In YDSE, the intensity of light at a point on the screen is I for a path difference `lambda` . The intensity of light at a point where the path difference becomes `(lambda)/(3)` is `(I)/(P)` . Find the value of P ?

Answer

4

Explanation

For `lambda` path phase difference = `2 pi`
`I = I_(0) cos^(2) (phi//2) = I_(0) cos^(2) ((2pi)/(2)) = I_(0)`
For `(lambda)/(3)` path difference phase difference
`phi^(1) = (2pi)/(lambda)xx(lambda)/(3)=(2pi)/(3)`
`I^(1) = I_(0)cos^(2) (phi^(1)//2) = I_(0) cos^(2)((2pi)/(3xx2)) = I_(0) cos^(2) ((pi)/(3))`
`I^(1)= (I_(0))/(4) = (I)/(4) = (I)/(P) implies P = 4`