Let's go a set up the ladder, to convex mirrors. In the hierarchy of optical complication, plane is low on the totem pole, followed by diverging (lenses or mirrors) and then converging (lenses or mirrors). With plane mirrors, the images are

*all*virtual, upright, the same size as the object, and the same distance behind the mirror as the object was in front of it. With converging lenses or mirrors, you can get upright, inverted, enlarged, diminished, same-size, real, virtual, or non-existent images.

Diverging mirrors and lenses have a more varied assortment of images than plane mirrors, but are less complex than converging lenses and mirrors.

Let's start with magnification, and let's start with some ray tracing. Students got into groups and looked at a variety of object placements, determining the locations and sizes of the images with ray diagrams. All of the images were upright, and all were virtual - that is, all of the refracted rays

*appeared*to come from locations that they, in fact, never came from. We couldn't project these images on a screen. It's nice when something's absolute!

Here's the magnification data:

That's not a super-clear trend; it's a complicated relationship, it seems. Two things can make it easier, though. Let's think about the vertical intercept. Remember that the question "what's the vertical intercept" is equivalent to the question "what is the vertical quantity's value when the horizontal quantity is zero?" In this case:

*"What is the magnification when the object is right on the mirror?"*

If we include that point, it can help us not only place the curve, but determine what kind of curve it is. Here, the image is the same size as the object (magnification: 1) when the object's right on the mirror. This means that it's not a straight inverse or inverse-square function, because those don't have vertical intercepts. Adding this point will help, but let's think about the other side of the graph, too.

What's the long-term behavior of the graph? The question here is: "what happens to the vertical quantity when the horizontal quantity gets very large?" Here,

*"What happens to the magnification when the object gets far away from the lens?"*

Looking at the furthest objects away, the ray tracings start to show the images getting smaller and smaller, closer to zero. There's a horizontal asymptote here, because the graph won't ever cross that axis. If it did, that means that the image will be upside down, but all of our images are upright.

Check out the graph, now with intercept value and another point further away:

Trend looking a bit more clear? ...it'll be something like this:

That function is approaching zero as that image gets really far away. Not linear, but not too bad.

The

*image distance vs. object distance*graph is a bit more complex. In the first place, the values are negative (that's the easiest mistake to make here!), because the images are virtual. Here's a bit of data:

The intercept? Well, what's the image position when the object's right on the mirror? It's not hard to try it out. When you do it, the image is right, there, too: at position 0 when the object is. That's a vertical intercept of zero, and it adds to the picture.

Looking at the furthest objects away, the ray tracings start to show the images getting closer and closer to the focus. What's it like when the vertical variable keeps getting closer and closer to some value as the horizontal one keeping increasing? That, my friends, is a horizontal asymptote, and we have one at

*f*. Remember that

*f*is a negative number for diverging mirrors (and lenses)!

Put it all together, and what do you get?

Should you be able to just draw that off of the top of your head? Probably not. Should you be able to

*reason*through it, step by step, and piece that together one bit at a time,

*thinking*about each step individually and logically?

**You bet.**

## No comments:

## Post a Comment