Question

3. [20 pts] Calcite (a negative uniaxial crystal) is cut into the shape of a plate with the optic axis parallel to the plate face. A linearly polarized light impinged on the face at normal angle. The electric field of the light is oscillating along a line 45o tilted from the optic axis. You want to make a half-wave plate with this. Determine the right thickness of the plate. To facilitate the fabrication, it needs to be at least 1 mm but beyond that point, the thinner the better. (Use Table 6.1) TABLE 6.1 Principal refractive indices of some optically isotropic and aniso crystals (near 589 nm, yellow Na-D line) Optically isotropic 1.510 2.417 1.434 to 1.309 1.5442 2.616 nto 1.658 1.669 2.29 Glass (crown) Diamond Fluorite (CaF2) Uniaxial-Positive Ice Quartz Rutile (TiO2) 1.3105 1.5533 2.903 Uniaxial-Negative Calcite (CaCO3) Tourmaline Lithium niobate (LiNbO) 1.486 1.638 2.20 Biaxial n2 1.5936 Mica (muscovite) 1.5601

Answer 1

Given that, calcite negative uniaxial crystal is cut the shape of plate with the optic axis the parallel to the plate face� path difference for half plate = [(wave length)/2] alpha (1 - refractive index of calcite)� path difference of half plate is = (thickness of half plate) * cos(45)� (thickness of half plate) * cos(45) = [(wavelength)/2] alpha (1 - refractive index of calcite)� the auction with the value of wave length of light and refractive index of calcite, using above relation u can get your value� (thickness of half plate) - {wavelength/2} * (1 - refractive index of calcite)}/(cos 45))