Calculating Refractive Index for Extraordinary Ray in Calcite Crystal

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SUMMARY

The discussion focuses on calculating the refractive index for the extraordinary ray in a calcite crystal when a light beam makes a 30-degree angle with the optic axis. The relative dielectric constants for light polarized parallel and perpendicular to the optic axis are given as 2.208 and 2.749, respectively. The correct refractive index for the extraordinary ray is determined to be 1.61. The participants consider using Snell's law and discuss the implications of total internal reflection in a polarizing beam-splitter.

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  • Familiarity with the concept of extraordinary rays in birefringent materials
  • Basic principles of total internal reflection and polarizing beam-splitters
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Homework Statement


A beam of light travels through a calcite crystal such that its wave vector makes an angle of 30degrees with the optic axis. Calculate the refractive index experienced by the extraordinary ray if the relative dielectric constants for light polarized parallel and perpendicular to the optic axis are 2.208 and 2.749, respectively.


Homework Equations


Perhaps Snell's law? : n1 sin(theta1) = n2 sin(theta2)
Should I set it as Brewsters angle?

The Attempt at a Solution



I've also tried square-rooting 2.749 which gives me a similar answer to the actual solution.

The actual solution should be 1.61
 
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Could this be to do with total internal reflection in a polarizing beam-splitter?
 

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