Question

A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength = 640.0 nm in vacuum). Assuming that the visible spectrum extends from 380 to 750 nm, for which visible wavelength(s) in vacuum will the film appear bright due to constructive interference?

Answer 1

let refractive index of oil is n.
wavelength of the light ? = 640 nm = 640*10^-9 m

condition for destructive interference ,

                 2nt = m?

here ,t = thickness of the film

         m = order of the spectrum

so ,

for least possible thickness m = 1,

hence ,

           2nt = ?  

           2nt = 640 nm   ................. (1)

wavelength range of visible spectrum is 380 nm to 750 nm.

condition for constructive interference ,

                 2nt = (m + 1/2)?

wavelength ? = 2nt / (m + 1/2)

                   ? = (640 nm) / (m + 1/2)     (from eq 1 , 2nt = 640nm)

here , m = 0 , 1 , 2 , 3 , .......

if m = 0 ,

wavelength ? = 640 nm /( 0 + 1/2)

                      = 1280 nm

if m =1 ,

wavelength ? = 640 nm /( 1 + 1/2)

                     = 426.67 nm

                      = 426.67 *10^-9 m

this is the only possible wavelength in the visible range.