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What is the thinnest film of a coating with \(n = 1.42\) on glass \((n = 1.52)\) for which destructive interference of the red component \((650nm)\) of an incident white light beam in air can take place by reflection?

Short Answer

Expert verified

The thinnest film of a coating is 114nm.

Step by step solution

01

Formulas used to solve the question

Destructive interference

2t=(m+12)λ

Snell’s law:

n1λ1=n2λ2

02

Determine the thickness in terms of λ

Given: nair=1.0

ncoating=1.42nglass=1.52λair=650nm=650*10-9m

Light beam reflects from a surface that has a greater index of refraction than the index of refraction of the first medium, the wave experiences a phase change. This means that the first ray experiences the phase change and the second ray also experiences a phase change.

From equation (1), at m = 0 for the thinnest coating

t=14λ (3)

03

Determine the thickness

From equation (2),

\({n_{coating}}{\lambda _{coating}} = {n_{air}}{\lambda _{air}}\)

\( \Rightarrow {\lambda _{coating}} = \frac{{{n_{air}}{\lambda _{air}}}}{{{n_{coating}}}}\)

Plug the given,

\( \Rightarrow {\lambda _{coating}} = \frac{{{\lambda _{air}}}}{{{n_{coating}}}}\)

Plug into (3),

\(t = \frac{{{\lambda _{air}}}}{{4{n_{coating}}}} = \frac{{650*10{}^{ - 9}}}{{4*1.42}} = 114nm\)

Thus, the thinnest film of a coating is \(114nm\).

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