Question regarding optics(prism)

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The discussion centers on determining the index of refraction, n, for a 90-degree triangular prism using Snell's law. The user initially struggles to relate the angles of incidence and refraction, particularly θ1 and θ4, while trying to eliminate variables. After some back-and-forth, they realize that using the relationship between cosine and sine can help simplify the problem. Ultimately, they successfully express n in terms of θ, indicating that the solution is achievable with the right approach. The conversation highlights the importance of revisiting fundamental relationships in optics to solve complex problems.
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Homework Statement



figure shows a 90 degrees triangular prism of refractive index n whereby light enters at point A with an incident angle of θ. The light then refracts at point B with an angle of refraction of 90 degrees.

determine index of refraction, n in terms of θ. refer to figure attached.


Homework Equations



using snell's law of refraction,

n.sinθ=n1.sinθ1

The Attempt at a Solution



first I used an snell's law to get an eqn

1.sinθ=n.sinθ1----------------------(1)

using properties of angles and triangles I got
θ2, θ3 and θ4.
θ1=θ3, θ2=θ4.

using snell's law again I got
n.sinθ4=1.sin 90 degrees
n.sinθ4=1

I tried using trigonometry to compute the thetas together and eventually eliminate θ1. But I was not able to do so. Am I missing something?
Since I got stuck, I thought if θ1=45 degrees, getting n in terms of θ is not a problem. But I reasoned to myself that it cannot be possible as I do not know θ4 and n. So making an assumption is not right. I think? I could be wrong though.. hmm..

help would be greatly appreciated and I am sorry if posted anything wrongly such as the format or thought process since this is the first time I am posting.

Thanks!

Syeh
 

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You have two equations

sinθ = n sinθ1
n sinθ4 = 1

Can you find a relation between θ1 and θ4?
 
hmm.. thanks for replying.. I went through that before.. basically 1=ncosθ1 since 90 degrees-θ4=θ1=θ3. and if i were to combine both eqns together, it will result in

sinθ=tanθ1.

the thing is I couldn't eliminate θ1 but I eliminated n instead. I even tried substituting other values of relevant θ into the equation and it got me more irritated. It feels like I am missing something right before my eyes and I cannot comprehend what is it that I missed. oh man. Anyway, this qns is the first part of a 4 parts 12 marks qns. So the solution shouldn't be too complicated. Thanks for replying again.
 
shakeysg said:
basically 1=ncosθ1

So cosθ1=1/n. What is sinθ1 in terms of n?

ehild
 
oh my. hahaa. thanks for making me rethink the qns. i managed to n in terms of theta. thanks again! : D
 
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