Is there a shortcut for finding theta4 using Snell's Law when theta2 is known?

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The discussion revolves around finding a shortcut to calculate theta4 using Snell's Law when theta2 is known, particularly in the context of a prism. Participants explore the relationship between angles theta2, theta3, and the apex angle of the prism, with some confusion about the geometry involved. It is clarified that the angles in the triangle formed by the light ray and prism sides must sum to 180 degrees, leading to the equation theta3 = A - theta2. Ultimately, the solution involves applying Snell's Law twice to derive theta4 from the known values. The conversation highlights the importance of understanding the geometric relationships in the prism to solve for the angles accurately.
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Snells Law prism - from theta2 -> theta3?

Hi, I was wondering if there's a formula to go straight from theta1 to theta 4, when the apex angle is known. Thanks[PLAIN]http://cord.org/cm/leot/course06_mod07/Fig3.gif

My problem is getting from theta2 to theta3.

Thanks

ps. This is not a homework question, i didnt even take physics in school when i had the chance!
 
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Consider the triangle with the angles theta2 and theta3, the one with two dotted sides and the third side made by the light ray in the prism.
The third angle in this triangle is 180-A. (To see that is so, consider the quadrilateral made by the two normals and the sides of the prism).
This will give you the relation between theta2 and theta3.
 
Thanks for getting back to me.

However, I'm still struggling to see how that third angle is 180-apex. I believe that the relationship between theta2 and theta3 is simply the apex = theta2 + theta3 but I can't see why!

Thanks again!
 
PLEASE DISREGARD - See next comment.

First off, by looking at the drawing, I am assuming the Normal lines are || to the rays and not perp. If they are perp then by definition theta 1 = theta 4 = 90.

theta 1 = theta2 + beta = theta 3 + gamma = theta 4

You don't need A. It does nothing for you. 180-A does not equal the third angle in the theta 2-3 triangle except for one frequency of the incoming ray. Sigma is dependent on multiple factors. The angle A is only one of those.

PLEASE DISREGARD - See next comment.
 
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Pyle said:
First off, by looking at the drawing, I am assuming the Normal lines are || to the rays and not perp.

Thanks!

Aren't the normal line perpendicular to the prism edges rather than parallel to the rays though?

Basically all i want to know is the exit angle for a set incident angle, n1 and n2.
 
glyon said:
I believe that the relationship between theta2 and theta3 is simply the apex = theta2 + theta3 but I can't see why!

Consider the triangle bounded by the sides of the prism at the top, and the light ray through the prism at the bottom. One of its angles is A. The other two angles aren't labeled, but they're related to \theta_2 and \theta_3 (how?). What do those three angles add up to?
 
Oops,
Wasn't paying attention.
180-A is the third angle in the theta 2-3 triangle.
Just run it through Snell's law twice.

Sorry nasu, I was hasty.
 
Thanks,

I think I figured it out now:

theta3 = A - theta2, then do snells law again for theta4.
 
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