In young's double-slit experiment, both the slits produce equal intensities on a screen. A 100% transparent thin film of refractive index `mu = 1.5` is kept in front of one of the slits, due to which the intensity at the point O on the screen becomes 75% of its initial value. If the wavelength of monochromatic light is 720 nm, then what is the minimum thickness (in nm) of the film?

Answer

240

Explanation

The path difference at the point O is:
`Deltax = (mu-1)t`
`I = 4I_(0)cos^(2) (phi/2)`
`0.75 (4I_(0)) = 4I_(0) cos^(2) (phi/2)`
`rArr phi_("min") = 60^(@)`
`phi_("min") = (2pi)/lambda Deltax = (2pi)/lambda (mu-1)t = pi/3`
`t = lambda/(6(mu-1)) = lambda_(0)/(6(1.5-1)) = 240 nm`