Comment Re:Journal Article (Score 1) 120
Oh, yeah, with the scanning beam it'd work fine. The image would still have to be captured from the perspective of the driver somehow and it wouldn't be relevant to anyone but the driver, but it would work. Since it didn't mention that I assumed the film itself was supposed to be a standalone night vision apparatus.
As far as the second paragraph, what you're talking about definitely works, but only when the lights are up close. In this case, the light right on top of an element will illuminate that element a whole lot more than the other lights because it's many times closer and the incident light is less tangential. When the lights are far away, say a couple hundred yards, the difference in angle of incident light from each light on an OLED element will be a tiny, tiny fraction of a degree; essentially zero. The difference in distance is also essentially zero. So all four lights are illuminating the element approximately equally. (I get what you mean with weaker flux density, but just saying less flux would be more accurate :) Light intensity can be described as a 2D density (e.g. W/m^2), and flux is multiplying that by an area to get a power.)
This goes back again to the frosted glass analogue. With a scanning visible light beam (akin to the scanning IR beam you suggested), you can project an image on it. Normally though, you can't see through it because of scattering. When lights are up close, you can differentiate them, but as they get farther away, they're harder to make out.
As far as the second paragraph, what you're talking about definitely works, but only when the lights are up close. In this case, the light right on top of an element will illuminate that element a whole lot more than the other lights because it's many times closer and the incident light is less tangential. When the lights are far away, say a couple hundred yards, the difference in angle of incident light from each light on an OLED element will be a tiny, tiny fraction of a degree; essentially zero. The difference in distance is also essentially zero. So all four lights are illuminating the element approximately equally. (I get what you mean with weaker flux density, but just saying less flux would be more accurate
This goes back again to the frosted glass analogue. With a scanning visible light beam (akin to the scanning IR beam you suggested), you can project an image on it. Normally though, you can't see through it because of scattering. When lights are up close, you can differentiate them, but as they get farther away, they're harder to make out.