In the arrangement shown in figure, light of wavelength `6000 Å` is incident on slits `S_(1)` and `S_(2)` have been opened such that `S_(3)` is the position of first maximum above the central maximum and `S_(4)` is the closest position where intensity is same as that of the ligth used, below the central maximum Point O is equidistant from `S_(1)` and `S_(2)` and O' is equidistant from `S_(3)` and `S_(4)`. Then intensity of inciden light is `I_(0)`

Find the intensity of the brightest fringe.

Options

A. `9 I_(0)`
B. `5 I_(0)`
C. `3 I_(0)`
D. `7 I_(0)`

Answer

a

Explanation

From the given condition,
` OS_(3) = (D lambda)/(d) = (1 xx 6 xx 10^(-7))/(3 xx 10^(-3)) = 2 xx 10^(-4) m`
Let the reaching from `S_(1)` and `S_(2)` to `S_(4)` and phase difference `phi` and intensity of incident light is `I_(0)`. Resultant intensity at `S_(4)` is
(i) ` I = 4 I_(0) cos^(2) ((phi)/(2))`
As `I = I_(0)`, hence
`(I_(0))/(4 I_(0)) = cos^(2) ((phi)/(2))`
`implies cos ((phi)/(2)) = (1)/(2) = cos 60^(@)`
(ii) ` phi = (2 pi)/(3), phi = (2 pi)/(3) Delta x implies Delta x =(lambda)/(3)`
(iii) As `Delta x = (lambda)/(3)` , using `(Delta x)/(d) = (y)/(D)`
`y = (Delta x D)/(d) implies OS_(4) = (D lambda)/(3d)`
Therefore, `S_(3) S_(4) = OS_(3) + OS_(4) = (4)/(3) (d lambda)/(d) = (8)/(3) xx 10^(-4)m`
Now, resultant wave coming out of `S_(3)` has intensity `4 I_(0)` and waves coming out of `S_(4)` have intensity `I_(0)`.
Phase difference at `S_(3) = 2 pi`
Phase difference at `S_(4) = 2 pi // 3`
These phase diffference ara ralative to the ligth incident on slit `S_(1)` and `S_(2)`.
Now, `S_(3)` and `S_(4)` are secondary sources of ligth.
Phase difference at O' `= 4 pi // 3`, equal to inital phase difference between the ligth reaching at O', i.e.,
`2 pi (2 pi)/(3) = (4 pi)/(3)`
Let intensity at O' be I'. Then
`I' = I_(0) + 4 I_(0) + 2 sqrt I_(0) sqrt( 4 I_(0)) cos ((4 pi)/(3))`
`= 5 I_(0) + 4 I_(0) cos (pi + (pi)/(3)) = 3 I_(0)`
For the brightest fringe, phase difference `= 2 n p, n = 0, +-1, +-2`,...
Let I'' be the intensity of the brightest fringe.
`I'' = I_(0) + 4 I_(0) + 2 sqrt I_(0) sqrt( 4 I_(0)) cos phi ` (where `cos phi = 1`).