In figure, parallel beam of light is incident on the plane of the slits of a Young's double-slit experiment. Light incident on the slit `S_(1)` passes through a medium of variable refractive index `mu = 1 + ax` (where 'x' is the distance from the plane of slits as shown), up to distance 'I' before falling on `S_(1)`. Rest of the space is filled with air. If at 'O' a minima is formed. then the minimum value of the positive constant a (in terms of l and wavelength `lambda` in air) is

Options

A. `(lambda)/(l)`
B. `(lambda)/(l^(2))`
C. `(l^(2))/(lambda)`
D. None of these

Answer

B

Explanation

When light passes through a medium of refractive index `mu,` the optical path it travels is `(mut).`
Therefore, berore reaching `O` light through `S_(1)` travels `(mul+b)` distance while that through `S_(2)` travels a distance `(l+b)`
`i.e.:` path difference `=(mul+b)-(l+b)=(mu-1)l.`
For a smalll element `'dx'` path difference `Deltax=[1(+ax)-1] dx =ax dx`
For the whole length ,
`Deltax=int_(o)^(l)ax dx =(al^(2))/(2)`
For a minima to be at `'O'`.
`Deltax=(2n+1) (lambda)/(2)`
`i.e.: (al^(2))/(2)=(2n+1)(lambda)/(2).`
For minimum `'a', n=0 rArr (al^(2))/(2)=(lambda)/(2) rArr a=(lambda)/(l^(2))`