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?

Step by step solution

01

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.

02

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$478.125nm$.

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