I'm having some troubles drawing down the main rays on a thick lens. In particular I don't understand if the parrallel ray which comes from the right will start to divert intercepting the MP1 or the MP2. How you can see in the picture below, two possibilities are drawn; I know that the right one is the B, however I can't give me an explanation of why it is. I know how to get analitically the position of the image given an object and a thick lens, however I'd like to learn also how to get the image in this graphical way. So, someone could explain why the fig. B is the right one?
closed as off-topic by Kyle Kanos, ZeroTheHero, Jon Custer, Aaron Stevens, user191954 Feb 9 at 15:56
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Let's concentrate on fig. B. There are several things which need correction.
First of all you left understood that you're using Gauss' approximation. Otherwise you had no focal or principal points. Incidentally (but this may be an idiosincracy of mine) I wouldn't speak of "ray tracing" in that approximation. I'd reserve the term for exact calculation of rays through an optical system.
Second: it's better to reason about a general (centred) optical system, not on your special example of a plano-concave thick lens. So it's easier to understand what focal and principal points are. If you have a general system you will not worry about the actual path a ray follows within the system. Using focal and principal points you can find how a given entering ray will exit on the other side.
As a side note, I don't understand why you drew arrows on rays, sometimes pointing left, other times pointing right. It's a general convention for dioptric systems (lenses) that light always comes from left and goes out to the right.
Still on conventions. F$_1$ is the first (front) focal point, F$_2$ the second (rear). This doesn't mean that you'll always find F$_1$ at the left, F$_2$ at the right. This is true for a positive (converging) system, but for a negative (diverging) one it's the opposite. This is correctly shown in your fig. B, where F$_1$ is placed after the lens, F$_2$ is placed before.
Now for rays and how they're are drawn. Consider first a ray parallel to the optical axis. It comes from left, reaches the first optical surface ... and there should be interrupted, since in general you don't know its exact path within the lens (in your special case you do know it, as the first lens surface is plane - this is why it's better to think of a general system). From then onwards it should be drawn dashed, until it intersects the first principal plane.
Let me call A$_1$ this point. I repeat: the real ray will not (in general) pass through this point, but it doesn't matter. Here we'll use the basic property of principal planes: a ray intersecting (as a geometrical straight line, not as the physical ray) the first principal plane in A$_1$ will exit the system as it would came from A$_2$, the point of the second principal plane at the same distance from axis. To show this you should draw segment A$_1$A$_2$ (parallel to the axis) in a different style, say dotted.
We know that the outgoing ray will pass through F$_2$. Then draw A$_2$F$_2$, but with two different styles: a full line outwards the lens, dashed within it and towards F$_2$ (in your case, where the actual ray doesn't pass through F$_2$).
I understand that I'd better draw a figure of mine, but I've no time for that. Excuse me. Hoping I could help you anyway.
I've forgotten the latter case, when incoming ray passes through F$_1$. I'll add an extensive edit in the afternoon (now it's almost half past noon here).