Calculating the flux from a flashlight

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Discussion Overview

The discussion revolves around the calculation of flux from a flashlight using the flux-luminosity relation and other methods. Participants explore whether this relation can be applied to non-spherical sources like flashlights, and how to accurately measure the flux at a distance.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the applicability of the flux-luminosity relation for a flashlight, suggesting it may not be valid for non-spherical sources.
  • Others propose using the power output of the flashlight and the solid angle of the reflector to calculate irradiance at a distance.
  • One participant emphasizes the importance of distinguishing between flux and flux density, noting that the formula may only be valid for point sources at sufficient distances.
  • There are discussions about the complexities involved in calculating flux, including factors like beam shaping, transmission losses, and detector characteristics.
  • Some participants mention the need for specific efficiency factors based on the type of light source, such as incandescent versus LED flashlights.
  • A few participants express uncertainty about finding reliable resources or links that address the complexities of distributed sources.

Areas of Agreement / Disagreement

Participants generally do not reach a consensus on the applicability of the flux-luminosity relation for flashlights, with multiple competing views and uncertainties remaining regarding the best approach to calculate flux.

Contextual Notes

Limitations include the dependence on the definitions of terms like flux and flux density, the complexity of the problem involving non-point sources, and the variability in efficiency based on different flashlight types.

Phys12
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Can you use the flux-luminosity relation, f=L/4*pi*d^2 to calculate the flux of a flashlight at a distance d? As I understand, we can use the relationship given above for a star since we assume that it emits radiation uniformly in all directions, we take the total energy emitted and then divide it by the area in which we receive that energy. But for something that's not spherical, can we still use this relationship? My guess is no. And if no, how do you measure the flux from a flashlight at a distance?
 
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If you know the power ## P ## that the flashlight puts out, and can estimate the full cone angle of the reflector ## \Delta \theta ##, there is a solid angle associated with that cone angle ## \Delta \Omega \approx \frac{\pi}{4}( \Delta \theta)^2 ## . The irradiance ## E ## (watts/m^2) at a distance ## s ## is then ## E=\frac{P}{\Delta \Omega \, s^2} ##. ## \\ ## (The solid angle is given in steradians, but just like the radian, it is dimensionless).
 
Phys12 said:
Can you use the flux-luminosity relation, f=L/4*pi*d^2 to calculate the flux of a flashlight at a distance d?
Not for a distributed source like a flashlight with a wide reflector the Inverse Square Law only applies to point sources (not even spherical ones). The formula would work 'near enough' once the distance is great enough to consider the source to be a point.
This subject is full of terms which should be used correctly. I think you may be confusing flux with flux density. The radiant flux the total amount of power radiated by an object. You are probably wanting the flux density. This Hyperphysics link could possibly help to get the right terms for th quantities involved (find and follow links from that page).
 
Phys12 said:
...we take the total energy emitted and then divide it by the area in which we receive that energy.

That gives you the average flux density. Should work for something like an LED flashlight that doesn't need a reflector.
 
CWatters said:
an LED flashlight that doesn't need a reflector.
Mine certainly has a reflector and it's focusable too.
 
Charles Link said:
If you know the power ## P ## that the flashlight puts out, and can estimate the full cone angle of the reflector ## \Delta \theta ##, there is a solid angle associated with that cone angle ## \Delta \Omega \approx \frac{\pi}{4}( \Delta \theta)^2 ## . The irradiance ## E ## (watts/m^2) at a distance ## s ## is then ## E=\frac{P}{\Delta \Omega \, s^2} ##. ## \\ ## (The solid angle is given in steradians, but just like the radian, it is dimensionless).

As pointed out, this is not a trivial computation. Overall, this is a 'radiometric' type of calculation (irradiance, radiance, flux, etc)- how is the radiant energy emitted and shaped by the projection lens? How is the radiant energy transmitted through the air, reflected off a target, re-transmitted through the air, and finally detected by a finite-sized detector?

You first have the source- an extended object that emits an angular-dependent radiance; an isotropic radiator is one limiting case, a Lambertian radiator is another. In general, the emitted radiance L depends on both angles L(θ,φ), but often the system is assumed to be axisymmetric so that L = L(θ). Often, this can be approximated to a more simple form, e.g. L(θ) ∝cos2(θ).

Then you have the beam-shaping elements; this will also include the rear reflector (if present). The efficiency of the projector system doesn't mean the fraction of light transmitted through, but the cone angle within which the emitted light is transmitted.

Then, for the full problem, you must account for the transmission losses through air (weather permitting) and the reflectivity of the illuminated object (which depends on relative angle). The reflected light is the re-transmitted though the lossy air and detected, so you finally need to account for the detector characteristics.

As you may have guessed, a lot of this was worked out for lighthouses and searchlights, many years ago. There's a great little book called "The range of electric searchlight projectors", I found a free PDF a while ago and can't quickly find the URL.

Good luck!
 
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@Phys12 You never mentioned the application for your flashlight or the accuracy you need. Just dividing the DC input power by the area of your projected spot and adding an arbitrary efficiency function (say 80 lumens per Watt) may be enough for you. The figure in practice can vary a bit with battery charge level.
 
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  • #10
sophiecentaur said:
@Phys12 You never mentioned the application for your flashlight or the accuracy you need. Just dividing the DC input power by the area of your projected spot and adding an arbitrary efficiency function (say 80 lumens per Watt) may be enough for you. The figure in practice can vary a bit with battery charge level.
This is not for a project. A person in my astronomy class posted an example problem and they suggested using the flux-luminosity relationship for a flashlight, which I didn't think was correct, so I wanted to know if that really was the case or not.
 
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I'm sure that @Andy Resnick could lay his hand on some intermediate level link that could do better than the point source assumption. All I can find is arm waving stuff about distributed sources.
 
  • #12
sophiecentaur said:
I'm sure that @Andy Resnick could lay his hand on some intermediate level link that could do better than the point source assumption. All I can find is arm waving stuff about distributed sources.

Unfortunately, I don't. Not sure if the reason is technical (the problem is too complex) or motivational (I haven't done much looking around). This kind of problem is typically lumped in with 'non-imaging optics' systems design, it's a huge field but mostly outside my comfort zone.
 
  • #13
The calculations in post 2, if you measure the full cone angle of the beam, will get you reasonably accurate numbers. ## \\ ## The inverse square law works quite well once the distance across the beam is more than about 10x the diameter of the reflector. i.e. when ## s \, \Delta \theta > 10 \, D ##, where ## D ## is the diameter of the source/reflector, and ## \Delta \theta ## is the full cone angle of the beam. ## \\ ## (Note: In post 2, I called it the "full cone angle of the reflector", but the "full cone angle of the beam" is a better designation for it). ## \\ ## For an incadescent source=a filament bulb, if you know the approximate wattage, you can assign an efficiency factor of about 10% of that wattage as being output power in the visible region of the spectrum. For an LED, this efficiency factor is likely to be 90% or higher.
 
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