How Does Birefringence Influence Optical Devices?

[unscientific]

Homework Statement

Part (a): Explain how birefringence adds phase difference.

Part(b): Explain how the concept of birefringence is used in a beam splitter.

Part (c): What's the orientation and retardance of retarder-1 to make both beams have same intensity?

Part (d): What's the orientation of polaroid-1 to observe maximum signal?

Part (e): What's the configuration of retarder-2 and polarizer-2 to correct the phase difference?

Part (f): With a ∏/2 phase difference, explain how this allows us to measure refractive index in the gas. Find an expression for ø₁.

Homework Equations

The Attempt at a Solution

Part (a)

For electric vector that is parallel to optic axis, it experiences refractive index $n_e$. For electric vector that is perpendicular to optic axis, it experiences refractive index $n_0$. the phase difference is therefore $\Delta \phi = \frac{2\pi}{\lambda}|n_0-n_e|L$.

Part (b)

Condition for internal reflection is ##\theta < \sin^{-1} (\frac{1}{n^*}). The higher the refractive index, the smaller the critical angle. We can set the angle of incidence to discriminate against the difference refractive indexes.

Part (c)

Since initially both horizontal and vertical components are present, the retarder-1 should be put 45 degrees to vertical to ensure the output has equal vertical and horizontal components.

Part (d)
The bouncing off of mirror introduces a -∏ phase change, so the horizontal and vertical components will be flipped. Therefore Polaroid-1 should be put at 135 degrees counter-clockwise from optical axis (z-axis).

Part (e)
Set the retarder such that it sets back the wave a phase difference of $\delta \phi$. Orientation is same as in part (d), 135 degrees.

Part (f)
If the phase difference is 90 degrees, this means that one will be a sine and the other will be a cosine, with equal amplitudes.

Not sure how to find an expression for $\phi_1$.

 

[unscientific]

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