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## Lenses

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# Object image and focal distance relationship (proof of formula)

## Video transcript

We've been doing a
bunch of these videos with these convex lenses
where we drew parallel rays and rays that go
through the focal point to figure out what the
image of an object might be. But what I want to
do in this video is actually come up with
an algebraic relationship between the distance of the
object from the convex lens, the distance of the image
from the convex lens usually on the other side
and the focal length. So let's see if we can do this. And just to save you
the pain of having to watch me draw straight lines,
I drew this ahead of time. And so we can imagine this green
thing right here is the object. This is the object, and
these two little pink points right here are the focal points. They're a focal length away. And I did what we always drew. I drew one parallel ray
from the tip of that arrow to the actual convex lens,
and then it gets refracted, so it goes through the focal
point on the right-hand side and goes all the way over there. And then I drew a ray that
goes through the focal point on the left-hand side. And then when it gets
refracted, it becomes parallel. And then it actually intersects
with that previous ray right over here. And so this gives us a sense of
what the image will look like. It's inverted. It's real. In this case, it is larger
than the actual object. What I want to do is come
up with a relationship with these values. So let's see if we
can label them here. And then, just do a
little bit of geometry and a little bit of
algebra to figure out if there is an algebraic
relationship right here. So the first
number, the distance of the object-- that's this
distance from here to here, or we could just label it here. Since this is
already drawn for us, this is the distance
of the object. This is the way we drew it. This was the parallel light ray. But before it got refracted,
it traveled the distance from the object to
the actual lens. Now, the distance from the
image to the lens, that's this right over here. This is how far this parallel
light ray had to travel. So this is the distance
from the image to the lens. And then we have the focal
distance, the focal length. And that's just this
distance right here. This right here is
our focal length. Or, we could view it
on this side as well. This right here is
also our focal length. So I want to come up
with some relationship. And to do that, I'm going
to draw some triangles here. So what we can do is--
and the whole strategy-- I'm going to keep looking
for similar triangles, and then try to see if
I can find relationship, or ratios, that relate these
three things to each other. So let me find some
similar triangles. So the best thing I
could think of to do is let me redraw this
triangle over here. Let me just flip it over. Let me just draw
the same triangle on the right-hand
side of this diagram. So this. So if I were to draw
the same triangle, it would look like this. And let me just be clear,
this is this triangle right over here. I just flipped it over. And so if we want
to make sure we're keeping track of the same
sides, if this length right here is d sub 0, or d
naught sometimes we could call it, or d0,
whatever you want to call it, then this length up here
is also going to be d0. And the reason why
I want to do that is because now we can do
something interesting. We can relate this triangle
up here to this triangle down here. And actually, we can see that
they're going to be similar. And then we can get
some ratios of sides. And then what we're
going to do is try to show that this
triangle over here is similar to this
triangle over here, get a couple of more ratios. And then we might be able to
relate all of these things. So the first thing we
have to prove to ourselves is that those triangles
really are similar. So the first thing to
realize, this angle right here is definitely
the same thing as that angle right over there. They're sometimes
called opposite angles or vertical angles. They're on the opposite side
of lines that are intersecting. So they're going to be equal. Now, the next thing-- and
this comes out of the fact that both of these lines--
this line is parallel to that line right over there. And I guess you could call
it alternate interior angles, if you look at the angles
game, or the parallel lines or the transversal of
parallel lines from geometry. We know that this
angle, since they're alternate interior
angles, this angle is going to be the same
value as this angle. You could view this
line right here as a transversal of
two parallel lines. These are alternate
interior angles, so they will be the same. Now, we can make that
exact same argument for this angle and this angle. And so what we see is
this triangle up here has the same three angles
as this triangle down here. So these two
triangles are similar. These are both-- Is really
more of a review of geometry than optics. These are similar triangles. Similar-- I don't have
to write triangles. They're similar. And because they're similar, the
ratios of corresponding sides are going to be the same. So d0 corresponds to this. They're both opposite
this pink angle. They're both opposite
that pink angle. So the ratio of d0 to d1--
let me write this over here. So the ratio of d0. Let me write this a
little bit neater. The ratio of d0 to d1. So this is the ratio of
corresponding sides-- is going to be the same thing. And let me make
some labels here. That's going to be the same
thing as the ratio of this side right over here. This side right
over here, I'll call that A. It's opposite this
magenta angle right over here. That's going to be the same
