Thin films

Thin films:-

               Thin films means a transparent layer of very small thickness in the range of 0.5 μm to 10 μm ; such as soap film or a layer of skin oil floating on water surface. When light is imcidein on a thin film, different colours are seen. This can be explained by considering interference between light reflected from the top and the bottom surface of a thin film. In case of thin films, interference occur due to i) reflected light and ii) transmitted light.

Interference in thin film of uniform thickness due to reflected light.

               Consider a thin transparent film of uniform thickness ‘t' and of refractive index ‘μ' be placed in air. Let a beam of light AB is incident on the upper surface l, at an angle' i as shown in figure A. At B, light suffers partial reflection along BE and partial refraction along BC. The angle of refraction is r. At C, it is again partially reflected along CD. At D it is partially refracted along DF.

               Incident beam AB suffers partial refraction at B  ( top surface )  and  at  C
 ( bottom surface). So we get a set of parallel beams  like  BE ,  DE.  etc.  These  beams originate from same incident beam. So these beams are coherent. When these reflected beams BE and DF superimpose, interference takes place.

               Let us calculate the path difference between the beams BE and BCDF. Let DM and BN be perpendicular to BE and CD respectively. Produce BH and DC to meet at G. Let t= BH is the thickness of the film.

By geometry ∠HGC = r ; ∠BDM =i ; ∠DBN= r
 .•.      BC = GC

The path difference between two reflected beams BE and DF=(BC+CD) in film-BM in air

          = μ ( BC + CD ) - BM
          = μ ( BC + CN + ND ) - BM ....(1)

We know that,

μ = sin i = BM / BD = BM
      sin r    ND / BD     ND

.•. BM = μ.ND

.•. Equation. (1) becomes, path difference
 
           = μ ( BC + CN + ND ) - μ. ND
           = μ ( BC + CN )
           = μ ( GC + CN )
           = μ . GN

In ∆ BGN,

cos r = GN = GN  ( •.• BG = BH+HG = t + t = 2t)
             BG     2t

.•. GN = 2.t cos r   .....(3)

Putting this value of GN in eq. (2) we get ,

Path difference =2.μ.t.cos r   .....(4)

               Eq. (4) does not represent a correct path  difference.  Because  whenever reflection takes place from the surface of optically denser medium a phase change of π radius ( or path difference λ/2) takes place.

                The film is optically denser than the  surrounding  medium  ( air ).
The ray BE originating  by reflection at the denser medium, suffers a phase change of  π radius due to reflection at B.

               ( No such change of phase occur for ray DF, as it is result of reflection at C, the surface of rarer medium ).

The correct path difference between reflected beams = 2μ.t.cos r + λ/2 = nλ

i) The firm appears bright if the path difference. = nλ

2μ.t.cos r + λ/2 = nλ

.•. 2μ.t.cos r = (2n - 1 ) λ/2

ii) The firm appears dark if path difference
     = (2n + 1) λ/2

.•. 2μ.t.cos r +  λ/2 = (2n + 1) λ/2


2μ.t.cos r  = nλ   where, n = 0,1,2,....