- [Voiceover] If I had been handed
this problem on a physics test
I probably would have freaked out.
This looks really intimidating
but it's actually not that bad.
It's a classic example of a two lens system
and overall, before we get lost in details here.
The overall idea of how we're going
to approach is this.
We've got two lenses.
This 1st lens is going to create
an image of this object over here.
It creates an image of this object.
Then the 2nd lens is going to create
an image of that image and so
what we're basically going to do
is we're going to use the thin lens formula.
We're first going to pretend like
this 2nd lens doesn't exist.
We're going to figure out, what image
does this 1st lens create?
Then, we'll pretend like the 1st lens
doesn't exist and we'll treat that image
that the 1st lens creates as if
it's the object for the 2nd lens.
We'll do another calculation, figure out
where this 2nd lens creates an image
of the thing it thinks is the object
but it's actually the image from the 1st one
and so we're going to figure out
first the object gets turned into an image
from the 1st lens, then the 2nd lens turns that image
into another image and that's what
our eye is going to see.
So that's the question we want to answer
is what image would our eye see
in this case if we had it over here
looking through these two lenses?
So let's do this.
First thing we got to do is I'm going
to pretend like the 2nd lens
doesn't exist so it doesn't confuse me.
We use the thin lens equation.
Alright here we go.
1 over f, so 1 over, okay here we go,
12 centimeters, do I make it positive or negative?
It's on the left.
Maybe I make it negative, nope.
The only thing you look at is what kind
of lens this is.
This is a convex lens.
I can tell by the way it's drawn.
Since this is a convex lens my focal length
is going to be positive 12 centimeters.
So that's what I plug in,
Positive 12 centimeters equals, alright,
object distance so 1 over the object distance
that's 24 right?
No, all distances are measured from
the center of the lens to the thing
you're looking at and so for an object distance
I'd have to go from the center of the lens
all the way to the object.
Don't just plug in any number they're giving you.
You've got to plug in that distance
and that's going to be 24 plus 12.
It's actually 36.
So I'd have 1 over and this is going
to be 36 centimeters.
Positive or negative?
Well that's going to be positive here.
We've got two lenses but for this first case
it's on the left hand side.
Everything's cool, it's on the opposite
side as my eye.
It's going to be a positive 36.
I'll show you in a minute how this
could possibly be negative in a second.
Plus 1 over the image distance.
I'll just label that with a di.
Alright so now we just got to solve.
If you solve this you going to go
1 over 12 centimeters minus 1 over 36 centimeters
equals 1 over di what you're going to
end up getting is 1 over 18 centimeters
on the left hand side equals 1 over di
but that's 1 over di, if you solve that
for di you'll get that di equals 18 centimeters
positive 18 centimeters.
And so what does that mean?
Where is this?
Well, alright, di, positive image distance.
Remember, positive means on the same
side as your eye for a lens.
So I'm going to have an image that's formed
over on this side of the lens
and 18 means 18 centimeters from where?
From the center of the lens.
So if I draw this.
I'm going to go 18 from here.
That's going to be past the focal point
because I know the focal point is only 12
so I'm going to be somewhere around here.
I'm going to label that and that's going to be
18 centimeters and that's where
the first images, so that's where
the image is going to be that
this 1st lens creates.
It's going to be right there.
Okay so now what do I do?
I told you, here's what we do.
We pretend, we'll bring the 2nd lens back.
We know the image that the 1st lens
created is right here.
This 2nd lens is going to think
that that's the object.
It's going to create an image of that image
as if that were an object.
So we do another thin lens formula
but this time we treat this positive 18
not at if it's the image, we treat it
like it's the object.
So this is going to be well, it's not going
to be equal to do because remember,
you have to measure everything from where?
From the center of the lens so for this
second calculation, my object distance
is going to be from the center of the lens
all the way to, let me not use that color.
Let me see, let's pick this one here.
From the center of the lens all the way to
where it's object is and it's object
is over here which is not 18.
That's why it gave you this distance between lenses.
The distance between lenses is 33 centimeters.
So if the distance from this 1st lens is 18
well the difference has to be equal to this length.
So 33 minus 18 is going to be 15, whoops.
15 centimeters will be where the "object"
for this 2nd lens is going to be.
So as we turn this positive 18 into
object distance of positive 15 centimeters.
Why positive?
Well it's still on this left hand side
opposite side as your eye.
That's where the objects are supposed to be.
Now here, this is where I told you
there's one case where you can get
negative object distances if our 1st lens
would have created an image of this object
way over here on the wrong side of the lens.
In other words, if this 1st lens created an image
that was closer to our eye than
the 2nd lens was, well that's just all messed up.
That's going to be a negative object distance.
Your not supposed to hold an object
between your eye and the lens.
So we'd count that, you'd do
your calculation just the same.
I mean everything would work out right.
It's not like the whole thing's going to break.
You'd get an image.
It's just when you do your calculation
for the second case.
If your image ended up on the wrong side
of the lens for the first image
that was created you'd have to treat it
as a negative object distance.
That didn't happen here.
Our image is still formed on this side.
We're going to pretend like this 1st lens didn't exist.
We got an object over here, we're going to
treat it like our object.
Everything's fine.
Eye looking through the lens at our object.
So we still treat this as a positive 15.
