InterviewSolution
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InterviewSolution
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White natural light falls on a system of two crossed Nicol prisms having between them a quartz plate 1.50 mm thick, cut parallel to the optical axis. The axis of the plate froms an angle of 45^(@ with the principle directions of the Nicol prism. The light transmitted through that system was split into the spectrum. How many dark fringes will be observed in the wavelength interval from 0.55 to 0.66mu m? The difference of refractve indices for ordinary and extraordianry rays in that wavelength interval is assumed to be equal to 0.0090. |
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Answer» Solution :As in the previous two problems the QUARTZ plate will introduces a phase difference `DELTA`. The LIGHT on passing through the plate will remain plane polarized only for `delta = 2kpi` or `(2k + 1)pi`. In the latter case the plane of polarization of the light incident on the plate will be rotated by `90^(@)` by its so light passing through the analyser (which was originally crossed) will be a MAXIMUM. Thus dark bands will be observed only for those `lambda` for which `delta = 2k pi` Now `delta = (2pi)/(lambda) (n_(e) - n_(0))d = (2pi)/(lambda) xx 009 xx 1.5 xx 10^(-3)m` `= (27pi)/(lambda)(lambda` in `mu m`) For `lambda = 0.55` we get `delta = 49.09 pi` Choosing `delta = 48pi, 46pi, 44pi, 42pi` we get `lambda = 0.5625mu m, lambda = 0.5870 mu m, lambda = 0.6136mu m` and `lambda = 0.6429mu m`. These are the only values between `0.55 mu m` and `0.66 mu m`. Thus there are four bands. |
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