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A quartz wedge with refracting angle Theta = 3.5 is inserted between two crossed Polaroids. The optical axis of the wedge is parallel to its edge and forms an angle of 45^(@) with the principal directions of the Polaroids. On transmittion of light with wavelength lambda = 550 nm through this system, an interference fringe pattern is formed. The width of each fringe is Deltax = 1.0 mm. Find teh difference of refractive indices of quartz for ordinary and extraodianry rays at the wavelength indicated above. |
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Answer» Solution :If a ray TRAVERSE the WEDGE at a distance `x` below the joint, then the distance that the ray moves in the wedge is `2x tan((Theta)/(2))` and this cause a phase difference `delta = (2pi)/(lambda) (n_(e) - n_(0)) 2x tan ((Theta)/(2))` between the `E` and `O` wave components of the ray. For a general `x` the resuling light is ellipically polarized and is nor completely quenched by the analyser polariod. The CONDITION fro complete quenching is `delta = 2K pi bar` DARK fringe That for maximum brightness is `delta = (2k + 1) pi-` bright fringe. The fringe width is given by `Deltax = (lambda)/(2(n_(e) - n_(0))tan((Theta)/(2)))` Hence `(n_(e) - n_(0)) = (lambda)/(2Delta xtan Theta//2)` `tan(Theta//2) = tan 175^(@) = 0.03055`, using `lambda = 0.55mu m` and `Delta x = 1mm`, we get `n_(e) - n_(0) = 9.001 xx 10^(-3)` 
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