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Fig. shows a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I_(0) is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima. |
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Answer» Solution :As is known, RESULTANT amplitude is the sum of amplitudes of either beam in perpendicular and parallel polarization, i.e.,`A = A_("perp".) + A_("parallel")` Now,`A_("perp".) = A_("perp".)^(1) + A_("perp".)^(2)` `= A_("perp".)^(0) sin (kx - OMEGA t) + A_("perp".)^(0) sin (kx - omega t + phi) ` Similarly,`A_("parallel") = A_("parallel")^(1) + A_("parallel")^(2)` `= A_("parallel")^(0)[sin(kx - omega t) + sin (kx - omega t + phi)` `:.`Inetensity `= A_("perp".)^(2) + A_("parallel")^(2) = [A_("perp".)^(0) + A_("parallel")^(0^(2))] [sin^(2)(kx - omega t) + (kx - omega t + phi)]` average `= [ A_("perp".)^(0^(2)) + A_("parallel")^(0^(2)) ]((1)/(2)) 2(1 + COS phi)` As `|A_("perp".)^(0^(2))|"average" = |A_("parallel")^(0^(2))|"average"`, THEREFORE, Without `P`, Intensity `= 2|A_("perp".)^(0^(2))| (1 + cos phi)`...(i) With polariser `P`, suppose `A_("perp")^(2)` is blocked. `:.` Intensity `= {A_("parallel")^(1) + A_("parallel")^(2)}^(2) + (A_("perp".)^(1))^(2) = |A_("perp".)^(0) |^(2) (1 + cos + phi) + | A_("perp".)^(0) |^(2) xx (1)/(2)`...(II) We are given that without polarizer, intensity of principal maximum is `I_(0) = 4 |A_("perp".)^(0)|^(2)`(iii) `:.` From (ii), intensity of principal maximum with polariser would be `I = |A_("perp".)^(0)|^(2) [(1 + 1) + (1)/(2)] = (5)/(2)|A_("perp".)^(0)|^(2)` Using (iii), we get`I = (5)/(2)((I_(0))/(4)) = (5)/(8)I_(0)` Again, intensity at first minima with polsrizer [From (ii)] `I' = |A_("perp".)^(0)|^(2)(1 - 1) + |A_("perp".)^(0)|^(2) xx (1)/(2) = (|A_("perp".)^(0)|^(2))/(2) = (I_(0)//4)/(2) = (I_(0))/(8)` |
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