 0.8x2x1045

Can someone help me proceed?

2. Apr 26, 2005

### Galileo

Out of those photons, how many will go to a surface of known area a given distance away?

3. Apr 26, 2005

### jdavel

Reshma,

2E45 is a lot of photons, right? What do you think the problem means when it says the star emits that many photons/second? What are you picturing?

4. Apr 26, 2005

### Reshma

Thank for looking into this-Jdavel and Galileo.

Jdavel-I'm picturing the number of visible photons here, but few will be detected by the night adapted eye.

Galileo- are you referring to the surface area of the pupil?

5. Apr 26, 2005

### learningphysics

The 2E45 photons form a shape... they are burst by the star every second.

What is the area of the shape when photons reach 1200 light years away.

Can you use the area of this shape and the area of the eye to calculate how many of the 2E45 photons hit the eye?

That is the required number of photons per second.

6. Apr 27, 2005

### Reshma

OK, the distant star is a sphere and HERE you take the radius(R) as 1200 light years, right? Using this assumption:
R=1.14x1019m. Area is given by $$4 \pi R^2$$
Can you clarify this?

7. Apr 27, 2005

### Reshma

Also area of puplil $$A= \pi (2\times 10^{-3})^2$$

8. Apr 27, 2005

### maverick280857

I would suppose that the elementary flux model we use will give the answer here: you know the fraction f of the total number of photons emitted by the source (N). A light year is the distance travelled by light at a velocity 299792458 m/s in vacuum in one year (which means a LOT of seconds: 86400s/day * 365 days or if you want to be ultraprecise (you can't be) then its 365.25 days). The source gives out fN photons per second but the source is apparently isotropic so it gives out fN photons over a spherical region of space. If you "cut in" somewhere with some radius r then the effective area over which the fN photons are "available" completely is $A_{T}=4\pi r^2$ so the number of photons arriving at the eye is actually $\frac{fN}{A_T}A_{eye}$. Whats wrong with that? :shy:

9. Apr 27, 2005

### SpaceTiger

Staff Emeritus
In addition to reading the above comments, you may want to check out this thread.

10. Apr 27, 2005

### learningphysics

Yes. Looks right.

(area of eye)/(area of sphere) = (no.visible photons that hit eye)/(total number of visible photons)

I made a mistake in my previous post. Be sure to use 1.6E45 (visible photons) instead of 2E45.

The number of visible photons that hit the eye is what you're looking for.

Last edited: Apr 27, 2005
11. Apr 28, 2005

### Reshma

Thanks for the help everyone. My final answer is: 12 visible photons per sec hit the eye.
SpaceTiger- Thank you so much for the link. Your post was extremely well written and gives me a clear picture in context to this question.

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