Photon Emission At A Distance

[hy23]

Homework Statement

A light bulb 10km away from you emits visible photons at a rate of 3x10¹⁸ photons/s. Assuming this is the only source of light (the whole place is dark), and your dark-adjusted retina has a diameter of 7mm, then how many photons/s hits your retina?

Homework Equations

Not really sure if any formulas are needed, but the ones presented in the chapter containing this question are
E=hf (Plancks)...lambda =h/p (matter waves)

r=dsin(a)=m(lambda)...path length difference ( for constructive interference)
y=Lsin(a)...(position of fringes on a a screen L distance away from slit)

The Attempt at a Solution

I've taken the light bulb to be point source since it's so far away. I believe the lightbulb also emits photons in all directions (i.e. 360 degrees). Thus the portion of its photon emission that actually travels in the direction of your retina should correspond to the fraction of the angle of incident light rays divided by 360 degrees.

So, using small angle approximation, this angle is a=0.007m/10000m

my final answer was 3x10¹⁸x a/360 = 5.8333 x 10⁹ photons/s

I doubt this is the correct answer because my initial assumption that the light bulb emits photons in all directions means that by assuming 360 degrees I have incorrectly simplified this problem to a 2D situation which is not what it is; in addition I have used none of the formulas in the chapter which worries me.


Thanks in advance for any help

 

[gneill]

The radiation is going in all directions (3D), so think in terms of surface areas. Your 3 x 10¹⁸ photons per second are going to be spread out over a surface of a shell 10km in radius...

 

[hy23]

ohhhhhhhhhhhhhhh! very clever!

so it should be the area of your retina divided by the surface area of the 10km shell, why didn't I think of that...thanks a lot

but overall is this way of approaching the problem correct? I have this fear that there may be some other formulas involved..

 

[gneill]

ohhhhhhhhhhhhhhh! very clever!

so it should be the area of your retina divided by the surface area of the 10km shell, why didn't I think of that...thanks a lot

but overall is this way of approaching the problem correct? I have this fear that there may be some other formulas involved..

I would be surprised if it were incorrect...

This is typically how things like solar irradiance at a given distance from the Sun are handled.

 

[rwisz]

.

I would say that your approach is a good start but there are a few things to consider. First, the light bulb does emit photons in all directions, but the intensity of the light decreases as you move further away from the source. This means that the fraction of photons that actually reach your retina will be less than what you calculated. Additionally, the size of the light bulb and its distance from your retina will also affect the number of photons that reach your retina.

To accurately calculate the number of photons that hit your retina, you can use the formula for the intensity of light at a given distance from a point source: I = P/4πr^2, where P is the power of the light source and r is the distance from the source. You can then use this intensity to calculate the number of photons that reach your retina by using the formula N = I/A, where A is the area of your retina.

Another important factor to consider is the wavelength of the photons emitted by the light bulb. This will determine the number of photons that can actually hit your retina, as only those with a wavelength within the visible spectrum can be detected by your retina.

Overall, the correct approach would involve taking into account the size and distance of the light bulb, the intensity of the light at your retina, and the wavelength of the photons emitted. This would give you a more accurate calculation of the number of photons that hit your retina.