Question

In order to introduce a phase difference of $\pi/2$ in a Quarter wave plate, the thickness of the crystal should have a value of... (Given: wavelength = $5893 \times 10^{-10} \, \text{m}$, refractive index for O-ray is $1.65836$ and refractive index for E-ray is $1.48641$)

• A. $8.57 \times 10^{-4} \, \text{mm}$
• B. $8.57 \times 10^{-4} \, \text{cm}$
• C. $8.57 \times 10^{-4} \, \text{m}$
• D. $8.57 \times 10^{-4} \, \text{nm}$

Hint:

Quarter-wave plates are used to create circular polarization from linear polarization by introducing a phase shift of π/2 between orthogonal polarization components.

Answer 1

The Correct Option is A

The thickness \(d\) needed to introduce a phase difference \(\Delta \phi = \pi/2\) in a quarter-wave plate is given by:
\[d = \frac{\lambda}{4 \times |n_e - n_o|}\]
where \(n_e\) and \(n_o\) are the refractive indices for the extraordinary and ordinary rays, respectively. Substituting the given values:
\[d = \frac{5893 \times 10^{-10} \, \text{m}}{4 \times |1.65836 - 1.48641|} \approx 8.57 \times 10^{-6} \, \text{m} = 8.57 \times 10^{-4} \, \text{mm}\]