So let's do our calculation.
We'll have for this lens it's going
to be 1 over.
Alright, focal length.
What do we look at for focal length?
We look at what type of lens it is.
This lens is a diverging lens.
Diverging lenses always get contributed,
always have a negative focal length
that they contribute into this equation.
So a negative 10 centimeters.
That's the focal length of the 2nd lens,
equals 1 over our object distance we just found.
It's not really an object but that's okay.
It's an image but this lens doesn't know that.
It thinks it's an object so 1 over positive 15
because it's still on the side that
that object was supposed to be on,
plus 1 over our image.
This is going to be a second image.
This is the second image that this
lens system is going to create
and so I'll just put di.
Alright, so we do the math, alright
1 over negative 10 centimeters
minus 1 over 15 centimeters equals 1 over di,
you solve that on the left hand side.
You flip it over, you're going end up
getting the di is, once you do that inversion
negative 6 centimeters.
That's what our second image is going to be.
So what does this mean?
Negative 6 centimeters.
This is always from the lens and negative.
Remember negative distance means
on the left hand side or the opposite
side as your eye.
So I'm going to come up here
and I'm going to draw a line that goes
from the center of my lens, 6 centimeters
to the left somewhere around there.
So that's going to be 6 centimeters right here.
Whoops, I'll label that.
That's going to be 6 centimeters
because that's our image distance
and that means our final image is right there.
Our eye is going to see an image
of something our arrow right at that spot
right there that's where our image
is going to be and just to recap what happened.
So this 1st lens created an image
of the object over at this point here,
this 15 centimeter point and then
the 2nd lens created an image of that image.
It treated it like it was an object
and it created an image at this white point
right here that's where we're going
to see the image but we don't know
what it's going to look like.
These calculations, this thin lens formula
only shows us where the image is going to be.
If we wanted to figure out how big it was
we have to use the magnification formula
so let's do that.
Over here, magnification equals negative di.
Remember it's negative di over do.
Alright, so what was di for this first case?
We're going to do it one step at a time.
For this first case, the di was positive 18.
So I'm going to put a positive 18 up here.
I'm going to leave off the units
because they're going to cancel.
Do was not 24 it was 36.
So what do I get?
I get negative 18 over 36 that's 1/2, negative 1/2.
What does that mean?
That means this 1st lens, the image
that it created is going to be an image
that's 1/2 as big as my object
and it's going to be negative magnification.
That means it's upside down so I'm going to
draw this first image.
I'm going to do it in orange
because that's what I labeled the image distance.
It's going to be upside down
and half as big as the object.
So if the object was that big.
My image is going to be upside down
and about half as big.
So this was my image one.
I'll label it image one.
That's what I'm treating as my second object.
So that's image one that's object two
essentially because that's what this
2nd lens is going to think is the object.
What magnification does it provide?
Well let's do that one.
Magnification equals negative di over do
so negative of my image distance for
this 2nd lens was, it was negative 6
remember right here?
That's what we solved.
The 2nd lens created an image distance negative 6.
I'll leave off the units because they cancel
and my object distance for
the 2nd lens was 15 centimeters.
That's what this was right here.
The distance from the lens to where
it's "object" was and that was positive 15
and what do I get?
If you take negative 6 over 15
you end up getting positive 2/5.
So this is the second magnification.
The magnification of the 1st lens was negative 1/2.
The magnification of the 2nd lens was positive 2/5.
That means at this point over here,
I'll make it white because that's where
I labeled this image distance.
I'm going to get an image that's
2/5 as big as what?
As the 1st image.
Now this is a magnification of the image
because that's what this 2nd lens is doing.
So I don't look back at my object over here.
I'm done with this guy for now.
Now I'm just looking at what this 2nd lens
thinks is the object.
So 2/5 as big as this object and positive
means right side up but in this case
this really just means not inverted.
So this started off our object for
the 2nd lens started off upside down.
This positive means that you maintain
the direction, you don't invert it.
So I leave this thing upside down.
That's what it means
I leave it upside down.
If it would have started right side up
I would have left it right side up
but in this case, I leave it upside down
and I make it 2/5 as big.
2/5 is a little under 1/2 and so I'm going
to draw this about like that.
That's where my image would be, right there.
That's what I would see.
I would see a very little image
that's upside down and it would be
right around there and so that's an example
of where you can do a two lens system.
You treat each lens separately
and you use these formulas accordingly.
The last thing I'll say is,
if you wanted to know why we had
two magnifications here, is there some way
to figure out the total magnification?
Some slick way of doing it?
Yeah and it's really easy it turns out.
You can just take your total magnification.
If all I wanted to know was if I had this object
what my total magnification was going to be.
What could I multiply this object's height by,
in other words, what factor could
I multiply it by to get both
the orientation and height of this final image?
Well you just take the magnification
of the 1st lens times the magnification
of the 2nd lens.
If I had more I'd just multiply them all.
That would give me the factor.
That's the overall magnification that I can
multiply this object's height by
to get the height of the final image
and the orientation of it.
It doesn't give you the position.
This just gives you the overall magnification,
That could be handy if your making
a microscope or something but it doesn't
tell you the position.
For the position you've got to use
these thin lens formulas.
So that's how you treat a multiple lens problem.