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Light with wavelength lambda falls on a system of crossed polarizer P and analyzer A between which a Babinet compesator C is inserted (Fig.). The compensator consists of two q2uartz 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 from an angle of 45^(@) with the optical axes of the compensator. The refracting angle of the wedges is equal to Theta(Theta lt lt 1) qand the difference of refractive indices of quartz is n_(e) - n_(0). The insertion of investigated birefringent sample S, with the optical axis oriented as shown in the figure, results in displacement of the fringe upward by deltax mm. Find: the width of the fringe Deltax, (b) the magnitude and the sign of the optical path difference of ordinary and extraordianry rays, which appears due to the sample S. |
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Answer» Solution :The light from `P` is plane polarized with its electric vector vibrating at `45^(@)` with the plane of the paper. At first the sample `S` is absent Light from `P` can be resolved into components vibrating in and perpendicuar to the plane of the paper. The former is the `E` ray in the left half of the Babinet components and the latter is the `O` ray. In the right half the nomenclature is the opposite. In the compenstor the two components acquire a phase difference which depends on the relative position of the ray. If the ray is incident at a distance `x` above the central line through the compensator then the `E` ray acquires a phase `(2pi)/(lambda) (n_(E) (l - x) + n_(0) (l +x))tan Theta` while the `O` ray acquires `(2pi)/(lambda) (n_(0)(l - x) +n_(E) (l + x)) tan Theta` so the phase difference between the two reays is `(2pi)/(lambda) (n_(0) - n_(E)) 2 x tan Theta = delta` we get dark FRINGES when ever `delta = 2k pi` because then the emergent light is the same as that coming form the polarizer and is quenched by the analyser. {If `delta = (2k +1) pi`, we get BRIGHT fringes beacuse in this case, the plane of polarization of th emergent hight has rotated by `90^(@)` and is therefore fully transmitted by the analyser.} If follows that the fringe width `Deltax` is given by `Delta x = (lambda)/(2|n_(0) - n_(E)|tan Theta)` (b) If the finges are displaced upwards by `delta x`, then the path difference intoduced by the sample between the `O` and the `E` rays must be such as to be exactly CANCELLED by the compensator. Thus `(2pi)/(lambda) [d (N'_(0) - n'_(E)) + (n_(E) - n_(0)) 2 delta x tan Theta] = 0` or `d(n'_(0) - n'_(E)) =- 2(n_(E) - n_(0)) delta x Theta` usng `tan Theta_(~=) Theta`.  
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