All Problems

Polarization of Light

Problem 5.188

Light with wavelength λ\lambda falls on a system of crossed polarizer PP and analyzer AA between which a Babinet compensator CC is inserted (Fig. 5.33). The compensator consists of two quartz wedges with the optical axis of one of them being parallel to the edge, and of the other, perpendicular to it. The principal directions of the polarizer and the analyser form an angle of 4545^{\circ} with the optical axes of the compensator. The refracting angle of the wedges is equal to Θ(Θ1)\Theta(\Theta \ll 1) and the difference of refractive indices of quartz is neno.n_{e}-n_{o} . The insertion of investigated birefringent sample S,S, with the optical axis oriented as shown in the figure, results in displacement of the fringes upward by δx mm\delta x \mathrm{~mm}. Find: (a) the width of the fringe Δx\Delta x; (b) the magnitude and the sign of the optical path difference of ordinary and extraordinary rays, which appears due to the sample SS

Reveal Answer
5.188.(a)Δx=1/2λ(nen0)Θ, (b) d(n0ne)==2(nen0)Θδx<0\begin{aligned} & 5.188 .(a) \Delta x=1 / 2 \lambda\left(n_{e}-n_{0}\right) \Theta, & \text { (b) } d\left(n_{0}^{\prime}-n_{e}^{\prime}\right)=\\ =&-2\left(n_{e}-n_{0}\right) \Theta \delta x<0 \end{aligned}