Chapter 19.1, Problem 7PP

To determine

The thickness of the thinnest soap film.

Answer to Problem 7PP

  $t=196\text{ nm}$

Explanation of Solution

Given:

Refractive index of air is ${\text{n}}_{air}=1.00$

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

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

Formula used:

There will be a phase inversion in the light rays reflected from the interface between air and soap film because of $n_{film}>n_{air}$ and also there is no phase inversion in the light rays reflected from the interface between film and the air because of $n_{air}<n_{film}$ .

The equation for destructive interference to get a black stripe is given by,

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

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, $m=1$

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

  $\text{2t = }\frac{\lambda }{n_{film}}$

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

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

  $\text{t = }\frac{\left(521\text{ nm}\right)}{2\times 1.33}=196\text{ nm}$

Conclusion:

The thickness of the thinnest soap film at which the black stripe can be seen is $196\text{ nm}$ .