thing as the ratio of that side to this side over
here, to side B. And once again, we
can keep track of it because side B is
opposite the magenta angle on this bottom triangle. So that's how we know that this
side, it's corresponding side in the other similar
triangle is that one. They're both opposite
the magenta angles. So we know d0 is to d1 as
A is to B. As A is to B. So that's interesting. We've been able to
relate these two things to these kind of
two arbitrarily lengths. But we need to somehow connect
those to the focal length. And to connect them to
a focal length, what we might want to do
is relate A and B. A sits on the same triangle
as the focal length right over here. So let's look at this
triangle right over here. Let me put in a better color. So let's look at this
triangle right over here that I'm highlighting in green. This triangle in green. And let's look at that in
comparison to this triangle that I'm also highlighting. This triangle that I'm
also highlighting in green. Now, the first thing
I want to show you is that these are also
similar triangles. This angle right over
here and this angle are going to be the same. They are opposite angles
of intersecting lines. And then, we can make
a similar argument-- alternate interior angles. Well, there's a couple
arguments we could make. One, you can see that this is
a right angle right over here. This is a right angle. If two angles of two
triangles are the same, the third angle also
has to be the same. So we could also say
that this thing-- let me do this in another
color because I don't want to be repetitive too
much with the colors. We can say that
this thing is going to be the same
thing as this thing. Or another way you
could have said it, is you could have said, well,
this line over here, which is kind of represented
by the lens, or the lens-- the
line that is parallel to the lens or
right along the lens is parallel to kind of the
object right over there. And then you could make the
same alternate interior argument there. But the other thing
is just, look. I have two triangles. Two of the angles in those
two triangles are the same, so the third angle
has to be the same. Now, since all three
angles are the same, these are also both
similar triangles. So we can do a similar thing. We can say A is to B.
Remember, both A and B are opposite the 90-degree side. They're both the hypotenuse
of the similar triangle. So A is to B as-- we could say
this base length right here. And it got overwritten
a little bit. But this base length
right over here is f. That's our focal length. As f in this triangle is
related to this length on this triangle. They are both opposite
that white angle. So as f is to this
length right over here. Now, what is this length? So this whole distance is
di, all the way over here. But this length is
that whole distance minus the focal length. So this is di minus
the focal length. So A is to B as f is to
di minus the focal length. And there you have it,
we have a relationship between the distance
of the object, the distance of the image,
and the focal length. And now we just have to do
a little bit of algebra. If this is equal to this
and this is equal to that, then this blue thing has to be
equal to this magenta thing. And now we just have
to do some algebra. So let's do that. So we got d naught, or d0
I guess we'd call it, to di is equal to the ratio
of the focal length to the difference of
the image distance to the-- the focal
length to the image distance minus the focal length. And now here, we just
have to do some algebra. So let's-- just to simplify
this, let's cross multiply it. So if we multiply d0 times this
thing over here, we get d0 di. I'm really just
distributing it, minus d0 f. I'm just distributing this
d0, just cross multiplying, which is really just the
same thing as multiplying both sides by both
denominators, or multiplying the denominators twice. Either way, that's all
cross multiplying is. That is going to be
equal to di times f. And now, we can add this
term right over here to both sides of this equation. I'm just going to switch
to a neutral color. So we get d0 di is
equal to-- I'm just adding this to both sides. When you add it to the left
side, it obviously cancels out. Is equal to di f, this
thing over here, plus d0 f. And then, let's see. We could factor out
an f, a focal length. So we get d0 di is equal
to f times di plus d0. And then, what can we do? We could divide both sides by f. So this will become over f. This becomes over f. Essentially, canceling it out. And then, I just don't want
to skip too many steps, so let me just rewrite
what we have here. So these cancel out. So we have d0 di over f
is equal to di plus d0. And now, let's divide
both sides by d0 di. So 1 over d0 di. Divide this side by d0 di. It cancels out over here. And so we are left with,
on the left-hand side, 1 over the focal length is
equal to this thing over here. And we can separate
this thing out. This thing over here,
this is the same thing-- we just separate
out the numerator. Is the same thing as di over
d0 di plus d0 over d0 di. But di over d0 di,
the di's cancel out. We just have a 1. Here, the d0's cancel out. You just have a 1. So this is equal to 1 over
the distance of the object. And this is plus 1 over
the distance the image. So right from the get-go, this
was a completely valid formula. We actually had
achieved what we wanted. But this is a neater formula. You don't have the di's
repeat and the f repeated. Right here, we have an
algebraic relationship for a convex mirror that
relates the focal length to the distance of the object
and the distance of the image. Anyway, I think that's
pretty neat how it came out to be at least a
pretty clean formula.