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 in nanometres, and the wavelength λ in nanometres of the light as measured in air. Where is missing, give the wavelength that is in the visible range. Where is missing, give the second least thickness or the third least thickness as indicated.

The bright colors reflected from thin oil on water and soap bubbles are the result of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colors 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)

$2L=m\frac{\lambda }{n_2}m=1,2,3,..$(Minima)

Where,$\lambda$is the wavelength of the light in air, $L$is its thickness, and role="math" localid="1663009742868" $n_2$is the film’s refractive index.

Here, in this case, light travels in a medium with $n_1=1.40$and incident on the thin layer whose refractive index is $n_1=1.60$and the reflected light has$180°$phase change as the light is reflected off the denser medium. And then, the refracted light gets reflected of the back surface $n_3=1.75$while traveling through the film. This results in$180°$phase change. The total phase difference between $r_1$and $r_2$is still zero. As a result, the condition for destructive interference or minimum intensity is

$\begin{array}{rcl}2L& =& \left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}\\ \lambda & =& \frac{4L{n}_2}{2m+1}\\ & & \end{array}$(Destructive)

The wavelength for the first few orders is as below.

For:role="math" localid="1663008879425" $m=0$

For:$m=1$

$\begin{array}{rcl}\lambda & =& \frac{4\left(210nm\right)\left(1.46\right)}{2\left(1\right)+1}\\ & =& 409nm\\ & & \end{array}$

As$409\text{nm}$ is in the visible range, the wavelength of the light is $409\text{nm}$.