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As shown in figure a two slit arrangement with a source (S) which emits unpolarised light. I_(0) is the intensity of principle maxima when no polariseer is present. Now a polarised 'P' is placed as shown with its axis whose direstion is not given. Then |
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Answer» the intensity PRINCIPAL MAXIMA is `(5)/(8)I_(0)` `A=A_(_|_)+A_(11)` `A_(_|_)=A_(_|_)^(1)+A_(_|_)^(2)=sin(kx-omega t)` `+A_(_|_)^(@)sin(kx-omega t+phi), A_(11)=A_(11)^((1))+A_(11)^((2))` `A_(11)=A_(11)^(@)[sin(kx-theta t)+sin(kx-omega t+phi)]` where `A_(_|_)^(@), A_(11)^(@)` care the amplitudes of either of the beam in `_|_` and 11 polarizations. `:. "Intensity" =` `= = {|A_(_|_)^(0)|^(2) + |A_(11)^(@)|^(2)}` `[sin^(2)(kx-omega t)(1+cos^(2)phi+2sin phi)` `+sin^(2)(kx-omega)sin^(2)phi]_("average")` `={|A_(_|_)^(0)|^(2) + |A_(11)^(@)|^(2)} ((1)/(2)).2(1+cos phi)` `=2|A_(_|_)^(@)|^(2).(1+cos phi)since |A_(_|_)^(@)|_("average") = |A_(11)^(@)|_("average")` With P: Assume `A_(_|_)^(2)` is blocked: Intensity `= (A_(11)^(1)+A_(11)^(2))+(A_(_|_)^(1))^(2)` `= |A_(_|_)^(@)|^(2) (1+cos phi) + |A_(_|_)^(@)|^(2). (1)/(2)` Given: `I_(0) = 4|A_(_|_)^(@)|^(2)` = intensity wityhout polariser at principal maxima. intensity principal maxima with polariser `= |A_(_|_)^(@)|^(2) (2+(1)/(2))` `= (5)/(8)I_(0)` Intensity at first minima with polariser `= |A_(_|_)^(@)|^(2) (1-1)+(|A_(_|_)^(@)|^(2))/(2) = (I_(0))/(8)` . |
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