Let's illustrate the use of these matrix techniques by doing an example that
we just did with the tabular technique, particularly that doublet lens.
So I have the paraxial equivalent of the doublet lens we derived in that example.
Remember we found that by using this little table of the surface curvatures and
refractive indices to find the equivalent paraxial power of surfaces 1, 2 and 3.
And we labeled all the distances that we knew, and
we were trying to find the image distance.
So let's do that now with the ABCD matrices.
So in the tabular method,
I chose to put into my table that I launched array from the object,
surface 0, with 0 height and an arbitrary ray angle of 0.01.
In the ABCD method, that will be,
I will choose an initial two element vector here,
0 and 0.01, and so that's right there.
Then I will transfer to the first surface,
refract across that surface, transfer to the second surface, etc.
Refract through the third or plain all surface and
finally transfer to the image plane, where I'll discover what
the ray height and what the ray angle are at the image point four.
So my job now is simply to write down all these matrices.
So I've copied the definition of the transfer and refraction matrices here, for
your reference.
Here, for example, is that first transfer matrix with the first distance in it, 300.
And note that if I simply multiply this two by two matrix times this two element
vector, that I will find a new two element vector with a first element Y of three.
That's exactly what I would expect for the ray height at Surface 1.
So I can take each of these, and I encourage you to do this,
multiply them out.
And it would fill out the table for me, which it should,
because they are equivalent methods.
Notice that when I get over here finally to the final transfer,
I have to leave d3 prime, the image back focal distance,
as a symbol, because I don't know it.
Which emphasizes that just like the tabular method, we can use these matrix
methods symbolically to derive equations that describe how our system works.
In this case, I'll leave that d3 prime.
I'll multiply all these matrices out, and of course there's handy programs like
Mathematica and Matlab that'll do that for you.
And I end up finding that the ray height, y4 and
the ray angle, u4 prime, which is equivalent
to u3 prime, are given by these expressions.
And notice that there's still this variable, d3 prime.
That means, as I move away from surface 3,
the ray height changes, and it changes as a straight line.
So this is an expression for the straight line, which is this ray.
And of course, as I did before, where I set the ray height,
excuse me, at the 0.4, at the image point d0.
I can do that here, so I can set this quantity equal to 0, solve for
that back focal distance and indeed get the same number.
So, the point is, it's the same thing.
And once you're comfortable with the tabular method, it's much faster,
because multiplying all these matrices up just automates the whole process.
And very rapidly allows you in basically one
somewhat tedious calculation to go from object to image.
