Modern Physics 7 Question 73
3. An electromagnetic wave is represented by the electric field
$\mathbf{E}=E _0 \hat{\mathbf{n}} \sin [\omega t+(6 y-8 z)]$. Taking unit vectors in $x, y$ and $z$-directions to be $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$, the direction of propagation $\hat{\mathbf{s}}$, is
(a) $\hat{\mathbf{s}}=\frac{3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}}{5}$
(b) $\hat{\mathbf{s}}=\frac{-4 \hat{\mathbf{k}}+3 \hat{\mathbf{j}}}{5}$
(c) $\hat{\mathbf{s}}=\frac{-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}}{5}$
(d) $\hat{\mathbf{s}}=\frac{4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}}{5}$
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Solution:
- Standard expression of electromagnetic wave is given by
$$ \mathbf{E}=E _0 \hat{\mathbf{n}}[\sin (\omega t-\mathbf{k} \cdot \hat{\mathbf{r}})] $$
Here, $\mathbf{k}$ is the propagation vector. Direction of propagation in this case is $\hat{\mathbf{k}}$.
Given expression of electromagnetic wave,
$$ \begin{array}{ll} & \mathbf{E}=E _0 \hat{\mathbf{n}} \sin [\omega t+(6 y-8 z)] \\ \Rightarrow \quad & \mathbf{E}=E _0 \hat{\mathbf{n}} \sin [\omega t-(8 z-6 y)] \end{array} $$
Comparing Eq. (ii) with Eq. (i), we get
$$ \begin{aligned} & & \mathbf{k} \cdot \hat{\mathbf{r}} & =8 z-6 y \\ \text { Here, } & & \hat{\mathbf{r}} & =x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ & \text { and } & \mathbf{k} & =k _x \hat{\mathbf{i}}+k _y \hat{\mathbf{j}}+k _z \hat{\mathbf{k}} \\ & \therefore & \mathbf{k} \cdot \hat{\mathbf{r}} & =x k _x+y k _y+z k _z \end{aligned} $$
From Eqs. (iii) and (iv), we get
$$ \begin{aligned} x k _x & =\text { zero } \Rightarrow k _x=0 \\ y k _y & =-6 y \Rightarrow k _y=-6 \\ z k _z & =8 z \Rightarrow k _z=8 \end{aligned} $$
Hence, $\mathbf{k}=-6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$
So, direction of propagation,
$$ \hat{\mathbf{s}}=\hat{\mathbf{k}}=\frac{\mathbf{k}}{|\mathbf{k}|}=\frac{-6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}}{\sqrt{6^{2}+8^{2}}}=\frac{-6 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}}{10}=\frac{-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}}{5} $$
