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Calculating Photon-Flux

  • Thread starter roman93
  • Start date
  • #1
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Homework Statement


The faintest objects that have been detected at optical wavelengths with the Hubble
Space Telescope have apparent magnitudes m  31. Calculate the flux from
an object of this magnitude, and, assuming that each photon has a typical optical
wavelength, convert your result into the number of photons per unit time per unit
area at Earth (this is the so-called photon flux).
[The Sun has apparent magnitude m = -26.8, the flux from the Sun is 1350Wm^-2]


Homework Equations


m_2 - m_1 = 2.5log( f_1/f_2) log with base 10 here

E = hf, λ = c/f


The Attempt at a Solution



well I worked out the flux of the object to be 1.02*10^-20 Wm^-2. (I'm pretty sure this is right) however I am not sure what the typical optical wavelength is or how to go about working out the "Photon Flux".

I thought if I could work out the energy of a photon with typical wavelength, I can then divide the calculated flux by energy of photon which will give me the number of photons...
I am not sure how I work out the number of photons per unit time though =/
 

Answers and Replies

  • #2
6,054
390
Typical optical wavelength would be those of visible light.
 
  • #3
537
1
typical optical wavelength just means a wavelength somewhere in the visible spectrum

since the visible spectrum goes from 390nm to 750nm, looks like 570nm would be a good choice, it's right in the middle

and so using that, you should be able to figure out the typical energy per photon

if you have

[itex]\frac{energy}{photon}[/itex]

and

[itex]\frac{energy}{seconds*area}[/itex]

then how would you get

[itex]\frac{photon}{seconds*area}[/itex]
 
  • #4
14
0
typical optical wavelength just means a wavelength somewhere in the visible spectrum

since the visible spectrum goes from 390nm to 750nm, looks like 570nm would be a good choice, it's right in the middle

and so using that, you should be able to figure out the typical energy per photon

if you have

[itex]\frac{energy}{photon}[/itex]

and

[itex]\frac{energy}{seconds*area}[/itex]

then how would you get

[itex]\frac{photon}{seconds*area}[/itex]
guess I just divide one by the other. How do I work out the energy given the flux? (The total energy would be luminosity right?) if so the formula f = L / 4(pi)d^2 is the diameter the object receiving or the one emitting?
 
  • #5
6,054
390
How do I work out the energy given the flux?
Energy of what, and, more importantly, what for?
 
  • #6
537
1
flux *is* energy

watts per unit area is energy per unit time per unit area

the flux tells you how much energy is moving through a surface per unit time
 
  • #7
14
0
flux *is* energy

watts per unit area is energy per unit time per unit area

the flux tells you how much energy is moving through a surface per unit time
Oh perhaps... I calculated the flux wrong then. Because using a wavelength of 570nm I get the energy per photon to be higher than the flux. so surely I can't have almost 0 photons reaching earth?
 
  • #8
14
0
Energy of what, and, more importantly, what for?
can you check that I have calculated the flux right?

m_1 = 31, m_2 = -26.8, f_1 = ? and f_2 = 1350.

31+26.8 = -2.5log(f_1 / 1350) so

10^(-57.8/2.5) = f_1/ 1350 so f_1 = 1350*10^(-57.8/2.5)
 
  • #9
6,054
390
so surely I can't have almost 0 photons reaching earth?
Almost is subjective. Your flux seems correct.
 
  • #10
14
0
Almost is subjective. Your flux seems correct.
okay E = hc/λ. so using a wavelength of 570nm I get that E = 3.49*10^-19 J.

so the number of photons = (1.02*10^-20)/3.49*10^-19 which is 0.03...

I think somewhere I went wrong but can't figure out where?
 
  • #11
6,054
390
It is indeed a very small number, that's why the source is considered very faint.
 
  • #12
14
0
It is indeed a very small number, that's why the source is considered very faint.
Ah okay, thank you both very much for your help!
 

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