In the figure shown, a 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 the plane of slits as shown ), upto a distance 'l' before falling on `S_(1)`. Rest of the speace is filled with air. If at O a minima is formed, then the minimum value of the positive constant a ( in term of l and wavelength `'lambda'` in air ) is:

Options

A. `(lamda)/(l)`
B. `(lamda)/(l^(2))`
C. `(l^(2))/(lamda)`
D. none of these

Answer

B

Explanation

Path difference occurs due to the medium of variable refractive index `=(mu-1)t`
`thereforeint_(0)^(A)[(1+ax)-1]dx=(al^(2))/(2)` ltbr. For minima at 0 the path difference should be `(lamda)/(2)` (for minimum value of)
`therefore(al^(2))/(2)=(lamda)/(2)`
`thereforea=(lamda)/(l^(2))`