Question

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 ${\mathbf{r}}_{\mathbf{1}}$andlocalid="1663664382127" ${\mathbf{r}}_{\mathbf{2}}$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 λ 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.

Short answer

The 3rd least thickness of the thin layer is $273\mathrm{nm}$.

Step-by-step solution

Interference in thin films

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$n_2$ is the film’s refractive index.

Determine the 3rd least thickness of the thin layer

Here, in this case, light travels in a medium with${\mathrm{n}}_1=1.32$ and incident on the thin layer whose refractive index is$n_2=1.75$ 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${\mathrm{n}}_3=1.39$ while traveling through the film. This results in no phase change. The phase difference between$r_1$ and$r_2$ is $180°$. As a result, the condition for constructive interference or maximum intensity is

$2\mathrm{L}=\left(\mathrm{m}+\frac{1}{2}\right)\frac{\mathrm{\lambda }}{{\mathrm{n}}_2}$ (Constructive)

The 3^rd least ($m=2$) thickness of the thin layer is

$\begin{array}{c}\mathrm{L}=\left(2+\frac{1}{2}\right)\frac{382\mathrm{nm}}{2(1.75)}\\ =273\mathrm{nm}\end{array}$

Hence the 3rd least thickness of the thin layer is $273nm$.