Using Snell's law to find the entering/exiting rays' angles

[anthesco]

Homework Statement

This is one of my homework questions. I think I have the right answer, but I don't understand how to figure out the "given" angle, so if someone could explain it to me, that would be great! Be sure to look at the picture attached...it's what's tricking me.

A prism whose cross section is shaped like an isosceles right triangle is made from a material with index of refraction, n = 1.31. Find the angle θ of the entering/exiting rays that travel parallel to the lower side (in degrees).

Homework Equations

Snell's law, n1*sin[itex]\theta[/itex]1=n2*sin[itex]\theta[/itex]2

The Attempt at a Solution

n1=1.31
n2=1
[itex]\theta[/itex]1=45°

1.31*sin[itex]\theta[/itex]1=1*sin[itex]\theta[/itex]2

[itex]\theta[/itex]2=67.867

What I need to know is how to figure out that the first angle is 45°.

 

[invertioN]

I don't see any attached pictures.

 

[Sunil Simha]

If θ₁ is given and you have to find the angle of the exiting ray w.r.t. the normal, then you only require some geometry here as the ray travels parallel to the base.

(Hint:Construct the normals on the faces until they meet)

 

[rude man]

What are the base angles?

Then use plane geometry to figure theta1, the angle between the normal to the left side of the prism and the horizontal ray inside said prism.

 

[anthesco]

I'm unsure of the base angles, it isn't given to us. I now realize that I use 45 deg because you're supposed to use the top angle of the triangle (so, where the right angle is) and divide that by two, but I don't understand why you use that angle over all the other ones...

 

[rude man]

Sum of angles = 180 and you're given the top one = 90. Considering the two sides are of equal length, don't you think you can come up with the base angles?

Then look at the figure you provided us and you should be able to figure out the angle between the normal to the left side and the flat beam section inside. If n1, n2 and sin(theta2) are given, you know how to compute theta1. Look at the angle between the flat inside beam and the left side of the prism. Look like the left base angle? So what is theta2, since the angle between the prism's left side and its normal is by definition 90 deg?

BTW I'm using "1" for air and "2" for inside the prism. It's the logical choice since a beam is coming FROM the air (1) TO the glass (2).

 

[anthesco]

That makes a lot more sense now! The issue I was having was figuring out which angle I needed to designate theta2. Man, it's been a long time since I've used geometry. Thank you!