Optics: Can a magnified real image be larger than the lens diameter?

1. Aug 6, 2013

arnoke

Hi all,

This is probably a silly questions, but I want to be sure :).

I'm wondering if a real image, created by a convex lens, can appear larger than the lens diameter itself.

As an example, I'm thinking about the following:

- I have a an object of height $h_0=7.76"$ (display size of my IPAD :p)
- Two convex fresnel lenses, each width diameter of $8"$ and focal length $12.5"$
- The lenses will be put back-to-back, so the combined focal length is $f=6.25"$
- I will place the IPAD at distance $d_0=1.2f=7.5"$ from the lens

Now I can calculate at which distance the real image will appear to float in front of the lens:
$\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_1}$
$\Rightarrow d_i = \frac{1}{1/f - 1/d_0} = \frac{1}{(1/6.25 - 1/7.5} = 37.5"$

Also I can calculate the magnification:
$\frac{h_i}{h_0}=-\frac{d_i}{d_0}$
$\Rightarrow h_i = -h_0\frac{d_i}{d_0} = -7.76\frac{37.5}{12.5} = -23.28"$

So the real object would be at $37.5"$ from the lens, inverted, and would be $23.28"$ high.

Now my question is; would I indeed see an IPAD of height $23.28"$ if my (fresnel) lens has the same size as the original Ipad? Or should my lens be at least of diameter $23.28"$, the same size as the magnified, real image?

Thanks a lot for your input!

2. Aug 6, 2013

voko

The diameter of the lens has nothing to do with magnification. It has everything to do with how much light the lens collects and how bright (or dim) the image will be. In your case, with the lens bigger than the subject and the subject very close to it, having a bigger lens would be just about useless.

3. Aug 6, 2013

arnoke

Just to be sure; that would mean that I would see a 'floating' (real) image that is larger than the lens?

4. Aug 6, 2013

voko

You can' see a real imagine directly; it must be projected upon some screen. But you can see a virtual image.

5. Aug 6, 2013

arnoke

I remember playing around with fresnel lenses a while ago. When I placed an object (e.g. my Ipad) at distance 2f behind the lens, I could see the inverted image, floating 'in thin air', in front of the lens.
Isn't that a real image then?

Also, theoretically, if I would have placed the object at a distance somewhere between f and 2f, it would be magnified. So my original question was then if I would actually see the complete enlarged object if my lens is only as small as the un-magnified original Ipad.

6. Aug 6, 2013

voko

I think what happened was that your eye and the lens combined into a single optical system, so the real image was formed on your retina.

Typically, TV screen magnifiers (iPad magnifiers by extension) are placed close to the lens, about one focal length away, and the eye is further away. Then you can see a magnified virtual image comfortably.

Are you trying to build something practical, or is this just a theoretical exercise?

7. Aug 6, 2013

arnoke

Currently it is just a theoretical exercise, but in the end I'm thinking about building something. Just want to order the right lenses, as they are not so cheap (at least at Edmund Optics).

I once build the image floating device that I described earlier. An example of this, but using a mirror and a single fresnel lens, in stead of two fresnel lenses, is described by this patent (obviously not mine :p): http://patentimages.storage.googleapis.com/US6375326B2/US06375326-20020423-D00000.png

At the bottom you can see an LCD display. The fresnel lens creates a real image, resulting in a hologram like object, floating in front of the box.
Here, the LCD display is placed at 2f, such that there is no magnification.

Now, if I would use an Ipad in stead of a 24" LCD display, the setup would be less heavy, and I would have a retina display. I would place the Ipad closer to the lens (somewhere between f and 2f), to obtain a magnified image, such that the floating image is still a 24" image.

That's the reason I was wondering if my lens would have to be 24", or just the size of my small Ipad.

8. Aug 6, 2013

Staff: Mentor

The same thing happens when the lens produces a virtual image, as with a simple magnifying glass. Any image on the retina has to be real.

When we talk about "seeing an image", we usually refer to the light before it enters the eye, not the light inside the eye. In that sense, we can definitely see the real image formed by a lens, if we stand far enough beyond the lens so that the image is in front of us as we look towards the lens.

