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Refraction in a prism, Snell's law: Please help! 
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#1
Mar2912, 12:00 AM

P: 37

1. The problem statement, all variables and given/known data
A horizontal beam of light enters a 459045 prism at the center of it's long side, as shown below. The emerging ray moves in a direction that is 34˚ below the horizontal. What is the index of refraction for the prism? 2. Relevant equations n[itex]_{1}[/itex]sinø[itex]_{1}[/itex] = n[itex]_{2}[/itex]sinø[itex]_{2}[/itex] 3. The attempt at a solution n1 x sin(i) = n2 x sin(r1): (1st refrection) and n2 x sin(r2) = n1 x sin(34˚): (2nd refraction) ^^Here n1 is the refractive index of air, n2 is the refractive index of prism, r1 and r2 are the angles of reflection at the two surfaces, and i is the incidence angle. r1 + r2 = 45˚ or r2 = 45˚  r1. Substituting the above value of r2 in the equation(2),we get n2 x sin(45  r1) = 1 x sin(34˚) ...and then I'm stuck! To simplify what I've figured out: first refraction: sin45 = n sin a second refraction: n sinb = sin34 and using geometry: a + b = 45 BUT HOW DO I FIND a and b?!?!? 


#2
Mar2912, 04:51 AM

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#3
Mar2912, 05:29 AM

P: 37

but I'm not sure where to take it from there... I tried setting up something like: sin(45)/sin(34) = sin(a)/sin(b) sin(45)/sin(34) = sin(a)/sin(45a) sin(45)/sin(34) = sin(a)/[sin(45)cos(a)  cos(45)sin(a)] but I don't even know if any of that ^^ is right, I could be going in a totally wrong direction. 


#4
Mar2912, 05:36 AM

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P: 26,148

Refraction in a prism, Snell's law: Please help!
Hi ObviousManiac!



#5
Mar2912, 07:16 AM

P: 37

...doesn't this just introduce a new variable? (n?) 


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