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 localid="1663142040666" $\mathit{L}$is missing, give the second least thickness or the third least thickness as indicated

Short answer

The thickness of the thin layer is$608\mathrm{nm}$

Step-by-step solution

Given Data

• 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 thickness of the layer is $L=285nm$.

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:

$2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}m=0,1,2,...$(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.

Determining the wavelength of light

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 thickness of this thin layer is$L=285nm$. Therefore, the light is out of phase. And then, the refracted light gets reflected of the third surface while travelling through the film. This does not result in phase change as the light is reflecting of a rarer medium. As a result, the condition for constructive interference or maximum intensity is

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

The wavelength of the reflected light is

$$\begin{gathered} m=0;\lambda =\frac{4L{n}_2}{2m+1}=\frac{4\left(285nm\right)\left(1.60\right)}{1}=1824\mathrm{nm} \\ m=1;\lambda =\frac{4L{n}_2}{2m+1}=\frac{1824nm}{3}=608\mathrm{nm} \end{gathered}$$

As 608 nm is in the visible range, hence the thickness of the thin layer is $608\mathrm{nm}$.