For convex mirrors, we’re going to follow the same sort of logic but there’ll be a few differences.
For a convex mirror, we could send one beam parallel to the axis again. But instead of that beam
bouncing and going through the focal point, it will bounce and go directly away from the focal point
since the focal point is on the opposite side now of the mirror. The second beam, we could send
directly towards the focal point and that beam would bounce parallel away from the mirror itself.
To find an image here, we’re going to have to do something slightly different knowing how these
two beams behave. What we’ll do is trace both beams back to again, where it appears that they’re
coming from. One more time. When our eyes see the beams, we think that it’s coming from directly away
So, we just trace those two beams back and look at where is the source, where is the apparent
source that it looks like this image is originating. In this case, you can see we have an image of a small
object behind the mirror which is just the image of our original object. This image is something that
we would call a virtual image because the light rays never actually intersect with each other.
It only appears as though they intersect with each other. In other words, this image that we’re seeing
behind the mirror is not something we could ever, for example, cast or project onto a screen if we
wanted to send this light onto some screen like a piece of paper or something and see it. We can
never do that with a virtual object because again, these light rays are never actually meeting
at the same point. We can now quantify the distances that we just discussed. For example,
we can discuss where is the object relative to the mirror. We can then ask where is the image
relative to the mirror. Finally, we can ask where is the focal point relative to the mirror. For each of these,
the object, the image, and the focus, we’ll give a different letter, the first letters of each word
in this case. We use a lowercase letter to represent the distance, the actual distance and units
of distance from the mirror’s center itself. So, we have an object distance, a smaller image
distance and in this case, an even smaller focal point distance. The way these three quantities
are related for mirrors is given by this lens equation. We have 1 divided by the distance of the object
plus 1 divided by the distance of the image is equal to 1 divided by the distance of the focus.
We have some conventions for how we define this though. We can use the equation we just
introduced to find, for example, the distance that the image will be if we know where the object is
and we know the focal point of our mirror. But we do have to be careful to follow these conventions
when we’re discussing the mirror. By convention, we say that the focal point is a positive number
for a concave lens and for a convex lens, it is a negative number. The object distance for mirrors
is always going to be a positive number because we can never put our object behind the mirror
and see an image of it. But the image for a mirror can be positive for a real image that is in front
of the mirror that we saw on our first example but will be a negative number for those virtual images
that we talked about that are behind the mirror. We also have finally a magnification equation.
So, when we have an object and we get an image from that object, we could ask ourselves which one
is bigger or what’s the ratio of their sizes. Are we magnifying the image and making it appear to be
a much bigger object or are we minimizing it and getting a smaller image as we have in this example here?
We define this again as a magnification which we use a letter m to describe. This is equal to
minus the distance of the image divided by the distance of the object. This negative magnification
will refer to inverted images. That’s the reason we have the minus sign. You can see that if we used
a negative sign for the image, again, that would mean that the image was behind the mirror.
It would be a virtual image. That virtual image would instead be upright. That happens
because, again, we will put any negative number for this i here. That negative would cancel
with the negative we have in the definition. So, that’s the reason it looks the way it does.
It’s important to notice again the difference between the real and virtual images in terms of whether
they’re upright or upside down. In this case, you can see in this picture here, we have an upside down
image. That’s always going to be the case with mirrors that your real images are upside down
and in front of the mirror. On the other hand, for the virtual images, they’re always going to be
upright. Again, they’re on the other side of the mirror. There’s one way to think about or
to remember how to use this magnification equation which is with this fun little device called
the magnification triangle. The way this works is you pick whichever variable that you’re looking for.
For example, suppose you’re trying to find the magnification itself. In that case, you cover the m
in this triangle because that’s the number you’re trying to find. Then the triangle sort of shows
you the equation for the magnification. By covering the m in this triangle, we would just see
i over o, the image over the object, which we can see from our magnification equation as exactly
the right equation. If we wanted to rearrange this, we could cover a different letter. For example,
if we wanted to find the location, the distance to the image, we could cover the i in this triangle.
In that case, we wouldn’t see the i anymore. We would just see the o times m, which would
if we rearrange the equation be the proper equation for the image distance. Of course, you do have
to be careful with the sign here because notice this triangle doesn’t have a place for a minus sign
anywhere in it. So, you could always introduce that on the outside yourself or just be careful
to remember the minus sign. For the magnification, we actually have one last way to define
this magnification as not just the ratio of the distances, and remember this i and o are distances
from the mirror, but as the heights of the image and which image is taller. So one of the objects,
for example, we could take the height of the object and give it an h sub o or we could talk about
the height of the image that we create, h sub i. The ratio of these two numbers would also be
the magnification. For example, suppose I showed an object to a mirror. I see an image
in that mirror. The image appears twice as tall as the object. In this case, I would have
a magnification of two. So, it’s sort of an intuitive definition for the magnification when you’re using
an object like a mirror. As a summary, what we have for objects with light being sent in towards
a mirror, if the object is infinitely far away or far away enough that all the rays coming towards
the mirror seem to be parallel to each other, then the image type would simply be an image
at the focal point because again, every single one of the parallel rays will go right through
the focal point. So, this wouldn’t be such a useful thing unless we have some sort of a small
camera, for example, located at the focal point itself. If the object is near the mirror but still outside
of the focal point, we end up with an inverted, real image in front of the mirror. If on the other hand,
the object is right at the focus, we can’t really get an image. The image would go to infinity
because any light rays going to the mirror from the focal point would leave parallel to each other.
Finally, we could put the object inside the focal point closer to the mirror than the focal point.
In this case, again drawing the two rays that we described, one of them going towards the mirror
and then bouncing back into the focal point, one of them heading away from the focal point
will bounce back to the focal point. We could trace those images, those light rays back and see
we would perceive an image to be behind the mirror and that image would be upright and virtual.
Finally, if we had a concave mirror, it turns out that no matter where we put our object,
since we can’t possibly put the object on the focal point or inside the focal point since the focal point
is on the opposite side of the mirror, we always get a virtual, upright image.