Thin Lenses and Virtual Images (Jeffrey Scott Nuttall)

Title: Thin Lenses and Virtual Images

Principle(s) Investigated: Optics: Thin lenses (convex and concave), real images, virtual images, thin lens equation

Standards : HS-PS4-5. Communicate technical information about how some technological devices use the principles of wave behavior and wave interactions with matter to transmit and capture information and energy.

Materials: Optic bench for each group (includes: meter stick, stands for meter stick, candle with mount, mount for index card/paper, two lenses (one convex and one concave) with mount for each); also, something to light the candles with

Procedure: Divide the students into groups (preferably no more than three or four students per group, but it may be necessary to have more if there aren't enough optic benches to go around).

Give an overview of thin lenses, how they relate to refraction, what it means for a lens to be "thin". Define the focal length of a convex lens (the distance from the lens at which parallel rays are focused).

Have the students place a lit candle at one end of their optic bench, a card at the other end, and the convex lens in between. Tell them to adjust the positions of the objects until a very clear image of the candle flame appears on the card. Ask them to try this for various distances.

Have all students enter in an online spreadsheet the object distance p and image distance q for each trial, in centimeters. Plot the data. Does the graph form a recognizable shape? It's clear that the larger p is the smaller q is, but aside from that the relationship isn't obvious. However, plotting 1/p vs. 1/q should show a clear linear graph: the equation is 1/p + 1/q = C for some constant C. We can figure out C by considering parallel rays... then p should be infinity and q should be equal to the focal length, f. Then solving the equation gives C = 1/f, so the full thin lens equation is 1/p + 1/q = 1/f.

Describe how the rays work with a convex lens (show ray diagram).

Now have the students try it with a concave lens. It should not be possible to get a good image. Does that mean there is no image?

(If time permitted, it would be nice for the students to do the experiment using two convex lenses, to see how the object of one lens can be the image of the other -- but given the short time and the fact we only have one convex lens per optic bench, we're going to have to pass on that for now.)

Have the students use both the concave lens AND the convex lens (with the concave lens closer to the flame). Now it should be possible to get an image. We can measure q for the convex lens, so we can (now that we know the convex lens's focal length) figure out where the object is that's causing it. The object is the image from the concave lens -- but it's on the same side of the lens as the flame! Explain this is a virtual image, and the image length is considered negative. Show a ray diagram to demonstrate how the virtual image works.

Now, knowing p and q for the concave lens, we can calculate f. The focal length for a concave lens turns out to be negative!

Student prior knowledge: It is assumed that students already know about refraction before beginning this experiment; however, it's not absolutely necessary.

Explanation: The thin lens equation, 1/p + 1/q = 1/f, holds for both concave and convex lenses, and both real and virtual images. This experiment is designed to show examples of each, and to give students a feeling for what it means to have a virtual image (or even a virtual object)! For a virtual image, q is negative; for a "virtual object" (when the "object" is an image from another lens which is on the opposite side of the lens in question), p is negative, and for a concave (diverging) lens, f is negative.

The position of the image can be found approximately by ray-tracing. To perform this procedure, plot, to scale, the position of the object and the focus of the lens, and then for a converging lens trace the following rays: one that goes directly through the center of the lens (and continues in a straight line), one that goes parallel to the axis (and on passing through the lens passes through the far focus), and one that goes through the near focus (and on passing through the lens becomes parallel to the axis). The image occurs where the three traced rays intersect. Note that it's really only necessary to trace two rays to find the intersection, but the third ray makes a handy check on the answer. Below are a few examples:

For a concave lens, the three rays traced are similar, but not quite the same. Again, a ray through the center of the lens will continue in a straight line. A ray that begins parallel to the axis, however, will on passing through the lens be diverted directly away from the near focus. The third ray is a ray that starts toward the far focus, and on passing through the lens goes parallel to the axis. Below, again, are examples:

Questions & Answers:

We've seen in the lab an example of a virtual image. Could there be such a thing as a virtual object? Explain.

(Answer: Yes! We've already seen an example of the image from one lens being used as the object of another. What happens if that image is on the far side of the second lens? Then the light rays don't really converge at that object! Here's an example of such a virtual object:

Note that just as q is negative for a virtual image, p is negative for a virtual object!)

What happens if you place the object directly at the focus of a lens? Will you see an image? If so, where will it be?

(Answer: If you place an object at the focus of a lens, then p = ±f. If the lens is concave, then p = -f, so 1/p + 1/q = -1/p. This becomes 1/q = -2/p, so q = -p/2. There will be a virtual image, appearing half the distance to the lens as the object. If the lens is convex, then p = +f, so 1/p + 1/q = 1/p. This reduces to 1/q = 0, so q = ∞. The rays will not converge, and no image will be seen.)

Two lenses placed right next to each other act as a single lens. Can you come up with a formula for the effective focal length of this lens, in terms of the original focal lengths?

(Answer: Consider where the final image will form. Let's call the focal lengths of the component lenses f1 and f2. Now, the image distance due to the first lens can be found by the thin lens equation: 1/p + 1/q1 = 1/f1, so q1 = 1/(1/f1 - 1/p). Now, that image acts as the object of the second lens... but with the sign flipped, since a real image will be on the far side of the lens and thus be a virtual object, and vice versa. So we can find the final image distance q2 by using the thin lens equation again: 1/q2 + (-1/q1) = 1/f2. But 1/q1 = 1/f1 - 1/p, so this becomes 1/q2 + 1/p - 1/f1 = 1/f2, which we can rearrange to 1/p + 1/q2 = 1/f1 + 1/f2. Now, if we treat the two combined lenses as a single lens, the equation becomes 1/p + 1/q2 = 1/feff (where feff is the effective focal length of the two lenses combined). Comparing the two equations, we see that 1/feff = 1/f1 + 1/f2, so feff = (1/f1 + 1/f2)-1.)

Applications to Everyday Life:

Lenses are in common use in many everyday phenomena. Here are just a few examples:

The "lens" of your eye is a literal lens that actually focuses incoming rays of light on your retina! In some people, the lens isn't exactly the right shape and doesn't focus the light just right, so a second lens is put in front of the lens of your eye to to adjust the effective focal length, just as we've seen the effect of combined lenses in this lab. That second lens would be glasses, or a contact lens. Cameras, too, use lenses to focus the light onto the surface that records the image, much like the lens of your eye.

Binoculars and telescopes also use lenses to focus light rays. In a very simple telescope (like, say, a pirate's spyglass—arr!), one lens (the objective lens) focuses near-parallel rays from a distant object onto the shared focus with a second lens (the eyepiece), which then makes the rays parallel again but closer together, effectively magnifying the image.

Microscopes work similarly, except that in that case the object isn't very distant, so the lens isn't focusing parallel rays. Rather, the object is positioned near the focus of the objective lens. Once again, though, the objective lens refocuses the light on the other side onto the shared focus of the eyepiece, which then sends the rays parallel toward your eye.

Virtual images in front of a lens can have spooky-seeming effects. A sort of box can be created with two concave mirrors (mirrors follow similar equations to lenses) so that if you look at it at the right angle a ghostly image of the box's content would seem to be floating above it! Here's an example: https://www.youtube.com/watch?v=Qh4nMYfGjMQ (You may hear this referred to as a "hologram", but that's not really accurate; a hologram is a different phenomenon entirely.)

Photographs:

Here you can see how the concave lens and convex lens together form an image. You can also see that I badly need a haircut.