The reflection of perpendicularly incident white light by a soap film in the air has an interference maximum at $\mathbf{600}\mathbf{}\mathbf{\text{nm}}$and a minimum at role="math" localid="1663024492960" $\mathbf{450}\mathbf{}\mathbf{\text{nm}}$, with no minimum in between. If $\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{33}$for the film, what is the film thickness, assumed uniform?

• The refractive index of the thin film is$n_2=1.33$
• The maximum interference is observed atlocalid="1663026501026" role="math" ${\lambda }_{\max }=600\text{nm}$
• The minimum interference is observed atlocalid="1663026510207" ${\lambda }_{\min }=450\text{nm}$

Bright colours reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

$2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}m=0,12..$(Maxima—bright film in the air)

where $\lambda$is the wavelength of the light in air, $L$is its thickness, and $n_2$is the film’s refractive index.

Here, in this case, the air is on the both sides of the soap film, therefore the conditions for constructive and destructive interference.

$2L=\frac{m{\lambda }_{\min }}{n_2}$ (Destructive)

$2L=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{\max }}{n_2}$ (Constructive)

For constructive interference, the value of thickness for the first few orders is

$\begin{array}{l}m=0;L=\left(0+\frac{1}{2}\right)\frac{600nm}{2\left(1.33\right)}=113\text{nm}\\ m=1;L=\left(1+\frac{1}{2}\right)\frac{600nm}{2\left(1.33\right)}=338\text{nm}\\ m=2;L=\left(2+\frac{1}{2}\right)\frac{600nm}{2\left(1.33\right)}=564\text{nm}\end{array}$

For destructive interference, the value of thickness for the first few orders is

$\begin{array}{l}m=1;L=\frac{\left(1\right)450nm}{2\left(1.33\right)}=169\text{nm}\\ m=2;L=\frac{\left(2\right)450nm}{2\left(1.33\right)}=338\text{nm}\\ m=3;L=\frac{\left(3\right)450nm}{2\left(1.33\right)}=254\text{nm}\end{array}$

.

The least thickness, which is common for constructive and destructive interference, is $338\text{nm}$. Hence the thickness of the soap film is $338\text{nm}$.