Question

Daylight gets reflected from a freestanding soap film and appears reddish (wavelength = 660 nm) when observed from a direction perpendicular to the film surface. Assume a refractive index of n = 1.33.

(a) What are the two smallest possible thicknesses of the soap film which fit to your observation?
(b) Which typical wavelength and related color do you observe when watching the film from an angle of 50° (relative to the surface normal) using the smallest possible thickness of the film from (a)?

Answer 1

Interference due to the Reflected Beam:

In this figure, i and r are incidence angle and refraction angle. t and µ is the thickness of the film and refractive index of the soap medium. The path difference between the waves BC and EF is δ = (BD+DE)_{in film} – (BM)_{in air} = 2µt cos r. A ray of light travelling in air and getting reflected at the surface of a denser medium, undergoes an automatic phase change of π (or) an additional path difference of λ/2.

According to Snell´s law, sin i = µ sin r. Therefore, the path difference will be

For the constructive interference, path difference δ = 2n.λ/2, where n = 0,1,2,3 and the film appears bright. So,

For the destructive interference, path difference δ = (2n+1).λ/2, where n = 0,1,2,3 and the film appears bright. So,

(a) For normal incidence, i = 0, If you consider the destructive interference for the first order, the thickness (t) will be

For normal incidence, i = 0, If you consider the constructive interference for the first order, the thickness (t) will be

Smallest possible thickness will be 124 nm.

(b) t = 124 nm, i= 50^o. Therefore, if you consider constructive interference, you will obtain the

if you consider destructive interference, you will obtain the

This is in ultraviolet region. Hence, the color will be 539.3 nm i.e. green color.