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 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 wavelength of the reflected light is $455\text{nm}$.

Step-by-step solution

Interference in thin films:

Light interference is the cause of the vivid colors that thin oil on water and soap bubbles reflect. These vivid hues can be seen because of the light reflection from the thin film's front and rear surfaces' constructive interference.

For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:(Maxima—bright film in the air)

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

$2L=m\frac{\lambda }{n_2}m=1,2,3,..$(Minima)

Where, $\lambda$is the wavelength of the light in air, $L$is its thickness, and $n_2$is the film’s refractive index.

Determine the wavelength of light:

Here, in this case, light travels in a medium with $n_1=1.32$and incident on the thin layer whose refractive index is $n_2=1.75$and the reflected light has phase change as the light is reflected off the denser medium. And then, the refracted light gets reflected of the back surface $n_3=1.39$while traveling through the film.

This results in no phase change. The total phase difference between $r_1$and $r_2$ is ${180}^o$.

As a result, the condition for constructive interference or maximum intensity is,

$$\begin{gathered} 2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2} \\ \lambda =\frac{4L{n}_2}{2m+1} \end{gathered}$$

The wavelength for first few orders is as below.

For : $m=0$

localid="1663011607370" $$\begin{gathered} \lambda =\frac{4\left(325nm\right)\left(1.75\right)}{2\left(0\right)+1} \\ =2275\text{nm} \end{gathered}$$

For:$m=1$

localid="1663011956572" $\begin{array}{l}\lambda =\frac{4\left(325\mathrm{mm}\right)(1.75)}{2\left(1\right)+1}\\ =758\mathrm{mm}\end{array}$

For :$m=2$

$$\begin{gathered} \lambda =\frac{4(325\text{mm})(1.75)}{2\left(2\right)+1} \\ =455\text{nm} \end{gathered}$$

As $455\text{nm}$is in the visible range, the wavelength of the light is $455\text{nm}$.