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Natural monochromatic light of intensity I_(0) falls on a system of two Polaroids between which a crystalline plate is inserted,cut parallel to its optical axis. The plate introduces a phase difference delta between the ordinary and extraodianry rays. demonstrate that the intensity of light transmitted through that sysytem is equal to I = (1)/(2)I_(0)[cos^(2)(varphi - varphi') - sin 2varphi. sin 2varphi' sin^(2)(delta//2)], where varphi and varphi' are the angles between the optical axis of the crystal and the principle directions of the Polaroids. In particular, consider the cases of crossed and parallel Polaroids. |
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Answer» Solution :Light emerging from the first polaroid is plane polarized with amplitude `A` where `N_(1)` is the principle direction of the polaroid and a vibration of amplitude can be resolved into two vibration : `E` wave vibration along the optic axis of amplitude `A cos varphi` and the `O` wave with vibration perpendicular to the optic axis and having an amplitude `A SIN varphi`. These acquire a phase difference `delta` on PASSING through the plane. The second polaroid transmits the components `A cos varphi cos varphi'` and `A sin varphi sin varphi'` What emerges from the second polaroid is a set of two plane polarized waves in the same direction and same plane of polarization but phase difference `delta`. They interfere and PRODUCE a wave of amplitude squared `R^(2) = A^(2) [cos^(2) varphi cos^(2) varphi' + sin^(2)varphi sin^(2) varphi' + 2cos varphi cos varphi' sin varphi sin varphi' cos delta]`, using `cos^(2) (varphi - varphi') = (cos varphi cos varphi' + sin varphi sin varphi')^(2)` `= cos^(2)varphi cos^(2)varphi' + sin^(2)varphi sin^(2) varphi' + 2cosvarphi cosvarphi' sinvarphi sin varphi'` we easily find `R^(2) = A^(2) [cos^(2)( varphi -varphi') - sin 2varphi sin2 varphi' sin^(2)((delta)/(2))]` Now `A^(2) = I_(0)//2` and `R^(2) = I` so the result is `I = (1)/(2)I_(0) [cos^(2)( varphi -varphi') -sin 2 varphi sin2varphi' sin^(2) ((delta)/(2))]` Special cases :- Crossed polaroids : Here ` varphi -varphi' = 90^(@)` or`varphi' =varphi - 90^(@)` and `2 varphi' = 2 varphi - 180^(@)` Thus in this case `I = I_(bot) = (1)/(2)I_(0)sin^(2)2 varphi sin^(2) ((delta)/(2))` Parallel polaroids : Here ` varphi =varphi'` and `I = I_(||) = (1)/(2)I_(0) (1-sin^(2)2 varphi sin^(2)((delta)/(2)))` with `delta = (2pi)/(LAMBDA)Delta`, the conditions for the maximum and minimum are easily found to be that shown in the answer. 
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