what is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 520 nmnm ? the index of refraction of the film is 1.33, and there is air on both sides of the film.

Answer :

The thinnest soap film that appears black when illuminated with light with a wavelength of 520 nm is about 195 nm.

A natural phenomenon known as "thin-film interference" occurs when light waves reflected by a thin film's upper and lower limits collide with one another, either enhancing or weakening the light that is reflected.

First, draw the diagram. From the diagram 1 is the first reflected ray and 2 is the second reflected ray. The red circle indicates the phase change. The first ray experience phase change but the second ray doesn't experience phase change.

Therefore, the thickness of the soap bubble is given by the thin-film interference formula,

[tex]\begin{aligned}2t& = m\lambda_{soap}\\t&=\frac{m\lambda_{soap}}{2}\\\lambda_{soap}&=\frac{2t}{m}\end{aligned}[/tex]

From Snell's law, we know that n₁λ₁=n₂λ₂, where λ is the wavelength and n is the index of refraction.

Then, for the soap film, [tex]n_{air}\lambda_{air}=n_{soap}\lambda_{soap}[/tex].

We know that n (air) is 1. From this,

[tex]\begin{aligned}\lambda_{soap}&=\frac{\lambda_{air}}{n_{soap}}\\\frac{2t}{m}&=\frac{\lambda_{air}}{n_{soap}}\\t&=\frac{m\lambda_{air}}{2n_{soap}}\end{aligned}[/tex]

Excluding the case of zero thickness, substitute m=1 for the first destructive interference.

Then, the smallest thickness is,

[tex]\begin{aligned}t&=\frac{\lambda_{air}}{2n_{soap}}\\&=\frac{520}{2\times1.33}\\&=\mathrm{195\;nm}\end{aligned}[/tex]

The answer is 195 nm.

To know more about thin-film interference:

https://brainly.com/question/17164449

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