Question

For a crystal, the refractive index n_o for the o-ray is 1.5442 for light of wavelength 

6×10^-5 m. The least thickness of the crystal used as a quarter wave plate is found to 

be 1.65×10^-5 m. Determine the refractive index ne for the e-ray in this crystal.

Answer 1

Refractive index for ordinary ray, $\mu_o=1.5442$

Wavelength, $\lambda=6\times10^{-5}\space m$

Thickness of the crystal, $t=1.65\times10^{-5}\space m$

Refractive index for o-ray = $\mu_e$

Thickness, $t=\dfrac{\lambda}{4(\mu_e-\mu_o)}$

$\mu_e=\mu_o+\dfrac{\lambda}{4t}$

$\mu_e=2.45$