Figure shown 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, calculte in the present case, the intensity of the principal maxima as well as the first minima.

Explanation

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)`