Calculating attenuation necessary to reduce number of photons

In summary: Thanks for the help!In summary, the attenuation necessary to reduce the number of photons in a beam to single-photon levels is N approximately 1 or 2.
  • #1
kevincb672
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3
Homework Statement
How are we to calculate the attenuation necessary to reduce the number of photons emitted from a laser beam to one photon per meter?
Relevant Equations
N = Plambda/hc(^2)
where N = photons per meter, P = power of beam, lambda = wave length, h = planck's constant, c = speed of light
E = hc/lambda
where E = energy of single photon
Hi there. I am attempting to do calculations for my own project, the question being what is the attenuation necessary to reduce the number of photons in a beam to single-photon levels. N approximately 1 or 2.

The laser in question is a 650nm 5mW laser.
I have solved the energy per photon 3.06*10(^-19) J.
I have solved the number of photons emitted from the laser beam per second 1.63*10(^16) photons of light per second.
I am struggling to find the photons per meter. I may have got the previous calculations wrong as well.
My calculations suggest 5.44662309×10(^12). A small part of me wants to be this wrong as this is if I am right entailing the need for the laser to be attenuated by 10x(^-12) so 12 orders of magnitude. If that is right of course.

I am very appreciative of any and all help. I've probably got this all wrong lol.

Thanks
 
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  • #2
kevincb672 said:
I am struggling to find the photons per meter.
Imagine you shoot the beam into space for one second. How long is the beam in meters and how many photons are in it?
 
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  • #3
Hello @kevincb672 ,
:welcome: ##\qquad##!
My calculations suggest 5.44662309×10(^12).
N/c is 5.46 107 according to my calculations ...

##\ ##
 
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  • #4
Drakkith said:
Imagine you shoot the beam into space for one second. How long is the beam in meters and how many photons are in it?
With a length to detector 2m is it correct to say the beam has 1.63398693×10(^16) photons? Apologies for my own ignorance.
BvU said:
Hello @kevincb672 ,
:welcome: ##\qquad##!
My calculations suggest 5.44662309×10(^12).
N/c is 5.46 107 according to my calculations ...

##\ ##
Hi that does look right! Do you mind running through this a little bit, just so I can grasp it entirely :nb), thanks again..
 
  • #5
It's your own ##N = P\lambda/(hc^2)##

When I see ##\ 1.63\times 10^{16} ## divided by ##3\times 10^8\ ##, I expect something like ##5.5\times 10^7## !

Always estimate before grabbing a calculator.​

And if you have 1 digit accuracy data (5 mW), there is no point in presenting 9 digits results (not even intermediate).

Finally: 1 second of beam is ##3\times 10^8\ ## meters long. That's the distance over which the ##\ 1.63\times 10^{16} \ ## photons are spread out...
 
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  • #6
kevincb672 said:
With a length to detector 2m is it correct to say the beam has 1.63398693×10(^16) photons? Apologies for my own ignorance.
Show your work. I don't know how you came to this solution.
 
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  • #7
Drakkith said:
Show your work. I don't know how you came to this solution.
1644504880401.png
my wrong workings out but corrected now, your logic of "Imagine you shoot the beam into space for one second. How long is the beam in meters and how many photons are in it?" went straight over my head but I get it now!
BvU said:
It's your own ##N = P\lambda/(hc^2)##

When I see ##\ 1.63\times 10^{16} ## divided by ##3\times 10^8\ ##, I expect something like ##5.5\times 10^7## !

Always estimate before grabbing a calculator.​

And if you have 1 digit accuracy data (5 mW), there is no point in presenting 9 digits results (not even intermediate).

Finally: 1 second of beam is ##3\times 10^8\ ## meters long. That's the distance over which the ##\ 1.63\times 10^{16} \ ## photons are spread out...
Ah okay, I see! That helps me out. I can't understand why I didn't put two and two together realize that 1 second of the beam travels 3x10^8m and just put this under the number of photons. Always trying to improve my maths and common sense :oldbiggrin:thanks sir :bow:
 
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  • #8
I still see a 5.45 1012 ?

How much is 650 nm in meters ?
Remember that nano means nine ...

##\ ##
 
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  • #9
BvU said:
I still see a 5.45 1012 ?

How much is 650 nm in meters ?
Remember that nano means nine ...

##\ ##
:headbang:Correcting that silly error gets me the right answer as you had. I will refer myself back to primary school.
 
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Related to Calculating attenuation necessary to reduce number of photons

1. How do you calculate the attenuation necessary to reduce the number of photons?

The formula for calculating attenuation is: I = I0 * e-μx, where I is the intensity of the photons after attenuation, I0 is the initial intensity, μ is the linear attenuation coefficient, and x is the distance the photons travel through the material.

2. What is the linear attenuation coefficient?

The linear attenuation coefficient is a measure of how easily a material can absorb photons. It is dependent on the type of material and the energy of the photons.

3. How does the distance traveled affect attenuation?

The distance traveled through a material directly affects the amount of attenuation. The longer the distance, the more photons will be absorbed and the lower the intensity will be.

4. What factors can impact the number of photons that need to be attenuated?

The number of photons that need to be attenuated can be impacted by the initial intensity of the photons, the type of material they are traveling through, and the distance they need to travel.

5. How can I determine the necessary attenuation for a specific experiment?

To determine the necessary attenuation for a specific experiment, you will need to know the initial intensity of the photons, the linear attenuation coefficient of the material, and the distance the photons need to travel. Then, you can use the formula I = I0 * e-μx to calculate the required attenuation.

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