Calculating Min Incident Angle for Ray Emerging from Prism

AI Thread Summary
The discussion focuses on calculating the minimum incident angle for a ray to emerge from a prism with an apex angle of 74.3° and a refractive index of 1.55. The user initially solved for the angle of deviation (delta) and found it to be 65.4 degrees, leading to an incident angle of 28.3 degrees. However, there is confusion regarding the necessity of calculating delta and the origin of the "/2" in the equations. Participants express uncertainty about the correct approach to finding the incident angle without determining the critical angle. The conversation highlights the complexity of applying the relevant equations to solve the problem effectively.
NikkiNik
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Homework Statement


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?



Homework Equations



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

delta = incindence angle + alpha/2

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
 
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Hi NikkiNik! :smile:

(have an alpha: α and a delta: δ and a phi: φ :wink:)
NikkiNik said:
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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