A thin film, with a thickness of$\mathbf{272}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{nm}$and with air on both sides, is illuminated with a beam of white light. The beam is perpendicular to the film and consists of the full range of wavelengths for the visible spectrum. In the light reflected by the film, light with a wavelength of$\mathbf{600}\mathbf{}\mathbf{nm}$undergoes fully constructive interference. At what wavelength does the reflected light undergo fully destructive interference? (Hint: You must make a reasonable assumption about the index of refraction.

• The thickness of the film is$L=272.7\text{nm}$.
• The maximum intensity occurs at wavelength ${\lambda }_{\max }=600\text{nm}$.

We can assume that the refractive index of the thin film must $2.42$be between role="math" localid="1663083992126" $1.00-2.42$. With $1.00$being of air and $2.42$being of the diamond (highest refractive index).

The light that strikes thin films normally reflects off of both their front and rear surfaces, causing reflected light interference. When constructive interference occurs, the bright reflected light is produced, and when fully destructive interference occurs, a dark region is observed.

The maximum intensity condition for the given scenario is

$\begin{array}{c}2L=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{\max }}{n_2}\\ n_2=\frac{\left(2m+1\right){\lambda }_{\max }}{4L}\end{array}$

Calculating the refractive index for the first few order number

$\begin{array}{l}m=0;n_2=\frac{\left(2\left(0\right)+1\right)\left(600nm\right)}{4\left(272.7nm\right)}=0.55\\ m=1;n_2=\frac{\left(2\left(1\right)+1\right)\left(600nm\right)}{4\left(272.7nm\right)}=1.65\\ m=2;n_2=\frac{\left(2\left(2\right)+1\right)\left(600nm\right)}{4\left(272.7nm\right)}=2.75\end{array}$

As $1.65$only lies in the range$1.00-2.42$ , hence, the refractive index of the thin film is$1.65$.

The minimum intensity condition for the given scenario is

$\begin{array}{c}2L=m\frac{{\lambda }_{\min }}{n_2}\\ {\lambda }_{\min }=\frac{2L{n}_2}{m}\end{array}$

Finding minimum wavelengths for the first few order numbers

role="math" $\begin{array}{l}m=1;{\lambda }_{\min }=\frac{2\left(272.7nm\right)\left(1.65\right)}{1}=900\text{nm}\\ m=2;{\lambda }_{\min }=\frac{2\left(272.7nm\right)\left(1.65\right)}{2}=450\text{nm}\\ m=3;{\lambda }_{\min }=\frac{2\left(272.7nm\right)\left(1.65\right)}{3}=300\text{nm}\end{array}$

As $450\text{nm}$lies in the visible range, hence, the fully destructive interference occurs at $450\text{nm}$.