I have a 4.8" LCD. Dimensions : 11cm x 7.5cm

I thought that maybe i can use a reflector with the almost same sizes of the lcd.

Here is what i meant :



Dimensions = 13cm * 7.5cm

Actually this is a "Fog Light" using in automobiles but i will use only the reflector part of course.

Fresnel lens FL = 12cm



What do you think?

(trying very hard not to sigh.)

The idea isn't new. it's been explored before, and really, doesn't work.

There are several reasons why, here's a few.

Let's start with the lamp itself. Someone quoted me in their signature as saying this, but it STILL amazes me how many people will figure that a 55W halogen lamp will match a 400W metal halide lamp. Even if you go to an HID Xenon lamp, you're still dealing with a fraction of the lumens. A small fraction at that. (There's a work-site on the freeway where I live, and the work crew has light fixtures with 2 400W MH lamps in it. I can tell you that there are NO 2 headlights which are going to match those for illumination factor.) But let's ignore that physical piece, and scale it up...

Okay, let's say that you do the same thing with a 400W MH lamp. What you're looking for then is a parabolic reflector. This is similar to a satellite dish, where it takes light from a source and collimates it. This works great for signals, not so great for illumination. Why? Inverse square law strikes again. Every point on your parabola will be a specified distance frmo the reflector. Only light from the reflector is collimated, and therefore only light from the reflector is useful. Take a look at any parabola, and the locus (The point where you can put a light or emitting source and get a collimated reflection) will be much closer to the center than it will to anything else. Dramatically so.



(Borrowed from wikipedia)


You can see here that a light source placed at F will reflect collimated light towards line L. The problem comes in that the strength of the reflected light is directly proportional to the strength of the light that hits points P1, P2, and P3. The problem comes that the distance from F-P1 is much less than from F-P3. In this case it will be nearly double. This means that the light intensity at Q1 will be over 4 times that of the intensity at Q3. Light directly radiated from the lamp would not be captured into our projection lens, and therefore is only a heat source, and not useful light at all.

There's also the matter that in this case, we would be solely relying on the efficiency of reflected light. There are always losses in reflection, more so than refraction (What lenses do)

Well, you COULD increase the distance to point F, but this will reduce the amount of overall light, in the same manner that using a higher FL collimator fresnel does, but this will reduce the overall light intensity.

I crunched some numbers, assuming in both cases that F=220mm (fresnel lens, and parabola)

If you could get 100% efficient reflection, you would start out with a base of 67% of the light intensity at the center. This is becasue with the fresnel, we can use a reflector to add to the radiated light. with a parabola, we cannot add the radiated light to the reflected.

But, for the sake of fairness, (though unrealistic) I will presume that we can get 100% efficient reflection and with the fresnel, we do not use any reflector at all.

Light in the center becomes of equal lux, but light at the corners of a 15" LCD panel are different values. For a fresnel, the corners are about 38% further away, which means that we get just a little over 50% of the brightness value of the center.

For the parabola... It's a different matter. The corners would be about 1.7 times as far away, so the light there would be 1/3 the intensity of the light in the center.

The shorter the distance that you make F, the greater this discrepancy becomes. Add in the losses that we get by not being able to add ratiated and reflected light together, and it becomes huge.

On top of that, we have to add in the fact that we don't have a super small light source, and the lamp itself will cast a shadow. (I originally missed this until Dazzla actualy did the experiments, and couldn't get rid of the lamp shadow.)

Basically, it's a dead end.

Thank you for detailed reply.

Dude, of course I'm not going to use halogen lamp in this reflector I know that. And because my lcd is so small(4.8") , I think 150w would be enough for me (almost 11000 lumen. thus nearly 700-900 lumen on the 60"-80" projection screen would be enough right?)

I know the square law but my lcd is not 15 inch. So i thought because of my lcd is so small and collimator fresnel's FL is so short(120mm) - thus the distances are so close - then square law should not matter so much.

And by the way i didn't understand why "Light directly radiated from the lamp would not be captured into our projection lens"
MH will be in the focus point of collimator fresnel lens. Therefore rays will be going parallel. Why its useless?

One probelm with your math.

With a projector we get only about 1/2 of 1% of the lamps lumens onto the projection screen. This is because the average Lcd only passes 5% of the light. Then you have each piece of glass/fresnels that absorb 10% each. Then the fact we only can use the direct rays from the lamp, which is just a small percent of the lamps total lumens. Etc, etc....

So with a 11,000 lumen lamp the most you will get on the screen is 55 lumens.

WOW
after reading these debates, how anyone would have questions is beyond me!
Bravo on the details!

This is becasue there is no such thing as a lens which selectively bends light.

If you have a single lens which will take collimated light and collect it into the objective lens, then only colimated light will be collected. In this case, with the parabolic reflector, that is light which is reflected. Light which leaves the lamp to the lens will have the wrong angle of incidence, and therefore will miss the objective lens. This light is useless to us, and in fact actively does harm in that it still heats up the LCD, and increases stray light on the front side of the projector, which can cause th image to get washed out.

If you have the normal double lens setup, this captures light directly from the lamp, but it will fail to capture the collimated light from the reflector.

A crude paint drawing:

Click to view attachment

The green lines represent light straight from the lamp, the blue lines are reflected light.

As you can see, if the green lines are collimated, the blue lines are not. Therefore the blue lines get bent again by the collector lens, and end up missing hte obhective lens. If instead you do not use a collimator fresnel, and the lens is the collector fresnel, it will gather the light from the blue lines to the objective lens, but the green lines will miss instead. You can't have it both ways, becase there no lenses exist which can handle both circumstances. Such a thing cannot exist.

If you're going to use a parabolic reflector, the only way that it makes sense is to use it instead of a collimator lens, and that has more problems than the regular collimator lens.

Take inverse square law. Back to the earlier incarnation of that diagram.

in pixels, the line segment:
F - P1 is 83 pixels
F - P2 is 105 pixels (11% more)
F - P3 is 140 pixels (41% more -- 34% as bright at Q3 as at Q1)
In comparison

F - Q1 is 146 pixels
F - Q2 is 164 pixels (11% more)
F - Q3 is 190 pixels (23% more -- 60% as bright at Q3 as at Q1)

From this we can safely conclude that the rate of vignetting with a parabolic reflector is much greater. Also, since we cannot use direct light, only reflected light, even with a 100% efficient reflector, we only stand to gain a small amount of brightness in the center, considering that we have a documented solution which increases the "direct" light output by 50% with fewer issues, it starts to become a losing proposition pretty quickly.