# Refraction within a prism

Discussion in 'Physics' started by Sparky49, Sep 20, 2011.

1. ### Sparky49 Thread Starter Well-Known Member

Jul 16, 2011
835
417
Hi all,

I'm trying to prove that if I were to shine a beam of light into a prism at an angle of incidence 35°, then the angle of incidence where the light leaves the prisim is 38°. The prisim having a refractive index of 1.55, and being an equilateral triangle.

However, I'm having abit of trouble with this.

I'm aware of the rule: n=sin i/sin r
and: 1/n=sin i/sin r

but I still can't figure it out.

Can anyone tell or show me the steps to achieve this?

Thanks.

2. ### davebee Well-Known Member

Oct 22, 2008
539
46
Don't you also have to specify the angle between the entrance and exit prism faces?

3. ### steveb Senior Member

Jul 3, 2008
2,433
469
So he said "equilateral triangle" which I assume is referring to the shape of the prism. This would make the faces with a 60 degree angle.

I'd like to see some work from the OP before helping. At least a drawing should be shown with the angles defined for clarity. Then an attempt to apply the refraction formula should be made so we can see where the conceptual problem is coming in.

Jul 16, 2011
835
417
Certainly.

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5. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, so i agree with your first calculation to get 21.72 degrees refraction angle.

I think your problem is that you are not defining the angle of incidence on the other face correctly. The angle you show is the refraction angle (kind of), not the incident angle. I said "kind of" because usually the exiting angle is defined relative to the normal of the face, not parallel to the face.

So basically, I think the angle of incidence you are looking for is inside the prism (what you call i2) and not outside. With this assumption, I calculate an angle of incidence of 38.28 degrees. Can you do it now?