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

A=Resultant amplitude
=A parallel `(A_(||))` + A perpendicular `(A_(bot))`
`implies A=A_(bot)+A_(||)`
Without P ` A=A_(bot)+A_(||)`
`A_(1)=A_(bot)^(1)+A_(bot)^(2)=A_(bot)^(0) sin (kx-omegat)+A_(bot)^(0) sin (kx-omegat+phi)`
`A_(||)=A_(||)^((1))+A_(||)^((2))`
`A_(||)=A_(||)^(0)[ sin (kx-omegat)+ sin (kx- omegat+phi)]`
where `A_(bot)^(0), A_(||)^(0)` are the amplitudes of either of the beam in perpendicular and parallel polarisations.
`:.` Intensity `={|A_(bot)^(0)|^(2)+|A_(||)^(0)|^(2)}[sin^(2) (kx-omegat)(1+cos^(2)phi+2 sin phi)+sin^(2)(kx-omegat)sin^(2) phi]`
`={|A_(bot)^(0)|^(2)+|A_(||)^(0)|^(2)}((1)/(2))2(1+ cos phi)`
`=2|A_(bot)^(0)|^(2)(1+cos phi)," since ", |A_(bot)^(0)|_(av)=|A_(||)^(0)|_(av)`
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)`
Given, `I_(0)=4|A_(bot)^(0)|^(2)`= intensity without polariser at principle maxima.
Intensity at principal 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)`.