How does one calculate the number of photons striking a surface per second?

  • #1
Danielk010
9
1
Homework Statement
I am given the wavelength, number of waves, the wattage, and efficiency of the light bulb. I am also given the width and length of square surface, and the distance between the bulb and said surface.
For this part of the question, it says
"Assuming the emitted photons to be distributed uniformly in space, how many photons per
second strike a x mm by y mm paper held facing the bulb at a distance of delta x m?"
From the previous part of the problem, I calculated the photons emitted per hour and per second.
Relevant Equations
# of photons per second = wavelength * E / hc
I am more so stuck on where to start with this problem. I know dividing the photons per second by the area gets me the photon per area, but I am not sure how the distance is related to this part of the problem. If anyone can help, thank you.
 
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  • #2
What if the piece of paper were held 100 million kilometres from the light bulb?
 
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  • #3
PeroK said:
What if the piece of paper were held 100 million kilometres from the light bulb?
There would be very few photons that hit that paper depending on placement of paper?
 
  • #4
Danielk010 said:
There would be very few photons that hit that paper depending on placement of paper?
Exactly.
 
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  • #5
PeroK said:
Exactly.
Ooohhh. Thank you. Am I on the right track with photons per area? If so, how would I correlate the distance with the values I have mathematically? Would it just be photons per area * (displacement)^2 since the units would cancel out?
 
  • #6
You are on the right track with (photons/area)/second. Imagine a sphere of radius R surrounding the source. All the energy emitted by the source per second has to go through the surface of that sphere. What portion of the energy goes through a rectangular area (x) × (y)?
 
  • #7
kuruman said:
You are on the right track with (photons/area)/second. Imagine a sphere of radius R surrounding the source. All the energy emitted by the source per second has to go through the surface of that sphere. What portion of the energy goes through a rectangular area (x) × (y)?
If the sphere covers the source, then no photons would be passed through, depending on the type of material the sphere is made out of. That means that no energy from the photons goes through the rectangle area, right? I am still confused on the problem.
 
  • #8
You missed my point. Put the x mm by y mm paper on the inside surface of the sphere. How many photons per second strike its surface?
 
  • #9
kuruman said:
You missed my point. Put the x mm by y mm paper on the inside surface of the sphere. How many photons per second strike its surface?
(photons per second / x * y) 4πr^2?
 
  • #10
Danielk010 said:
(photons per second / x * y) 4πr^2?
Try some limiting cases. For example, if the paper gets really small, say, ##x \to 0##, does your answer give the result you'd expect? What if ##r \to \infty##? You said earlier that if the paper is far from the light source, only a few photons would strike the paper. Again, is that the result your formula predicts?
 
  • #11
vela said:
Try some limiting cases. For example, if the paper gets really small, say, ##x \to 0##, does your answer give the result you'd expect? What if ##r \to \infty##? You said earlier that if the paper is far from the light source, only a few photons would strike the paper. Again, is that the result your formula predicts?
oh ok thank you. Would switching the x * y with the 4πr2 be accurate since as x * y grows the number of photons striking the surface would increase and as r decreases the number of photons striking the surface would decrease? Thank you for help.
 
  • #12
Danielk010 said:
oh ok thank you. Would switching the x * y with the 4πr2 be accurate since as x * y grows the number of photons striking the surface would increase and as r decreases the number of photons striking the surface would decrease? Thank you for help.
This principle is the basis of so much in physics that I'm surprised you haven't previously encountered it.

For example, you could calculate how much of the Sun's energy reaches the Earth's surface, based on the distance and cross sectional area of the Earth.
 
Last edited:
  • #13
PeroK said:
This principle is the basis of so much in physics that I'm surprised you haven't previously encountered it.

For example, you could calculate how much of the Sun's energy reaches the Earth's surface, based on the distance and cross sectional area of the Earth.
The due date for the assignment has passed. I looked up the example you mentioned and I got the concept. Thanks for the help
 

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