Images and Parallax

A teaching scenario by Donald E. Simanek.
A demo/lecture scenario is most effective when it develops ideas in a logical sequence, introducing new concepts building upon ones already grasped. This requires a good deal of thought and planning. It requires extreme care in choice of words and manner of expression, not only for precision, but to focus students' attention on what you want them to notice, observe, and think about.

[Watch this document for the addition of more diagrams and expanded discussion. Jan 10, 1997]

I find that students have a lot of trouble with virtual images and virtual objects, and part of the problem is conceptual. They use imprecise language in describing images and don't really grasp the relation between the real situation and the lens equation. This scenario attempts to give them more useful concepts. This isn't yet polished, and anyone who wants to try something like this is well advised to practice first, and adapt it to one's own teaching style. The script is a bit lengthy because I don't have diagrams that would save words. I find this can be completed in a 50 minute class period, but could be split into two parts.


Use a large diameter lens (4" or more, for bright image) of a focal length about a foot or more, on a 1 meter optical bench with the usual light box with + shaped mask. The + has an arrow pointing up, and a dot on the right. Use a frosted (ground-glass) screen (not an opaque screen). Have enough working room so that the bench itself may be moved to different positions on the lecture table. You may need to provide some protection against scratching the table with the legs of the heavy bench. In some parts of this demo you will have the bench parallel to the students' line of sight. Sometimes you will point it directly at them.

The 3d drawing above shows the setup from the viewpoint of a student in the the classroom. The object (O) is seen in the picture at the left rear. The lens (L) projects an image (I) onto a screen at the near right. This stereoscopic drawing has the image for the right eye in the middle. Those who use parallel (wall-eyed) viewing must use the left two drawings. Those who can do cross-eyed free viewing may use the right two drawings, or, they may use the enlarged inverted stereo pair shown below. Those who need instruction in 3d viewing may consult my 3d illusions document. The drawing is in perspective, so even if you can't view stereo, the arrangement of components should be apparent.

Place the screen at the far end of the bench. Cast a real image of the object onto the screen. Go through the usual moves to show that for a given object-screen distance you can get a large or a small image, depending on lens position. Show that a slight change of screen position in either direction makes the image become blurry and less distinct. When that happens, we say that what we see on the screen is 'out of focus'. Notice that I didn't say that the image is out of focus on the screen, for what we see on the screen isn't the image. The image is not on the screen when what we see on the screen is 'out of focus'.

Obtain the best focus. Ask students what sort of image we have found. [Again, I avoid saying 'image on the screen'.] They respond "Real." If they don't also say "inverted" remind them of that distinction. Position the screen so that they are looking directly at the side of the screen on which the light is incident. You are standing on one side of the bench, say to the students left. Ask whether the image is also inverted left-right. They may hesitate (which is good). Swing the object box so it faces them, and note that the dot is to their right, the arrow is up. Swing the box back, so a clear picture of the image is seen on the screen, and note that the arrow is up and the dot is to their right.

"Does this mean that the image is inverted up-down, but not right-left?" you ask with straight face. "The whole apparatus is symmetric about the optical axis, so why should up-down behave differently from right-left?"

After someone suggests that the question wasn't fair (good point!) because you shifted the reference frame for left-right when you rotated the object-box so they could see it, note that from the students' point of view the dot on the object-box is on their left, but the dot is on their right on the screen. So the image has been inverted when you keep the reference frame for left-right constant. Also note that, as you stand beside the bench, "The dot on the object box is on my side of the bench, but the dot on the screen is on the other side of the bench. The arrow is up on the object-box but down on the screen. Clearly both are inverted."

Swing the entire apparatus around so the students face the object box, and see the back side of the ground-glass screen. From this reference frame, they see the dot on the object-box on their right, but the dot is on their left on the screen.

You may wish to mention the question "Why does a mirror invert right/left but not up/down." But if this has not already been discussed in class, I think it would be distracting to discuss it here in the middle of this scenario. Besides, the mirror question has extra dimension to it. [The mirror image is not flat, but three-dimensional.]

You might point out that a real image seen without a screen is called an aerial (in the air) image. But the image is there, and of the same character, whatever we choose to name it!

When these points are made move to the next concept.


Position the bench perpendicular to the students' line of sight. This might be a good time to move the screen a significant way forward or back, so what they see on the screen is slightly 'out of focus' but still easily recognizable. Tell them you are moving the screen. "Now where is the image?" If they say "On the screen" they haven't yet gotten the point. Perhaps it is best not to resolve that question, but proceed to the next step, then return to this one.

