I want to make a handmade LED-flash for my project. I need to know an angle size of a beam from typical flashlight. Let's say from Samsung WB50F for example. It's a Xenon one, but what is the angle of its beam? I can't find online an answer.
-
By "flashlight", it seems you mean the light from a built-in flash on a compact camera? You can assume that it won't be much wider than the wide angle end of that camera's lens, possibly even a bit narrower, with fall-off at the edges at the widest end to provide better "reach" for zoomed shots.– Lowell MontgomeryApr 7, 2018 at 8:13
-
@LowellMontgomery this already sounds like an answer. ;-)– floliloApr 7, 2018 at 11:39
1 Answer
An on-camera flash has to cover at least the area seen by the camera objective. The more expensive ones can adjust the covered angle (either automatically or manually), and the angle is expresses in mm focal length, on a full-frame camera.
The example you give (Samsung WB50F) is a camera, with a 24-300 mm zoom (FF equivalent). Looking up 24mm lenses tells us they cover an angle of 84° (over the image diagonal). That means that for on-camera use, your flash has to cover at least that angle if you don't want dark corners (vignetting).
If you plan to use the flash off-camera, you'll have to take into account the distance between flash and object: if you move the flash closer, you'll need a wider beam, farther away, and you can use a narrower beam.
But keep in mind that a wide beam will be less efficient than a narrow beam: it will deliver less light per m² of your subject area, so the narrowest beam you can use is the most efficient (which is why some flashes allow adjustment of the beam).
-
-
1If the angle is not automatically zoomed, with such a camera (and it doesn't really look like a flash zoom is part of that), I suspect that the full 84º of the widest angle is not quite covered by flash (i.e. some compromise at the edges of 24mm to have better reach for the "300mm" end) Apr 7, 2018 at 8:31
-
@LowellMontgomery: probably, and the beam probably isn't circular either. But for the purpose of the question, it's probably precise enough (there's also the question of beam homogeneity).– remcoApr 7, 2018 at 8:52