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Figure showna two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose directionis not given. If I_0 is the intensity of the principal maxima when no polariser is present, calculte in the present case, the intensity of the principal maxima as well as the first minima. |
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Answer» <P>`(I_(0))/(8)` `A=A_(bot)+A_(||)` `A_(bot)=A_(bot)^(1)+A_(bot)^(2)=A_(bot)^(0)"sin"(kx-wt)+A_(bot)^(0)"sin"(kx-wt+phi)` `A_(||)=A_(||)^(1)+A_(||)^(2)=A_(||)^(0)["sin"(kx-wt)+"sin"(kx-wt+phi)]` where `A_(bot)^(0),A_(||)^(0)` are the amplitudes fo either of the beam in `bot` and `||` polarizations. ` :. ` intensity `={|A_(bot)^(0)|^(2)+|A_(||)^(0)|^(2)}[sin^(2)(kx-wt)` `(1+cos^(2)phi+2sin phi)+"sin"^(2)(kx-wt)"sin"^(2)phi]_("average")` `={|A_(bot)^(0)|^(2)+|A_(||)^(0)|^(2)}((1)/(2)).2(1+cosphi)` `= 2|A_(bot)^(0)|^(2)(1+cos phi),"since"|A_(bot)^(0)|_("average")=|A_(||)^(0)|_("average")` With P: Assume `A_(bot)^(2)` is blocked: Intensity`=(A_(||)^(1)+A_(||)^(2))^(2)+(A_(bot)^(1))^(2)` `=|A_(bot)^(0)|^(2)(1+cos phi)+|A_(bot)^(0)|^(2).(1)/(2)` `(I_(0)=4|A_(bot)^(0)|^(2)=` Intensity without polariser at principal maxima). Intensity at first maxima with polariser `=|A_(bot)^(0)|^(2)(2+(1)/(2))=(5)/(8)I_(0)` Intensity at first minima with polariser, `|A_(bot)^(0)|^(2)(1-1)+(|A_(bot)^(0)|^(2))/(2)=(I_(0))/(8).` |
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