Chapter 19.1, Problem 8PP

To determine

The thickness of the thinnest soap film.

Answer to Problem 8PP

  $t=97.9\text{ nm}$

Explanation of Solution

Given:

Refractive index of the soap film is $n_{film}=1.33$

Wavelength of light is $\lambda \text{ = 521 nm}$

Formula used:

The equation for constructive interference is given by,

  $\text{2t = }\left(m+\frac{1}{2}\right)\frac{\lambda }{n}$ $\left(1\right)$

Where, $\text{t }$ is the thickness of the film $\lambda$ is the wavelength of the light in the film $n$ is the refractive index of the film, In this problem $n=n_{film}=1.33$ $m$ is the order of the fringe,

Calculation:

For the thinnest film, $m=0$

Substituting for $m$ and $n$ in equation $\left(1\right)$ ,

  $\text{2t = }\left(\frac{1}{2}\right)\frac{\lambda }{n_{film}}$

  $\Rightarrow \text{t = }\left(\frac{1}{4}\right)\frac{\lambda }{n_{film}}$ $\left(2\right)$

Substituting the numerical values in equation $\left(2\right)$ ,

  $\text{t = }\frac{\left(521\text{ nm}\right)}{4\times 1.33}=\text{ 97}\text{.9 nm}$

Conclusion:

The thickness of the thinnest soap film for which light of wavelength $\text{521 nm}$ constructively interfere with itself is $\text{97}\text{.9 nm}$ .