# How do those cheap shoebox-phone projectors throw any “decent” image?

You've probably seen them; shoebox phone projectors.

They throw a very dim image, but I'm surprised it throws any image at all, since the lightsource is completely diffuse.

Why does it still somewhat work?

My theories are:

• The lens is mounted far from the screen, essentially eliminating most non-parallel rays.
• The phone has a screen that is most bright perpendicular to the screen.
• The lens automagically corrects the image? (Does it?)

If I hold one of these large LED lenses in front of my phone;

it throws an image too, so I'm questioning point 1.

That leaves point 2 and 3. If I watch my phone from the side I can still clearly make out the image on screen, so I cannot believe this is such a big factor.

That leaves point 3; every projector out there denounces projection using diffuse sources. I've seen only one instance of a DIY projector using a tight packed LED matrix as light source.

Can anyone explain why these shoebox projectors (somewhat) work?

• Your eye is a lens. How do you see anything if virtually every light source is diffuse? – probably_someone Nov 16 '17 at 23:27
• Cameras work: these are cameras run backwards. – tfb Nov 16 '17 at 23:28
• Ah, I see. So basically diffuse light is never used in projectors solely because of lack of efficiency? – RobotRock Nov 16 '17 at 23:31
• @RobotRock Diffuse light is used in enlargers for film, for instance. – tfb Nov 16 '17 at 23:47
• @RobotRock Yes, enlargers with condensers (which use directed light) do have higher contrast that diffusion enlargers. I am not sure when the change happened: I think that anything (relatively) modern is diffusion, and in particular variable-contrast ('multigrade') papers are all designed for use with diffusion heads. I don't think you should delete the question: it's a reasonable one I think. I am not sure if I have a good enough answer other than 'this other thing works as well so it must work' though. – tfb Nov 17 '17 at 7:44

If it is diffuse... how can you see anything on the screen when you hold the phone? Okay, I don't know how your phone is, but if it is similar to mine, it shows a nice image. It has not really good definition, but I can clearly read whatsapps and watch videos.

It seems you're confused about the term "diffuse". A point light source (if existed) would also emit in all directions, but you would be able to create an image of it. The fact that your phone emits in all directions does not mean it is diffuse. In fact, emmiters use to show a pattern which approximately follos Malus Law.

Diffuse would mean that every point emits in a totally different direction, so your eye couldn't tell if there is something behind that light spot. Your eye cannot extrapolate where the image comes from.

However, altough the phone is a large object, rays leave it with certain angle. For the lens, however, it is LIKE IF the rays came from much farther from an infinitesimal emmiter. The image, consequently, sould be nice enough (besides optical aberations).

The fact that you see it dimmer is because of energy loss through the lens and before it.

• There is literally a diffuser behind the phone screen. – RobotRock Nov 17 '17 at 0:22
• If it's behind, it doesn't affect. I don't see any diffuser in the middle, what do you mean? – FGSUZ Nov 17 '17 at 12:17

If the entire screen of a phone is set to white, the screen 'appears' approximately as a diffuse light source. But each individual pixel acts as a point source - i.e. if one pixel is set to white and the rest of the screen is black, then the lit pixel acts approximately like a point source - radiating light in all directions. So you could replace that pixel with a tiny candle and the emitted light would look/behave approximately the same. Now if you have a thin lens with focal length $$f$$ and a candle at a distance $$p$$ to one side of the lens, then the lens will focus the candle's light onto a point at distance $$q$$ to the other side (where $$1/p+1/q=1/f$$). If the candle is at height $$h$$ above the centre axis of the lens then the image will be at height $$-H$$, where $$H = (q/p)*h$$. ($$q/p = m$$, where $$m$$ is the magnification of the system). See the figure below. This is all 'standard' thin lens optics that you would likely be introduced to in an elementary physics course.

If you now add a second candle at the same distance behind the lens but a different height $$h'$$, then its light would be focused the same distance away, but also at at a different height $$H'$$. So if you replace the candles by the pixels of a phone (all at the same distance $$p$$ but different heights (and different sideways displacements too) then each pixel will have it's light focused at distance $$q$$ on the other side, and with its height and displacement magnified by the ratio $$m=p/q$$.