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 ${\mathit{r}}_{\mathbf{3}}$ and ${\mathit{r}}_{\mathbf{4}}$ interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 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 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.

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 is $n_3=1.80$.

The thickness of the layer is $L=200nm$.

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, as $n_1>n_2$and $n_2<n_3$the two transmitted rays have zero phase angle difference because the ray $r_4$ will be shifted by localid="1663145171476" $\frac{\lambda }{2}$ twice on two reflections.

Therefore, the condition for constructive interference is,

$$\begin{gathered} 2L=m\frac{{\lambda }_{\max }}{n_2} \\ {\lambda }_{\max }=\frac{2L{n}_2}{m} \end{gathered}$$

Calculating the wavelength for first few order numbers as follow.

For $m=1$:

$$\begin{gathered} {\lambda }_1=\frac{4\left(200nm\right)\left(1.40\right)}{1} \\ =1120nm \end{gathered}$$

For $m=2$:

$$\begin{gathered} {\lambda }_2=\frac{4\left(200nm\right)\left(1.40\right)}{2} \\ =560nm \end{gathered}$$

For $m=3$:

$$\begin{gathered} {\lambda }_3=\frac{4\left(200nm\right)\left(1.40\right)}{3} \\ =373nm \end{gathered}$$

As $560nm$ lies in visible range, hence the wavelength with maximum intensity of transmitted light is $560nm$.