91Ó°ÊÓ

The electric and magnetic fields for a monochromatic plane wave of amplitude is ${\mathrm{E}}_0$.

The frequency is $\mathrm{Ó\neg }$.

The phase angle is zero.

The expression to calculate the electric field is given as follows,

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{E}}_{\mathbf{o}}\mathbf{cos}\mathbf{(}\mathbf{k}\mathbf{â‹\dots }\mathbf{r}\mathbf{-}\mathbf{Ó\neg t}\mathbf{+}\mathbf{δ}\mathbf{)}\widehat{\mathbf{n}}$ …… (1)

Here,$\mathbf{δ}$ is the phase angle.

The expression to calculate the magnetic field is given as follows,

The expression to calculate the magnetic field is given as follows,

$\mathbf{B}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{0}}}{\mathbf{c}}\mathbf{cos}\mathbf{(}\mathbf{k}\mathbf{â‹\dots }\mathbf{r}\mathbf{-}\mathbf{Ó\neg t}\mathbf{+}\mathbf{δ}\mathbf{)}\mathbf{(}\widehat{\mathbf{k}}\mathbf{×}\widehat{\mathbf{n}}\mathbf{)}$ …… (2)

The wave is travelling in the negative$x$ direction. Therefore, the wave vector is expressed as,

$k=−\frac{\mathrm{Ó\neg }}{c}\widehat{x}$

Calculate the dot product of$k$ and$r$ as,

$\begin{array}{l}kâ‹\dots r=\left(−\frac{\mathrm{Ó\neg }}{c}\widehat{x}\right)â‹\dots (x\widehat{x}+y\widehat{y}+z\widehat{z})\\ kâ‹\dots r=−\frac{\mathrm{Ó\neg }}{c}x\end{array}$

Calculate the expression for$\widehat{k}×\widehat{n}$ as,

$\begin{array}{l}\widehat{k}×\widehat{n}=−\widehat{x}×\widehat{z}\\ \widehat{k}×\widehat{n}=\widehat{y}\end{array}$

Calculate the expression for the electric field,

Substitute$−\frac{\mathrm{Ó\neg }}{c}x$ for$k.r$ ,$\hat{z}$ for $\hat{n}$and $0$for$\mathrm{δ}$ into equation (1).

$\begin{array}{l}\mathrm{E}(\mathrm{x},\mathrm{t})={\mathrm{E}}_0\cos \left(−\frac{\mathrm{Ó\neg }}{\mathrm{c}}\mathrm{x}−\mathrm{Ó\neg t}+0\right)\widehat{\mathrm{z}}\\ \mathrm{E}(\mathrm{x},\mathrm{t})={\mathrm{E}}_0\cos \left(\frac{\mathrm{Ó\neg }}{\mathrm{c}}\mathrm{x}+\mathrm{Ó\neg t}\right)\widehat{\mathrm{z}}\end{array}$

Calculate the expression for the magnetic field.

Substitute$−\frac{\mathrm{Ó\neg }}{c}x$ for $kâ‹\dots r$,$\widehat{y}$ for$\widehat{k}×\widehat{n}$ and$0$ for$\mathrm{δ}$ into equation (2).

$\begin{array}{c}\mathrm{B}(\mathrm{x},\mathrm{t})=\frac{{\mathrm{E}}_0}{\mathrm{c}}\cos \left(−\frac{\mathrm{Ó\neg }}{\mathrm{c}}\mathrm{x}−\mathrm{Ó\neg t}+0\right)\widehat{\mathrm{y}}\\ \mathrm{B}(\mathrm{x},\mathrm{t})=\frac{{\mathrm{E}}_0}{\mathrm{c}}\cos \left(\frac{\mathrm{Ó\neg }}{\mathrm{c}}\mathrm{x}+\mathrm{Ó\neg t}\right)\widehat{\mathrm{y}}\end{array}$

Sketch of wave for the electric and magnetic field is shown below.

