Dear Community,
It would be very kind if someone could explain the column - Z Normal in the Single Ray Trace window (settings = Direction cosines).
Is the Z-normal a Direction Cosine or is it the Z-direction vector ?
Why is the value in this column always negative regardless of the type of the surface and direction of ray ?
Best answer by David.Nguyen
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+2
Hi Jacob,
I think the normal is expressed as a direction cosine because if you select a different Type in the settings, these columns disappear. The Z-normal must be the cosine of the angle of the normal with the Z axis. The other two components of the normal direction cosine are given by the other two columns (X-normal, and Y-normal). Its not a proof, but you can check that in every row, the square root of the sum of the squared values add to unity.
I hope this helps.
Take care,
David
Hi
Thanks for the reply and bring in more clarity. I see now that the RANZ() value and the Z- Normal column values are same, so it must be the Z- direction Cosine.
Now the second part of my question. All the values in the Z- Normal column are between 0 and -1, that means all the angles between Z-axis and the Surface normal (at the ray surface interface) is an angle between 90 and 180 degrees. But in the design there is a situation where the ray is travelling from left to right, upwards and meets a negative radius surface at the top half of the lens and is refracted. Even in this situation the RANZ() value is negative.
Am I understanding the fundamentals wrong here. Sorry i can not attach the Zemax files.
+2
Hi Jacob,
If the Z direction cosine of the normal is close to minus one. Then, the normal is close to being parallel with the Z axis but in opposite direction. So, the situation is more like I tried to depict below:
Does that make sense? Notice the angle the normal makes with the ray is the same either way.
Take care,
David
Hi
You are right. The way you define the direction of Normal is the right one.
I defined the direction of Normal as a vector moving away from the centre of the curvature.
Thanks again.
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