Please show all work, Thank you in advance A thin film of oil, noil = 1.24, is l

Question

Please show all work, Thank you in advance

A thin film of oil, noil = 1.24, is laying on a flat piece of glass, nglass = 1.83. When viewed through air normal to the oil film, constructive interference produces a band with a bright blue color, lambda = 505nm. What is the phase shift or phase change (in degrees or in fractions of a wavelength - specify air or film) for the reflected light off the air-oil interface? What is the phase shift or phase change (in degrees or in fractions of a wavelength - specify air or film) for the reflected light off the oil-glass interface? What is the minimum thickness of the oil film in nm?

Explanation / Answer

we have refractive index (n1) of air =1; refractive index of oil (n2)=1.24; refractive index of glass (n3)=1.83;

so n1<n2<n3;

(a) for reflaction through air oil interface

light is reflected from denser medium as n1<n2 so light wave will sufer a pie phase change or 180 degree phase change

but as pie phase change is equivalent to lemda/2 path diffrence so path diffrence is lemda/2

(b) for reflaction through oil glass interface

similarly in this case n2<n3 so path diffrence is lemda/2 and phase change is 180 degrees

(c) since both the waves are reflected at 180 degree phase change so they are in phase with each other

let d be the thickness of oil film

and theta be the angle between ray and normal ; as it is viewed normally so theta = 0 degrees

let w be wavelength of light; w=505 nm

formula for constructive interference is

2*n*d*cos( theta)=m*w; where m is +integer;

for minimun thichness m=1;

hence

2*1.24*d*cos(0)=1*505nm

or d=505/(2*1.24) nm

or d=203.62 nm

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