A thin film of soap with n 142 hanging in the air reflects
A thin film of soap with n = 1.42 hanging in the air reflects dominantly red light with = 635 nm. What is the minimum thickness of the film?
Now this film is on a sheet of glass, with n = 1.46. What is the wavelength of the light in air that will now be predominantly reflected?
Solution
The path difference between the reflections from the top and bottom surfaces must be one wavelength
Wavelength = 2nt + (wavelength / 2)
635*10^-9 =2*1.42*tmin + (635*10^-9)/2 => tmin = 111.7 nm
For this film on the glass,
both reflected waves having a phase change and so they cancel out and the path difference is
=wavelength = 2nt
=2*1.42*111.7*10^-9 = 317nm
b)
635*10^-9 =2*1.46*tmin + (635*10^-9)/2 => tmin = 108.7 nm
For this film on the glass,
both reflected waves having a phase change and so they cancel out and the path difference is
=wavelength = 2nt
=2*1.46*108.7*10^-9 = 317nm
