Consider a ray of light incident on a thin film of thickness dd (in the sub-nanometer to micron range) as in the figure below.
We are required to find minimum thickness of a thin film required for constructive interference in the reflected light.
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The ray travels from a medium of refractive index n_1n1 to a medium of refractive index n_2.n2. As light strikes the surface of thin film at AA at the upper surface, part of it is transmitted and remaining reflected. Part of light which is transmitted reaches the bottom surface and hits it at BB, it may once again be transmitted and/or reflected depending on the refractive index of the medium below the thin film.
The occurrence of constructive or destructive interference between the two reflected light waves depends on their phase difference. This difference depends on
1. the thickness of the film layer,
2. the refractive index of the film, and
3. the angle of incidence of the original wave on the film.
Additionally, a phase shift of 180^@180∘ or piπ radians occurs at the boundary depending on the refractive indices of the materials on either side of the interface. This may happen both at points A and BAandB. This phase shift occurs at AA if n_1 < n_2n1<n2.
Armed with information we know that to get constructive interference, the two reflected rays have to be shifted by an integral multiple of wavelength lambdaλ.
For minimum thickness of thin film, it must be =lambda=λ. Therefore, we must choose the three media so that phase of 180^@180∘ or piπ radians occurs at AA, and remaining 180^@180∘ or piπ radians by distance traveled by the transmitted ray down and up through the thin film. No phase shift at BB
(This has been taken care of by choosing the media in the figure above. Light bends towards the normal when traveling from a rarer to denser medium, n_1 < n_2n1<n2).
It implies that, the thin film must introduce integral multiple of 1//41/4 wavelength phase one way. For minimum thickness we must have the phase as lambda//4λ/4
Also the minimum dd the incident ray must strike at the interface at AA with theta_1=0θ1=0. The transmitted part of the ray travels 2d2d (down and up) and meet the reflected part of ray at AA to be in phase.
We know that wavelength in vacuum is related to wavelength in a medium of refractive index nn as
lambda_"film"=lambda_"vacuum"/nλfilm=λvacuumn, inserting given values
lambda_"film"=600/1.5=400nmλfilm=6001.5=400nm,
Equating with condition for constructive interference and solving for dd
d=400/4d=4004
d=100nmd=100nm
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Detailed derivation
From the figure,
Path traveled by reflected ray at A=ADA=AD, in medium n_1n1
Path traveled by reflected ray at B=AB+BCB=AB+BC, in medium n_2n2
Path difference=n_2(AB+BC)-n_1AD=n2(AB+BC)−n1AD .....(1)
Now AB=BC=d/(cos theta_2)AB=BC=dcosθ2 .....(2)
Also AC=2d.tan theta_2AC=2d.tanθ2
and (AD)/(AC)=sin theta _1ADAC=sinθ1
:.AD=2d.tan theta_2 .sin theta_1
Changing the expression in terms of theta_2, by using Snell's law n_1\sin\theta_1=n_2\sin\theta_2
=>sintheta_1=n_2/n_1\sin\theta_2, inserting in for AD, we get
AD=2d.tan theta_2 .n_2/n_1\sin\theta_2 .....(3)
Substituting from (2) and (3) in (1)
Path difference=n_2 (2d)/(cos theta_2)-n_1. 2dtan theta_2 .n_2/n_1\sin\theta_2
= n_2\frac{2d}{\cos\theta_2} - 2d\tan\theta_2.n_2\sin\theta_2
= 2dn_2(\frac{1-\sin^2\theta_2}{\cos\theta_2})
= 2dn_2\cos\theta_2
For constructive interference this Path difference should be equal to integral multiple of wavelength, nlambda.
For minimum thickness of a thin film this must equal lambda. Equating the two we obtain
2dn_2\cos\theta_2=lambda
Additional path difference introduced by reflection at A=lambda/2
Condition changes and hence net path difference required =2dn_2\cos\theta_2= lambda/2
Solving for d
d=1/4lambda/(n_2\cos\theta_2)
Maximum value of costheta=1 which gives minimum value of d
Hence d_"min"=lambda/(4n_2)
Inserting given values
d_"min"= 600/(4xx1.5)
=100nm
