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 ${\mathbf{r}}_{\mathbf{3}}$ (the light does not reflect inside material 2) and ${\mathbf{r}}_{\mathbf{4}}$(the light reflects twice inside material 2). The waves of ${\mathbf{r}}_{\mathbf{3}}$ and ${\mathbf{r}}_{\mathbf{4}}$ 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 ${\mathbf{n}}_{\mathbf{1}}\mathbf{,}{\mathbf{n}}_{\mathbf{2}}\mathbf{and}{\mathbf{n}}_{\mathbf{3}}$, the type.

Of interference, the thin-layer thickness $\mathbf{L}$ in nanometres, and the wavelength $\mathbf{\lambda }$ in nanometres of the light as measured in air.

Where $\mathbf{\lambda }$ is missing, give the wavelength that is in the visible range.

Where $\mathbf{L}$is missing, give the second least thickness or the third least thickness as indicated?

Short answer

The third least thickness is 478.125 nm.

Step-by-step solution

Introduction

Reflection is the process by which electromagnetic radiation is returned either at the boundary between two media (surface reflection) or at the interior of a medium (volume reflection), whereas transmission is the passage of electromagnetic radiation through a medium.

Find the third least thickness is

$$\begin{gathered} n_1=1.55,n_2=1. \\ L=\left(2+\frac{1}{2}\right)\left(\frac{612\text{nm}}{2\left(1.60\right)}\right) \\ =478.125\text{nm} \\ 60n_3=1.33 \\ {n}_2>n_1{n}_2>n_3 \end{gathered}$$

We use condition for destructive interference which gives the minimum reflection.

$n_1=1.55,n_2=1.60$and $n_3=1.33$

$n_2>n_1$ and $n_2>n_3$

Wavelength of light $\lambda =612$

The condition of minimum reflection or destructive interference is

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

Where $m=0,1,2,......$

For third least thickness occurs when $m=2$

$$\begin{gathered} L=\left(2+\frac{1}{2}\right)\left(\frac{612\text{nm}}{2\left(1.60\right)}\right) \\ =478.125\text{nm} \end{gathered}$$

Hence, the third least thickness is$478.125nm$.