9. Aug 6, 2013

voko

Hmm. I have no first hand experience with "image floaters" and the idea that one can just look at a real image without a screen sounds unusual, to out that mildly.

That said, again, the size of the lens does not matter for image formation, it only affects its brightness.

10. Aug 6, 2013

arnoke

You can get the same effect with a concave mirror in stead of a convex lens.
The old Sega "Time Traveler" game is an example of this: http://en.wikipedia.org/wiki/Time_Traveler_(video_game)

It creates a real image of a CRT monitor by reflection in a concave mirror. The result is something that looks like a 'hologram'; floating little guys, on top of a stage.

I just want to use a convex (fresnel) lens because they are cheaper and smaller :)

11. Aug 6, 2013

Staff: Mentor

Our physics department has one of these gadgets that uses two concave mirrors to produce a "floating" image. It comes with a small plastic pig to use as the object, so it's often called the "floating pig" or "air pig" illusion. Don't be misled by the word "hologram" on the page. It's just a plain old real image produced by mirrors, not a hologram.

http://www.optigone.com/m2000.htm [Broken]

It would make a nice exercise to use the dimensions given on that page to calculate the locations of the intermediate image from the first reflection, and the final image.

The size of the lens does affect the ease of viewing a real image directly, because the visible part of the image is restricted to the lens diameter (as seen from your viewing location). It can't extend beyond the edge of the lens.

Last edited by a moderator: May 6, 2017
12. Aug 6, 2013

voko

I do not understand how that could happen. When projecting, a real image can be significantly larger than the lens. I understand that in the case of image floating, the real image would be formed exactly where a screen would be, except there is no screen and the viewer is on the other side of the non-existent screen.

13. Aug 6, 2013

Staff: Mentor

Imagine using a translucent screen such as a thin sheet of paper, so you can see the image on it from the rear as well as from the front. In this case, looking at it from the rear, you see the entire image. When you remove the screen, while keeping your eye at the same location, you see only the portion of the image that is directly in front of the lens.

With the screen in place, the screen itself scatters some of the light from the outer portions of the image towards your eye, so you can see it. Without the screen, there's nothing to make the light from those points reach your eye.

Here's another way to think of it: the light that arrives at the image has to come from the lens. If there's no screen, each light ray from the lens has to continue along a straight line through the image, without deflection. In order for you to "see" the light represented by the ray, it has to enter your eye.

14. Aug 6, 2013

voko

OK, this is quite clear, thanks for the explanation.

What is not clear, however, is why the size of the lens would affect that. The rays that I cannot see are still there, I cannot see them because of the geometry, not because the lens is not big enough.

15. Aug 6, 2013

arnoke

Interesting.
For some reason I find it more difficult to understand this for light coming from an LCD screen, then for light rays coming from, for instance, a projector/beamer. A projector casts a diverging beam of light, that is only diffused/scattered when it hits the projection screen. An LCD on the other hand contains a diffusion material itself, so the light coming from the LCD is not a coherent beam of light; In stead, the light rays just go everywhere, which is why you can still see the display from an angle.

16. Aug 7, 2013

Philip Wood

Arnoke: You are likely to get confused if you use 'diverging' like this. In everyday speech it's perfectly reasonable to describe the beam going from the projector lens to the screen as 'diverging'. But in optics we use the term only for rays which originate from a single point on the object. If that point is, say, a speck of dust on a slide in a projector, then the projector lens will collect light diverging from the speck, and converge it to a single point on the screen. Thus, in the sense used in optics, the projector lens produces convergent light. It produces an infinite number of 'pencils' of convergent light, each originating from a different point, P, on the slide, and converging - through the action of the lens - to a different point, Q, on the screen.

[If you draw a straight line from P through the centre of the lens and continue it, it will go through Q - rather as if the lens were replaced by a pinhole at its centre. The cleverness of the lens is that it takes rays diverging from P and converges them all - even those which would miss the pinhole - to Q.]

Last edited: Aug 7, 2013