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 wavelength of reflected light is$560\mathrm{nm}$

Step-by-step solution

Given Data

• The refractive index of first medium is $n_1=1.60$.
• The refractive index of the thin film is$n_2=1.40$
• The refractive index of the third medium$n_3=1.80$
• The thickness of the layer is $L=200nm$.

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 refractiveindex.

Determining the wavelength of light

Here, in this case, light travels in medium with$n_1=1.60$ and incident on the film whose refractive index is$n_2=1.40$ and the thickness of the layer is$L=200nm$ the reflected light has no phase change. And then, the reftracted light gets reflected of the third surface$n_3=1.80$ while travelling through the film. This results in$180°$ phase change as the light is reflecting of a denser medium. As a result, the condition for destructive interference or minimum intensity is

$2L=\frac{m\lambda }{n_2}$ (destructive)

The minimum intensity is observed for wavelength,

role="math" localid="1663143094524" $m=1;\lambda =\frac{2L{n}_2}{m}=\frac{2\left(200\mathrm{nm}\right)\left(1.40\right)}{1}=560\mathrm{nm}$

As $560\mathrm{nm}$is in the visible range, hence the wavelength of the reflected light be $560\mathrm{nm}$.