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 ${\mathit{n}}_{\mathbf{1}}$, localid="1663139751503" ${\mathit{n}}_{\mathbf{2}}$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 thickness of the thin layer is $161\mathrm{nm}$.

Step-by-step solution

Given Data

• The refractive index of first medium is $n_1=1.68$.
• The refractive index of the thin film is$n_2=1.59$
• The refractive index of the third medium $n_3=1.50$
• The minimum intensity occurs at wavelength $\lambda =342nm$.

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:

$2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}m=0,1,2,...(\mathrm{Maxima}—\mathrm{bright}\mathrm{film}\mathrm{in}\mathrm{the}\mathrm{air})$

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 2nd least thickness for this arrangement of materials.

Here, in this case, light travels in medium with $n_1=1.68$and incident on the film whose refractive index is $n_2=1.59$And then, the light gets reflected of the third surface $n_3=1.50$while travelling through the film. A minimum intensity is observed at $\lambda =342nm$and thickness of the film is to be determined at 2nd order of intensity which $m=1$in this case. As a result, the condition for destructive interference or minimum intensity is

$2L=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{342}}{n_2}$ (Destructive)

The 2^nd least thickness is

$\begin{array}{rcl}2L& =& \left(1+\frac{1}{2}\right)\frac{342\mathrm{nm}}{1.59}\\ L& =& \frac{3\left(342\mathrm{nm}\right)}{4\left(1.59\right)}\\ & =& 161\mathrm{nm}\end{array}$

Hence the thickness of the thin layer is $161\mathrm{nm}$.