Move the screen back to the position of best focus. "I've just found, by moving the screen for the clearest and most distinct picture on the screen, approximately where the image is." Pointing to the screen, you say "The real image is very near here." [Note: don't say "on the screen" simply point to the screen.] Remove the screen. "Now where is the image," you ask.

[We are trying to get across that what they see on the screen is not the image, it is a pattern of light scattered from the screen that looks very much like the image if the screen is close enough to the image position.]

Some may say "Nowhere," or "There is no image." Respond "Well, the image must be somewhere. Look at the lens equation 1/f = 1/p + 1/q; it has a solution, q, for any value of p. Can this equation be lying to us?" If you are a 'ham' pantomime looking around for the 'lost' image. "Maybe it fell under the table."

Some may say "The image is still where it was before." Respond "How do you know?" Someone says "Put the screen back and find it." Do it, but say, "Good answer--sort of, but not complete. That's just one way to find the image, an image which, by the way, is there whether we look at it or not, and whether a screen is there or not." Comment on the stupid semantic question "If a tree in the forest falls and no one is there, does it make a sound?" if you are philosophically inclined.

Many students think that the real image is the 'picture' they see cast on a screen. This exercise is intended to get them over that naive 'concrete' view, and to encourage more precise language.

Now I think it best to put the screen back in place and relocate the lens so that it forms a small image on the screen, no larger than a few cm. [You may need to change lenses, but stick with a converging lens so you don't complicate the scenario.] Note that the screen must be near the image. Take the screen away. "The image is still there, we now agree. Now let's find it another way."

Place a wooden pencil point-up on a lens-carriage just below the image. Move it forward and back along the bench till it is precisely at the image location. Move your head sideways until there's no parallax (no relative motion) between pencil and image. You can position the pencil point at a height so the image seems 'glued' to it. Explain in detail what you are doing as you do it.

Now have the students come up one by one (it's worth the time spent) to look at the image on the pencil-point as you did. Insist they move their heads left/right and up/down to experience the parallax. I never cease to be amazed how many of them sincerely think this is wonderful to see. Suggest that those waiting in line hold their left and right hand index fingers at different distances from their eye, close one eye, and experiment with parallax. Suggest they try to touch the image, which seems so 'real'.

Now position yourself to observe the image again. Move your head toward it. "I can observe this image from different distances, let me try to get close. The image is getting a bit blurry when I am closer than 25 cm to it." [Explain the term 'near-point'.] "Now, when I'm too close, the image is very unclear, I can barely make it out." Remove the pencil and move closer. "When the image is right on my eyeball I can't make out its shape at all." Move closer to the lens. "I can't see any image from here. Where is it? Is it still there?" "Yes, behind your eye" students respond. Turn your head around and look backward as if trying to find it. [Hamming it up again!]


Now use this arrangement to locate some virtual images! To introduce virtual images, I like to digress and explain the implications of the lens equation graphically.

The graph shows the image distance vs. the object distance for a converging lens. Much can be learned from this. It depicts the lens equation 1/p + 1/q = 1/f, using the standard sign convention of freshman texts. p is the object distance, q is the image distance, measured from the lens. The curve is an hyperbola. The hyperbola has two branches. One branch is entirely in the first quadrant, and corresponds to real objects and real images. The other branch falls in the second and fourth quadrants.

In the second quadrant, the object is real, the image is virtual. This happens when the object distance is less than f.

In the fourth quadrant, the object is virtual, but the image is real. This happens when the object distance is negative.

In the spirit of our data analysis admonition: "When you can transform the data to a linear relation, you should do so." Here we easily can. Plot 1/q vs. 1/p. The intercepts of this straight line will both be 1/f. This has transformed the hyperbola to a straight line, re-emphasizing that it was one curve, after all.

"The graph predicts that we should get an image with negative image distance q when p is less than f. Let's do that." Move the object box close to the lens, within its focal length. Move the screen, hunting for a real image. None can be found, even moving the screen beyond the bench and walking it all the way to the opposite wall of the classroom. Give up. "Looks as if we are looking in the wrong place."

Set the screen aside. Move the lens to the far end of the optical bench, where the screen was. Move the object box close to the lens, within the lens' focal length. Now look through the lens, back toward the object box. "I see an image. I know that what I see isn't merely the object box, for it appears much larger than that. But where is this image? I move my head to observe parallax. The image seems to be toward the other end of the bench. It's farther away from me than the object box."

