Optical Prism: Refractive Index & Light Rays

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Homework Help Overview

The discussion revolves around the behavior of light as it interacts with a 45°-90°-45° prism made of crown glass, specifically focusing on the refractive index required for the surrounding medium to achieve certain optical outcomes based on wavelength. The original poster seeks to determine the refractive index that allows longer wavelengths to exit the prism while reflecting shorter wavelengths.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of Snell's law in relation to the refractive index and the behavior of light at different wavelengths. There are inquiries about the necessary conditions for light to exit or reflect based on its wavelength.

Discussion Status

The conversation is ongoing, with participants providing insights into the relationship between wavelength and refractive index. Some have noted the importance of understanding how light of a specific wavelength behaves in relation to the prism's geometry and refractive properties, while others are seeking clarification on the definitions and implications of terms used in the discussion.

Contextual Notes

There are mentions of missing information regarding the full context of the problem, including the absence of a figure and specific angles of incidence. Participants are encouraged to share relevant equations and their attempts at solutions, indicating a collaborative effort to explore the problem further.

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MODERATOR'S NOTE: THIS THREAD HAS BEEN MOVED HERE FROM ANOTHER FORUM, SO THERE IS NO TEMPLATE

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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