Reflection by thin layers. In Fig. 35-42, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays ${\mathit{r}}_{\mathbf{1}}$and ${\mathit{r}}_{\mathbf{2}}$interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction ${\mathit{n}}_{\mathbf{1}}$, ${\mathit{n}}_{\mathbf{2}}$and ${\mathit{n}}_{\mathbf{3}}$, the type of interference, the thin-layer thickness $\mathit{L}$in nanometres, and the wavelength $\mathit{\lambda }$in nanometres of the light as measured in air. Where $\mathit{\lambda }$is missing, give the wavelength that is in the visible range. Where $\mathit{L}$is missing, give the second least thickness or the third least thickness as indicated.

• The refractive index of first medium is$n_1=1.55$
• The refractive index of the thin film is $n_2=1.60$
• The refractive index of the third medium $n_3=1.33$
• The maximum intensity occurs at wavelength role="math" localid="1663150278048" $\lambda =612\mathrm{nm}$.

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:

$2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}m=0,1,2,...$ (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, light travels in medium with $n_1=1.55$and incident on the film whose refractive index is $n_2=1.60$. And then, the light gets reflected of the third surface $n_3=1.33$ while travelling through the film. As a result, the condition for constructive interference or maximum intensity is

$2L=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{612}}{n_2}$ (Constructive)

The 3^rdleast thickness is

$\begin{array}{rcl}2L& =& \left(2+\frac{1}{2}\right)\frac{612\mathrm{nm}}{1.60}\\ L& =& \frac{5\left(612\mathrm{nm}\right)}{4\left(1.60\right)}\\ & =& 478\mathrm{nm}\end{array}$

Hence the thickness of the thin layer is $478\mathrm{nm}$.