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) ∝cos
2(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!