Optical Prism: Refractive Index & Light Rays

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A ray of white light is incident on a 45o – 90o – 45o prism (ray 1 in the figure on the left). The prism is

constructed from crown glass with a refractive index of n2 = 1.5205 at the wavelength of the Fraunhofer C line (656.3 nm).

What would be the required refractive index of the surrounding medium (n1) to allow light at longer wavelengths than 656.3 nm to exit the prism along path 2 while light at shorter wavelengths is reflected along path 3.
 
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Bystander said:
No figure, no prism orientation, no incidence angle --- we do need the rest of the question.
Screen Shot 2015-09-27 at 12.57.53 PM.png
 
haruspex said:
That's progress. Next, you need to post any equations you have been taught that you deem may be relevant, and an attempt at solution.
  • I have snell's law : n sin theta 1 = n2 sin theta 2
 
princemartin1 said:
  • I have snell's law : n sin theta 1 = n2 sin theta 2
Ok, but the refractive index depends somewhat on wavelength. You are given the refractive index for a particular wavelength, and you want longer wavelength light to pass through but shorter wavelengths to be internally reflected. That being the case, where do you think light of the given wavelength will go?
 
haruspex said:
Ok, but the refractive index depends somewhat on wavelength. You are given the refractive index for a particular wavelength, and you want longer wavelength light to pass through but shorter wavelengths to be internally reflected. That being the case, where do you think light of the given wavelength will go?
It will exit through 2 and also the normal has to pass through the point 2 and 3
 
princemartin1 said:
It will exit through 2 and also the normal has to pass through the point 2 and 3
Longer wavelengths than 656.3nm will exit along paths like path 2, while those shorter will take path 3. But what will a wavelength of exactly 656.3 do?

By the way, you need to understand that 'path 2' is only illustrative. All wavelengths longer than 656.3 will exit that face of the prism, but at different angles depending on the wavelength. Shorter wavelengths, on the other hand, will follow exactly path 3.

I didn't understand your remark about the normal. Normal to what? '2' and '3' are not points, they are paths.