Question

Consider a retarder with refractive indices \(n_e = 1.551\) and \(n_o = 1.542\) along the extraordinary and ordinary axes, respectively. The thickness of this retarder for which a left circularly polarized light of wavelength \(600 \, \text{nm}\) will be converted into a right circularly polarized light is ........... µm. (Round off to 2 decimal places).

Hint:

A half-wave plate introduces a phase shift of \(\pi\) between ordinary and extraordinary rays, converting left circularly polarized light to right circularly polarized light.

Answer 1

Correct Answer: 33.32

Step 1: Retardation condition.
For conversion between opposite circular polarizations, phase difference = \(\pi\) radians. \[ \Delta \phi = \frac{2\pi (n_e - n_o) t}{\lambda} = \pi \]

Step 2: Solve for thickness \(t\).
\[ t = \frac{\lambda}{2(n_e - n_o)} = \frac{600 \times 10^{-9}}{2(1.551 - 1.542)} = \frac{600 \times 10^{-9}}{0.018} = 33.33 \times 10^{-6} \, \text{m} \] \[ t = 33.33 \, \mu\text{m} \]

Step 3: Conclusion.
Hence, the thickness of the retarder = 33.33 µm.