Index of refraction of the prism

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Homework Statement


5) A beam of monochromatic light enters an equilateral triangular prism at an angle of incidence of 30.0º. If the index of refraction of the prism is 1.53, at what angle with respect to the surface of the prism does the beam emerge from the other side of the prism?


Homework Equations





The Attempt at a Solution


i tried drawing it out, but i have absolutely no idea.
 
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{smile} said:

Homework Statement


5) A beam of monochromatic light enters an equilateral triangular prism at an angle of incidence of 30.0º. If the index of refraction of the prism is 1.53, at what angle with respect to the surface of the prism does the beam emerge from the other side of the prism?


Homework Equations





The Attempt at a Solution


i tried drawing it out, but i have absolutely no idea.

Show us a drawing and your work. A "relevant equation" might be _____'s Law, eh?
 


Use Snell's Law: [itex]n_1\sin{\theta_1} = n_2\sin{\theta_2}[/itex]
 


i don't know how to draw on here. :( and Snells law right? i cannot figure out how to do that. if Sin(feta)1/Sin(feta)2 = N2/N1, i only have two of the variables! don't i at least need three?
 


{smile} said:
i don't know how to draw on here. :( and Snells law right? i cannot figure out how to do that. if Sin(feta)1/Sin(feta)2 = N2/N1, i only have two of the variables! don't i at least need three?

What are your unknowns?
 


Assume that the ray of light is initially in air. The index of refraction in air is approximately one. Also, the symbol [itex]\theta[/itex] is called theta, not feta.
 


we don't know Sin(feta) 1 or N1. we only know the angle of refraction and the index of refraction.
 


ok let me try this again,

Homework Equations


N1sinθ=N2sinθ


The Attempt at a Solution


The light will enter the prism and refract, cross the prism, hit the other side, exit and refract. so there will be lots of angles for me to find, i just don't know how to solve for them except for this one.

N1sinθ=N2sinθ
Θ2=sin-1((1.0003)(sin30)/1.53)
Θ= 19 degrees
 


hey {smile}

you need to sketch so you can understand it well, you need to do it twice [Snell's Law] so you can get the angle of the emerged beam with respect to the surface of the prism [it’s theta 3], try to work out the other theta illustrated in the figure.


http://img197.imageshack.us/img197/1777/13527457.jpg
 
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Note that, in the above drawing, θ1 and θ3 are defined in such a way that Snell's Law should be written in terms of their cosines on the "air side" of the ray.
 


kuruman said:
Note that, in the above drawing, θ1 and θ3 are defined in such a way that Snell's Law should be written in terms of their cosines on the "air side" of the ray.


you’re right [we’ll get the same answers], but according to Snell's law, these angles are measured with respect to the normal lines which is perpendicular to the boundaries [that’s why I drew them, but then forgot about them:-p], anyhow, here’s the exact one;

http://img30.imageshack.us/img30/1359/10703724.jpg
 
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Great. Now see if you can find what the sum of the two internal angles is

θ2 + θ = ?

This will help you connect the first angle of refraction θ2 to the internal angle of incidence θ.
 


jgens said:
Also, the symbol [itex]\theta[/itex] is called theta, not feta.

Theta, feta what's the difference - it's all Greek anyway. :smile:
 


kuruman said:
Great. Now see if you can find what the sum of the two internal angles is

θ2 + θ = ?

This will help you connect the first angle of refraction θ2 to the internal angle of incidence θ.

was that directed to me? :biggrin:

I'm only helping the OPer
 


drizzle said:
was that directed to me? :biggrin:

I'm only helping the OPer

It was directed to {smile} who I believe is trying to solve this.