Since the wave is travelling in the direction from $(1,1,1)$with polarization to $xy$plane, therefore the wave vector is given by.

$k=\frac{\mathrm{Ó\neg }}{c}\left(\frac{\widehat{x}+\widehat{y}+\widehat{z}}{\sqrt{3}}\right)$

The normal vector is given by,

$\widehat{n}=\frac{\widehat{x}−\widehat{z}}{\sqrt{2}}$

Calculate the expression for $kâ‹\dots r$as,

$\begin{array}{l}kâ‹\dots r=\frac{\mathrm{Ó\neg }}{c}\left(\frac{\widehat{x}+\widehat{y}+\widehat{z}}{\sqrt{3}}\right)â‹\dots (x\widehat{x}+y\widehat{y}+z\widehat{z})\\ kâ‹\dots r=\frac{\mathrm{Ó\neg }}{\sqrt{3}c}(\widehat{x}+\widehat{y}+\widehat{z})(x\widehat{x}+y\widehat{y}+z\widehat{z})\\ kâ‹\dots r=\frac{\mathrm{Ó\neg }}{\sqrt{3}c}(x+y+z)\end{array}$

Calculate the expression for $\widehat{k}×\widehat{n}$as,

$\begin{array}{l}\widehat{k}×\widehat{n}=\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\widehat{x}& \widehat{y}& \widehat{z}\\ 1& 1& 1\\ 1& 0& −1\end{array}\right|\\ \widehat{k}×\widehat{n}=\frac{1}{\sqrt{6}}(−\widehat{x}+2\widehat{y}−\widehat{z})\end{array}$

Calculate the expression for the electric field,

Substitute $\frac{\mathrm{Ó\neg }}{\sqrt{3}c}(x+y+z)$for $kâ‹\dots r$, $\frac{\widehat{x}−\widehat{z}}{\sqrt{2}}$for $\widehat{n}$ and $0$for $\mathrm{δ}$into equation (1).

$\begin{array}{c}\mathrm{E}(\mathrm{x},\mathrm{y},\mathrm{z},\mathrm{t})={\mathrm{E}}_0\cos \left(\frac{\mathrm{Ó\neg }}{\sqrt{3}\mathrm{c}}(\mathrm{x}+\mathrm{y}+\mathrm{z})−\mathrm{Ó\neg t}+0\right)\frac{\widehat{\mathrm{x}}−\widehat{\mathrm{z}}}{\sqrt{2}}\\ \mathrm{E}(\mathrm{x},\mathrm{y},\mathrm{z},\mathrm{t})={\mathrm{E}}_0\cos \left(\frac{\mathrm{Ó\neg }}{\sqrt{3}\mathrm{c}}(\mathrm{x}+\mathrm{y}+\mathrm{z})−\mathrm{Ó\neg t}\right)\frac{\widehat{\mathrm{x}}−\widehat{\mathrm{z}}}{\sqrt{2}}\end{array}$

Calculate the expression for the magnetic field.

Substitute$\frac{\mathrm{Ó\neg }}{\sqrt{3}c}(x+y+z)$for $k.r$,$\frac{1}{\sqrt{6}}(−\widehat{x}+2\widehat{y}−\widehat{z})$for$\widehat{k}×\widehat{n}$and$0$for$\mathrm{δ}$into equation (2).

$\begin{array}{l}B(x,y,z,t)=\frac{E_0}{c}\cos \left(\frac{\mathrm{Ó\neg }}{\sqrt{3}c}(x+y+z)−\mathrm{Ó\neg }t+0\right)\frac{1}{\sqrt{6}}(−\widehat{x}+2\widehat{y}−\widehat{z})\\ B(x,y,z,t)=\frac{E_0}{c}\cos \left(\frac{\mathrm{Ó\neg }}{\sqrt{3}c}(x+y+z)−\mathrm{Ó\neg }t\right)\frac{1}{\sqrt{6}}(−\widehat{x}+2\widehat{y}−\widehat{z})\end{array}$

Sketch of wave for the electric and magnetic field is shown below.