Optical Prism: Refractive Index & Light Rays

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A 45o – 90o – 45o prism made of crown glass with a refractive index of 1.5205 is discussed in relation to the behavior of light at different wavelengths. For light longer than 656.3 nm, it will exit the prism along path 2, while shorter wavelengths will be internally reflected along path 3. The refractive index is wavelength-dependent, affecting how light behaves at the specified wavelength. The conversation emphasizes the need to understand the implications of Snell's law and the concept of the normal in relation to light paths. The discussion concludes that while longer wavelengths will exit the prism, the behavior of light at exactly 656.3 nm remains a point of inquiry.
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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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princemartin1 said:
(ray 1 in the figure on the left).
No figure, no prism orientation, no incidence angle --- we do need the rest of the question.
 
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
 
princemartin1 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.
 
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.
 
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