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 ${\mathit{r}}_{\mathbf{1}}$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.

Short answer

The 2nd least thickness of thin layer $339\mathrm{nm}$.

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:

$2\mathrm{L}=\left(\mathrm{m}+\frac{1}{2}\right)\frac{\mathrm{\lambda }}{{\mathrm{n}}_2}\mathrm{m}=0,1,2,...$(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 2nd least thickness of the thin layer

Here, in this case, light travels in a medium with${\mathrm{n}}_1=1.60$ and incident on the thin layer whose refractive index is${\mathrm{n}}_2=1.40$ 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${\mathrm{n}}_3=1.80$ 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 constructive interference or maximum intensity is

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

The thickness of the thin layer is

$\begin{array}{c}\mathrm{L}=\left(1+\frac{1}{2}\right)\frac{632\mathrm{nm}}{2\left(1.40\right)}\\ =339\mathrm{nm}\end{array}$

Hence the 2nd least thickness of the thin layer is $339nm$.