Quote: best measured with the sensor as close to the eye lens as possible, before it has a chance to spread out too big
Quote:Can this information be used to calculate transmission? Transmission Loss?Use An example from the data list12x50 binocular actual aperture 49mmactual magnification is 12.2xactual exit pupil is 4mmLux In 200Lux Out 700Lux = Lumens per square meter.Anybody?edz
MarkLeica 8x20; Nikon Action 7x35; Vixen Apex Pro 8x42; Orion 15x63; Docter Nobilem 15x60WO Megrez II 80 FD / APM 107mm f/6.5 / Mewlon 210 on DM-6 + Berlebach Planet
Quote:I think the answer to both those questions is yes, but I'm open to comment. The sensor is about 10mm dia, so it's large enough.Well, the ratio of aperture area to exit pupil area is 156/1, but I haven't got a clue how that might relate to Lux In/Out = 200/700.edz
Quote: In fact Ed has not given the size of the detector, and it appears to read light/area, not total light or energy.