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What is the thinnest soap film (excluding the case of zero thickness) that appears black when...

What is the thinnest soap film (excluding the case of zero thickness) that appears black when viewed by reflected light with a wavelength of 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.
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Concepts and reason

The concepts required to solve this problem is Interference in thin films.

First, rearrange the destructive interference condition for thin films to solve for the thickness of the film. Then, calculate the minimum thickness other than zero for the destructive interference by using the expression for the thickness.

Fundamentals

Interference is the variation of the intensity of the wave observed due to the superposition of light waves. The destructive interference is the case of interference where the resultant intensity is least or zero that is object appears black.

The condition of destructive interference for thin film is,

2nt=mλ2nt = m\lambda

Here, nn is the refractive index of soap film, tt is thickness of the soap film, mm is a integer that is 0,1,2,…, and λ\lambda is wavelength of incident light.

Use the condition of destructive interference.

Rearrange the condition of destructive to solve for the thickness tt of the soap film.

2nt=mλt=mλ2n\begin{array}{c}\\2nt = m\lambda \\\\t = \frac{{m\lambda }}{{2n}}\\\end{array}

Use the expression of thickness of the soap film.

Substitute 11 for mm, 480nm480{\rm{ nm}} for λ\lambda , and 1.331.33 for nn in the equation t=mλ2nt = \frac{{m\lambda }}{{2n}} to calculate the thinnest soap film that appears black.

t=(1)(480nm)2(1.33)=180.45nm\begin{array}{c}\\t = \frac{{\left( 1 \right)\left( {480{\rm{ nm}}} \right)}}{{2\left( {1.33} \right)}}\\\\ = 180.45{\rm{ nm}}\\\end{array}

Convert nm to m by multiplying with (109m1nm)\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right).

t=180.45nm(109m1nm)=180.45×109m\begin{array}{c}\\t = 180.45{\rm{ nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)\\\\ = 180.45 \times {10^{ - 9}}{\rm{ m}}\\\end{array}

Ans:

The thinnest soap film that appears black when viewed by reflected light is 180.45×109m180.45 \times {10^{ - 9}}{\rm{ m}}.

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