So now let's use the refraction and transfer equations to

solve a simple optical problem with multiple lenses.

As a typical problem, let's think about a finite conjugate microscope,

words that will make more sense as we get into the course.

The first lens in a microscope, the objective,

is typically a very short focal length lens.

And in a finite conjugate microscope, the cell, or

whatever is our object, is imaged to the end of the microscope tube.

This is older language, but it's the path length of the microscope itself.

And then, an eyepiece re-images that intermediate image here at plane 2,

off for acceptance by an eye, and I've drawn the cornea of the eye here.

So the question that we want to answer is if we have

an eye here that's going to be focused at infinity,

that's how our eye is going to look at the microscope.

Then that means that sort of thinking backwards here,

that this ray must be parallel although it's going towards the eye

because my eye is focused as if it's looking at infinity.

Then where is the object?

What is the working distance, d0 prime, of this microscope in that condition?

You can see that that's not entirely a straightforward question because

the object is not at the front focal plane of the objective because

the objective is being used in this finite conjugate geometry.

And we could use our single lens imaging equation to just find this distance.

But let's use the y-u technique to see how it works.

So what you do is you start out making a table.

You label every interesting surface.

The object should be Surface 0.

Then every lens,

I've stopped here and put a surface at this intermediate image plane.

You wouldn't have to, but I thought it would be interesting, so I did.

And since we know that a condition we want is that this final ray comes out parallel,

I stopped with the eyepiece.

I could have gone on through the cornea and to my retina, but I didn't have to.

So I'll make a table with one column for

each of the interesting surfaces that we want to trace the rays to.

The first rows of our table should be the power of each element

that we find at that surface, and the distance primed,

that is the distance to the next surface.

So we start out by writing the table, and then filling in everything we know.

Which usually means quite a lot about the optical system.

In this case, one of the things we don't know is one of the distances.

So the first surface 0, has 0 power because there's no lens there.

Surface number 1 has a power of 1 over 8.

I'm working in inverse millimeters here because I can scale to whatever I like.

So it has a power of 1 8th millimeter, inverse millimeter.

The distance to surface 2 is the tube length 160 plus 8 millimeters.

So here's d1 prime, 168.

Surface 2 has no power.

It's two millimeters farther because I'm the distance of the eyepiece

away from surface 3, so 2 millimeters between surface 2 and 3.

The eyepiece has 2 millimeters, so there's a one half power eyepiece.

And the other thing I know is that, let's see, I started out.

I'm going to start out launching an axial ray, because if I want to find where a ray

intersects the axis, that seems like the way to launch.

So for my problem statement,

I'm going to give the ray height y here at surface 0 to be 0.

And when I get all the way to the back of the system and coming out surface 3,

u3 prime, all set to 0 because that's what I said I wanted.

And now I just want to find everything in the table.

So let's go through that one step at a time.

So in this case to maybe exercise our brains,

I'm actually going to work backwards.

I'm going to start with this ray that's parallel to the axis.

Its angle is 0.

And I'm going to give it a height and

I'm going to walk backwards through the system.

And that's just to show you, you can run this forwards or backwards.

We'll do an example here in a minute where we go forwards.

So I'm just going to choose some height here and I might as well make life simple,

so I chose it to be 1, but

the way my diagram shows it's a minus 1 because of our sign convention.

So I now know everything about the ray at surface 3,

right after surface 3 formerly, and now I can work backwards.

So I'm going to use now my refraction equation to get across surface 3,

to get from here, the ray height at surface 3, to here,

what was the angle of the ray going into surface 3, or,

if you will, coming out of surface 2.

Well, to do that I write down my refraction equation with everything I

know in it.

I know that the ray angle coming out is 0.

I know that my ray height is minus 1.

I know that the power of this lens is 0.5.

And the only thing I don't know is what was the ray angle going in.

So I write that down and solve for this number,

which I also could have just gotten by observation here because I

know my geometry, but we're doing a simple problem.

Well, let's just keep going now, and notice I've kind of arranged this table so

that everything feeds from step to step.

So now C here is, I'm going to do a transfer.

This is just to show you the system works.

The height of my array at surface 2 is the thing I don't know.

I fill in the height of the ray at this surface here.

That was minus 1.

The angle of the ray minus 1f, the distance I'm going to, and

these are just from the equation.

And then the thing I don't know is the height of the ray at y2.

That turns out to be 0.

That's convenient, it should have been.

And now I can simply step through the system.

I do a refraction to get the ray angle at d.

I do another transfer.

I do another refraction.

And finally, I've got that the ray angle here at surface 0 is 10.

And of course, all this scales linearly.

So if I had chosen a different arbitrary height of my ray, at surface 3,

I would get a proportionally different ray angle here.

But the point is, I've now transferred this ray through, its height and

its angle at every surface.

And I've finally gotten back to what was the angle here.

And of course, I know what the height was here, so

I can do one more transfer, and that will give me this distance.

And now what I have to do is solve for the thing I didn't know.

So the point is it's a very mechanical procedure, and it allows you to walk

rays through a system without ever talking about object and image distance.

Instead we talk about ray height and ray angle.