Chapter 24, Problem 44P

To determine

Thickness of the foil.

Answer to Problem 44P

  $9.05\mu m$

Explanation of Solution

Given:

Wavelength, $\lambda =670nm$

Let order be $m=27$

Formula used:

Dark bands will occur when $2t=m\lambda ,m=0,1,2,\mathrm{...}$

The light rays are perpendicular to the plates, the extra path length equals to $2t$ , where $t$ is the thickness of the air gap at any point.

Where,

  $m$ is an integer.

  $t$ is the thickness of the film.

  $\lambda$ is wavelength.

Calculation:

According to the question,

  $\begin{array}{l}\varphi =\left(\frac{2t}{\lambda }\right)2\pi +\pi =\left(2m+l\right)\pi ,m=0,1,2,\mathrm{...},or\\ t=\frac{1}{2}m\lambda ,m=0,1,2,\mathrm{...}\end{array}$

Because $m=0$ corresponds to the edge where the glasses touch, $m+1$ represents the number of the fringe.

The thickness of the foil:

  $t=\frac{1}{2}\left(27\right)\left(670nm\right)=9.05\times {10}^{3}nm=9.05\mu m$

Conclusion:

The required value is $9.05\mu m$ .