Writing an expression for the scalar value of the electric field $E$

Question

I am currently studying Optics, fifth edition, by Hecht. In chapter 2.7 Plane Waves, the following example is given:

An electromagnetic plane wave is described by its electric field $E$. The wave has an amplitude $E_0$, an angular frequency $\omega$, a wavelength $\lambda$, and travels at speed $c$ outward in the direction of the unit propagation vector

$$\hat{\mathbf{k}} = (4\hat{\mathbf{i}} + 2\hat{\mathbf{j}})/\sqrt{20}$$

(not to be confused with the unit basis vector $\hat{\mathbf{k}}$). Write an expression for the scalar value of the electric field $E$.

Solution

We want an equation of the form

$$E(x, y, z, t) = E_0 e^{i\hat{\mathbf{k}} \cdot (\vec{\mathbf{r}} - \omega t)}$$

Here

$$\vec{\mathbf{k}} \cdot \vec{\mathbf{r}} = \dfrac{2\pi}{\lambda} \hat{\mathbf{k}} \cdot \vec{\mathbf{r}}$$

and

$$\vec{\mathbf{k}} \cdot \vec{\mathbf{r}} = \dfrac{2 \pi}{\lambda \sqrt{20}}(4\hat{\mathbf{i}} + 2\hat{\mathbf{j}}) \cdot (x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}})$$

$$\vec{\mathbf{k}} \cdot \vec{\mathbf{r}} = \dfrac{\pi}{\lambda \sqrt{5}}(4x + 2y)$$

Hence

$$E = E_0e^{i \left[ \dfrac{\pi}{\lambda \sqrt{5}}(4x + 2y) - \omega t \right]}$$

I have the following questions:

1. Shouldn't $E(x, y, z, t) = E_0 e^{i\hat{\mathbf{k}} \cdot (\vec{\mathbf{r}} - \omega t)}$ be $E(x, y, z, t) = E_0 e^{i(\hat{\mathbf{k}} \cdot \vec{\mathbf{r}} - \omega t)}$?

2. Where did $\vec{\mathbf{k}} \cdot \vec{\mathbf{r}} = \dfrac{2\pi}{\lambda} \hat{\mathbf{k}} \cdot \vec{\mathbf{r}}$ come from? What is the reasoning for this?

I would greatly appreciate it if people would please take the time to clarify these points.

Answer 1

1. Shouldn't $E(x, y, z, t) = E_0 e^{i\hat{\mathbf{k}} \cdot (\vec{\mathbf{r}} - \omega t)}$ be $E(x, y, z, t) = E_0 > e^{i(\hat{\mathbf{k}} \cdot \vec{\mathbf{r}} - \omega t)}$?

Yes. Just check the dimensions.

2. Where did $\vec{\mathbf{k}} \cdot \vec{\mathbf{r}} = \dfrac{2\pi}{\lambda} \hat{\mathbf{k}} \cdot \vec{\mathbf{r}}$ come from? What is the reasoning for this?

The hat indicates a unit vector. Hence $\hat k = \vec k /|\vec k|$. as you probably know the wavenumber (=magnitude of the vector $\vec k$) is given by $k=2\pi/\lambda$