Have a student place a finger near where you think the image is, then move the finger forward and back at your direction. "There, now you are pointing directly at it!" you announce.

In this process, emphasize that you are looking at the image through the lens, but looking at the student's finger directly (not through the lens). Ask the student to place a long rod in a carriage at the position you've found. The rod must stick up above the lens so you can see it. To save time, make the final adjustment to eliminate the parallax between image and rod by slightly adjusting the lens position. "My arm isn't long enough to reach the rod, you explain."

Now have students file by to see it themselves. Explain that at the location of this image, there's no light from the object box. So obviously we can't put a screen there and find it." Actually put a screen there to demonstrate that, if you think your students are a bit slow to get this point. "Our eye thinks there's a source of light at the image position because of the direction the rays enter our eye. Our brain triangulates this to calculate where those rays seem to come from That's what defines a virtual image. To see it one must look through the lens toward the light source.

You may wish to use overheads of appropriate ray diagrams here, and also at other points during this scenario.

This might be a good point to say that all images are defined by the rays emerging from the lens. If they are convergent, the image is real at the plane of convergence. If they are divergent, the image is at the apparent plane they diverge from.

I'll not follow this development further here. But you see how this sort of demo-scenario could help students realize that virtual images are just as important as real images, and really not all that mysterious.


"Is there any other way we could find an image?" you ask.

Return to the real aerial image case you began with, but perhaps use a longer bench, or a shorter focal length lens, so that there's working room.

Put up the overhead of this case. "Note that the rays we looked at when we could see the real image were divergent from the image after passing through it. Our eye is a lens system (cornea and lens) and a screen (retina). Let's substitute a lens and screen for the eye."

Put a lens and screen so that the image from the lens on the bench (Lens 1) becomes the object for this new lens (Lens 2) and forms a real image on the screen.

When this is achieved, I like to take a 50 cm long rod and clamp it to the lens holder of lens 2 (near the carriage) and likewise to the screen holder, to hold both in fixed relative position. The rod can be long enough so that a pointer can be placed at the position of the image formed by lens 1 (which is the object position for lens 2). "Now whenever there's anything giving off light at the pointer position, it will be sharply imaged on this screen. Even if it's an image at the pointer position, like it is right now, we'll see another image of it on the screen. We have made an 'image-finder' (lens 2, screen, rod and pointer)."

Demonstrate this for various positions of the real image, and confirm each time that the image is at the pointer by placing a ground glass screen there.


Now you may wish to repeat this scenario with a diverging lens. Note that when the object is real, the image is always virtual. "Can a diverging lens form a real image?" You ask.

"Let's look at the equation." Graph the equation, as you did before. "It is an hyperbola. Notice that it has two branches." [Don't outrage the mathematicians by calling it 'two hyperbolae' for it is mathematically considered a single curve, with two branches.] "Notice from the graph that we can get positive q (real image) only when we have negative p. What does negative p mean? It means the object is on the other side of the lens from the cases we've been doing so far. How can we get the image on the other side of the lens? Remember how we defined image, by the convergence or divergence of rays coming from the lens? We define object the same way, by the convergence or divergence of rays entering the lens."

"In the previous cases the objects were real, emitting divergent rays that entered the lens. To get a negative p, we have to have rays convergent into the lens. How can we do that?" Students say, "Use another lens to put a beam of convergent rays into the lens." You ask "How will we know where the image is?" Students respond "It's where they converge toward." [Don't you wish all students were so astute?]

Emphasize that we are studying lens 1 and trying to test the lens equation as applied to it. The other lens (lens 2) and the object-box are merely tools for producing convergent rays. To take measurements, we note the position of the screen, the position of lens 1, and then remove lens 1 and use a screen to find where the rays from lens 2 converge to. These three numbers are all we need to confirm the lens equation for lens 1. We do not need to know the position of lens 2, nor the position of the object box.

Now turn the students loose to apply this knowledge to the usual 'thin lens' laboratory. Insist that they study a converging, and a diverging lens, and produce data for all possible images within the span of the bench, real and virtual, and both real and virtual objects.

I've been a little less detailed here near the end, for the 'moves' are in the same spirit as before and get a bit repetitious to relate.

As you all know, wrong answers from the class can often be exploited for learning value. I once had an optics class so good that I had to conspire in advance with one of the best students to supply some wrong answers now and then. I haven't had to do that in a long time.

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