Figure 35-27a shows the cross-section of a vertical thin film whose width increases downward because gravitation causes slumping. Figure 35-27b is a face-on view of the film, showing four bright (red) interference fringes that result when the film is illuminated with a perpendicular beam of red light. Points in the cross section corresponding to the bright fringes are labeled. In terms of the wavelength of the light inside the film, what is the difference in film thickness between (a) points a and b and (b) points b and d?

• The width of the thin film increases downward.
• The film is illuminated with a beam of red light.

The expression to calculate the wavelength inside the film is given as follows.

$\mathit{L}\mathbf{=}\mathit{m}\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{n}}$

Here, n is the refractive index of the medium, $\mathit{\lambda }$is the wavelength, L is the thickness of the firm, and m is the order of the interference and it can take values 0,1,2,3........

The bright fringes in the thin films produce when 2L is equal to multiple of wavelength.

$2L=m{\lambda }_t$

Substitute$\frac{\lambda }{n}$for${\lambda }_t$into above equation.

$$\begin{gathered} 2L=m\frac{\lambda }{n} \\ L=m\frac{\lambda }{2n} \end{gathered}$$

............(i)

Between points, a and b, the order of the interference is 1. Therefore,

m=1

Substitute 1 for m into equation (i).

$$\begin{gathered} L=\left(1\right)\frac{\lambda }{2n} \\ L=\frac{\lambda }{2n} \end{gathered}$$

Hence, the difference in the thickness between points a and b is$\frac{\lambda }{2a}$.

Since between points b and d, the order of the interference is 2. Therefore,

m=2

Substitute 2 for m into equation (i)

$\begin{array}{rcl}L& =& \left(2\right)\frac{\lambda }{2n}\\ L& =& \frac{\lambda }{n}\\ & & \end{array}$