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Incident angle

  • Thread starter NikkiNik
  • Start date
1. The problem statement, all variables and given/known data
If the apex angle of a prism is f = 74.3° (see figure below),

http://i32.photobucket.com/albums/d2/NikkiNik88/apex.gif

what is the minimum incident angle for a ray if it is to emerge from the opposite side (i.e., not be totally internally reflected), given n = 1.55?



2. Relevant equations

sin(alpha + delta)/2 =nsin(alpha/2)

delta = incindence angle + alpha/2

3. The attempt at a solution

delta = min angle of deviation
alpha = apex angle

I solved for delta and got 65.4 deg

Then I used the second equation to solve for the incident angle and got 28.3 deg

I'm not even sure if I should go so far as to solve for delta. I'm thinking there is a simpler way
 

tiny-tim

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Hi NikkiNik! :smile:

(have an alpha: α and a delta: δ and a phi: φ :wink:)
sin(alpha + delta)/2 =nsin(alpha/2)

delta = incindence angle + alpha/2
Sorry, I've no idea what you're doing. :redface:

Where does "/2" come from?

And what are your first incident angle, refracted angle, and second incident angle? :confused:
 
I found in the equations in my book

(1) (sin(α + δ))/2 = n*sin(α/2)

All I'm given is the apex angle and n and I'm not sure where to go from there. Since I'm not supposed to find the critical angle I'm not sure how to go about finding the incident angle.
 

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