Chapter 24, Problem 77GP

To determine

The minimum thickness for the air layer between two flat glass surfaces. Also, when the glass is to appear bright

Answer to Problem 77GP

  $160nm$ and $320nm$ .

Explanation of Solution

Given:

Wavelength, $\lambda =640nm$

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, the half cycle phase change at the lower surface means that max destructive interference will occur when the thickness $t$ will be $2t=m\lambda ,m=0,1,2,\mathrm{...}$

Calculation for destructive interference, $m=1$ (to find the non zero value of $t$ ) by using the formula,

  $t=\frac{1}{2}m\lambda =\frac{\lambda }{2}=\frac{640nm}{2}=320nm$

Calculation for constructive interference, $m=0$ (to find the smallest value) by using the formula, (Also, max constructive interference will be occur when $2t=\left(m+\frac{1}{2}\right)\lambda ,m=0,1,2,\mathrm{...}$ )

  $t=\frac{1}{2}\left(m+\frac{1}{2}\right)\lambda =\frac{1}{4}\lambda =\frac{640nm}{4}=160nm$

Conclusion: Wavelength for destructive interference is $320nm$ and for constructive interference is $160nm$ .