00:01
So we want to demonstrate when two lenses are close enough that the distance between them can pretty much be neglected, that we can treat those two lenses combined as they were one lens, and that the total of the focal length of those combined lenses is going to be related to this equation, which represents one over the total of the focal length, is equal to the sum of one over the focal length of the first lens plus one over the focal length of the second lens, which in the terminology of the problem, f effective, or f, eff, is f total.
00:51
So let's first start off with how the light ray will be affected when it hits the first lens, in which it would have, if not for the other lens, bent at a certain direction down, which if we send our second ray straight through this system, assuming it can be treated like a compound lens, we would have the image form here from lens.
01:29
We would also have that the distance to the image is q1 and to the object is p1.
01:47
So let's go ahead and write out the lens equation as 1 over p1 plus 1 over q1 is equal to 1 over f1.
02:00
And we could have stopped there if it was not for the second lens bending the light further the moment the ray intersected the second lens.
02:11
And so the actual light path is different in which you have the actual image form right here, which we'll name i t for i total.
02:34
And it's going to have a position.
02:40
And let's go ahead and label this q total, t for total.
02:47
Okay, let's start writing out the second lens equation for the second lens.
02:54
In which we have one over, we're going to introduce in a moment, and one over q total is equal to one over the second focal length.
03:16
So what do we put right here? that's when we notice that we use the image formed from the first lens for this position.
03:33
So let's go ahead and place that...