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 ${\mathbf{r}}_{\mathbf{1}}$and${\mathbf{r}}_{\mathbf{2}}$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 ${\mathbf{n}}_{\mathbf{1}}$,${\mathbf{n}}_{\mathbf{2}}$, and${\mathbf{n}}_{\mathbf{3}}$, 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$\mathbf{L}$is missing, give the second least thickness or the third least thickness as indicated.

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:

$2\mathrm{L}=\left(\mathrm{m}+\frac{1}{2}\right)\frac{\mathrm{\lambda }}{{\mathrm{n}}_2}\mathrm{m}=0,1,2,\mathrm{...}$(Maxima—bright film in the air)

$2\mathrm{L}=\mathrm{m}\frac{\mathrm{\lambda }}{{\mathrm{n}}_2}\mathrm{m}=1,2,3,\mathrm{..}$(Minima)

where$\mathrm{\lambda }$ is the wavelength of the light in air, $L$is its thickness, and${\mathrm{n}}_2$ is the film’s refractive index.

Here, in this case, light travels in a medium with${\mathrm{n}}_1=1.50$ and incident on the thin layer whose refractive index is${\mathrm{n}}_2=1.34$ and the reflected light has no phase change as the light is reflected off the rarer medium. And then, the refracted light gets reflected of the third surface${\mathrm{n}}_3=1.42$ while traveling through the film. This results result in$180°$ phase change. The phase difference between$r_1$ and$r_2$ is $180°$ As a result, the condition for destructive interference or minimum intensity is

$2\mathrm{L}=\mathrm{m}\frac{\mathrm{\lambda }}{{\mathrm{n}}_2}$ (destructive)

The wavelength of the reflected light is

$\begin{array}{c}\mathrm{m}=1;\mathrm{\lambda }=\frac{2{\mathrm{Ln}}_2}{\mathrm{m}}=\frac{2\left(380\mathrm{nm}\right)\left(1.34\right)}{1}=1018\mathrm{nm}\\ \mathrm{m}=2;\mathrm{\lambda }=\frac{2{\mathrm{Ln}}_2}{\mathrm{m}}=\frac{1018\mathrm{nm}}{2}=509\mathrm{nm}\end{array}$

As$509\mathrm{nm}$ is in the visible range, the wavelength of reflected light is $509nm$.