In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is `lambda=l`. The intensity of light at a point where the path difference becomes `lambda//3` is

Options

A. `(I)/(4)`
B. `(I)/(2)`
C. `(I)/(3)`
D. I

Answer

A

Explanation

Forpath differnce, `lambda`
Phase diff `= 2pi`,
the intensity at this point is I
i.e., `I=sqrt(I_(1)I_(2)) 2 cos (alpha_(1)-alpha_(2))`
Let `I_(1)=I_(2)=I'" "`(say)
`:.I=I'= +I' +2I' =4I " "...(i)`
For path diff .`(lambda)/(3)`, phase diff. `=(2pi)/(lambda) (lambda)/(3)=(2pi)/(3)`
The intensity at this point is I''
`:. I''=I_(1)+I_(2)+sqrt(I_(1)I_(2)) 2 "cos"(2pi)/(3)=I`
But frome quation (i)`I'=I//4`
`:. I''=I//4`