Question

57 through 68 64, 65 59 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 second least thickness is 329 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 second least thickness

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

$n_1=1.5,n_2=1.34$ and $n_3=1.42$

$n_2<n_1$and $n_2<n_3$

Wavelength of light $\lambda =587$

The condition of maximum 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 second least thickness $m=1$

$$\begin{gathered} L=\left(1+\frac{1}{2}\right)\left(\frac{587\text{nm}}{2\left(1.34\right)}\right) \\ =329\text{nm} \end{gathered}$$

Hence, the second least thickness is$329nm$.