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.

Options

A. `(I_(0))/(8)`
B. `(3)/(4)I_(0)`
C. `(I_(0))/(16)`
D. `(2)/(5)I_(0)`

Answer

A

Explanation

Without P:
`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).`