Question

Transmission through thin layers. In Fig. 35-43, 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.) Part of the light ends up in material 3 as ray ${\mathit{r}}_{\mathbf{3}}$(the light does not reflect inside material 2) and ${\mathit{r}}_{\mathbf{4}}$(the light reflects twice inside material 2). The waves of and interfere,${\mathit{r}}_{\mathbf{3}}$and ${\mathit{r}}_{\mathbf{4}}$here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction ${\mathit{n}}_{\mathbf{1}}\mathbf{,}{\mathit{n}}_{\mathbf{2}}$and ${\mathit{n}}_{\mathbf{3}}$the type of interference, the thin-layer thickness $\mathit{L}$in nanometers, and the wavelength $\mathit{\lambda }$in nanometers 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 with minimum intensity of transmitted light is $680\text{nm}$.

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 $L=285\text{nm}$.

Interference of light through thin films.

Light that is incident normally on thin films is reflected from both the front and back surfaces, causing interference of the reflected light. When constructive interference happens, it produces bright reflected light, and when entirely destructive interference occurs, it produces a dark region.

The interference of the transmitted rays is similar to the interference of the reflection of light. Here in this case, the phase difference between the transmitted rays is zero. Therefore, the condition for destructive interference is

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

Calculating the wavelength for first few orders number,

$\begin{array}{l}m=0;{\lambda }_1=\frac{4\left(285\mathrm{nm}\right)\left(1.60\right)}{2\left(0\right)+1}=1824\mathrm{nm}\\ m=1;{\lambda }_2=\frac{4\left(285\mathrm{nm}\right)\left(1.60\right)}{2\left(1\right)+1}=608\mathrm{nm}\\ m=2;{\lambda }_3=\frac{4\left(285\mathrm{nm}\right)\left(1.60\right)}{2\left(2\right)+1}=365\mathrm{nm}\end{array}$

As $680\text{nm}$lies in visible range, hence the wavelength with minimum intensity of transmitted light is $680\text{nm}$.