Answer to Question #202880 in Optics for anuj

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

$(n_e-n_o)d=\frac{\lambda}{4},$

$n_e=n_o+\frac{\lambda}{4d},$

$n_e=1.5442+\frac{6}{4\cdot 1.65}=2.4533.$