How Is the Angle of Deviation Calculated in a Wollaston Prism?

[amph1bius]

Hello,

I have a question concerning the angle of deviation of the rays that exit out of the wollaston prism. (A wollaston prism is basically composed of two different crystal orientated blocks cemented together, each having a different index of refraction depending on the optical axis)

My homework question is, calculate the angle of deviation given the transverse index of refraction as 1.4864 and the imminent optical axis index of refraction as 1.6584. The wedge angle is 45°.

The ray is incident on the prism as:

------->| \ |

cHEERS,

Andrew

 

[vincentchan]

how hard could this problem be? apply snell's law 2 times... the rest is a tiny little bit of geometry......
tell me which part you get stuck, the snells law? or the geometry?

 

[wais]

Hello Andrew,

The angle of deviation for a Wollaston prism can be calculated using the equation:
𝛿 = (𝛼 - 𝛽)/2
Where 𝛼 is the angle of incidence and 𝛽 is the angle of refraction.

In this case, the angle of incidence is 45° and the angle of refraction can be calculated using Snell's law:
𝛽 = sin^-1 (n1/n2 * sin 𝛼)
Where n1 is the transverse index of refraction (1.4864) and n2 is the imminent optical axis index of refraction (1.6584).

Plugging in the values, we get 𝛽 = sin^-1 (1.4864/1.6584 * sin 45°) = 33.26°.
Substituting 𝛼 and 𝛽 into the angle of deviation equation, we get 𝛿 = (45° - 33.26°)/2 = 5.37°.

Therefore, the angle of deviation for this Wollaston prism is approximately 5.37°. I hope this helps answer your question.