Question

An electromagnetic wave is represented by the electric field $\overrightarrow{E}={{E}_{0}}\widehat{n}\sin \left[ \omega t+\left( 6y-8z \right) \right]$. Taking unit vectors in x, y and z directions to be $\widehat{i,}\widehat{j},\widehat{k}$ the directions of propagation $\widehat{s}$is:
$\begin{aligned} & A.\text{ }\widehat{s}=\dfrac{4\widehat{j}-3\widehat{k}}{5} \\\ & B.\text{ }\widehat{s}=\dfrac{3\widehat{i}-4\widehat{j}}{5} \\\ & C.\text{ }\widehat{s}=\left[ \dfrac{-3\widehat{j}+4\widehat{k}}{5} \right] \\\ & D.\text{ }\widehat{s}=\left[ \dfrac{-4\widehat{k}+3\widehat{j}}{5} \right] \\\ \end{aligned}$

Answer 1

First we will compare given formula to actual electric field formula $\overrightarrow{E}={{E}_{0}}\widehat{n}\left[ \omega t-\widehat{s}k \right]$after that we will convert the unit vectors 6y – 8z into direction vector then after we can find $\widehat{s}$after comparing equation.
Formula used:
$\overrightarrow{E}={{E}_{0}}\widehat{n}\left[ \omega t-\widehat{s}k \right]$

Complete answer:
It is given that the electric field,
$\overrightarrow{E}={{E}_{0}}\widehat{n}\sin \left[ \omega t+\left( 6y-8z \right) \right]....\left( 1 \right)$
In order to find direction of propagation $\widehat{s}$we have to compare above equation to actual electric field equation and that equation is
$\overrightarrow{E}={{E}_{0}}\widehat{n}\left[ \omega t-\widehat{s}k \right]....\left( 2 \right)$
Where,
${{E}_{0}}$ = initial electric field.
ω = angular velocity
t = time
$\widehat{s}$= propagation vector
k = resultant vector.
Now in order to compare both equations we have to convert equation (1) into direction vector. Now it is given that direction of y is $\widehat{j}$ vector and z is represented by $\widehat{k}$ vector so that,
$6y-8z=6\widehat{j}-8\widehat{k}$
Now equation (1) can be written as,
$\begin{aligned} & \Rightarrow \overrightarrow{E}={{E}_{0}}\widehat{n}\sin \left( \omega t+\left( 6\widehat{j}-8\widehat{k} \right) \right) \\\ & \Rightarrow \overrightarrow{E}={{E}_{0}}\widehat{n}\sin \left( \omega t+(-\left( -6\widehat{j}+8\widehat{k} \right) \right) \\\ & \therefore \overrightarrow{E}={{E}_{0}}\widehat{n}\sin \left( \omega t-\left( 8\widehat{k}-6\widehat{j} \right) \right)......(3) \\\ \end{aligned}$
Now comparing equation (2) and (3) we can get
$k\widehat{s}=8\widehat{k}-6\widehat{j}......\left( 4 \right)$
Here k is resultant vector to find k we have to use below formula
$\begin{aligned} & \Rightarrow k=\sqrt{{{\left( x\widehat{i} \right)}^{2}}+{{\left( y\widehat{j} \right)}^{2}}+{{\left( z\widehat{k} \right)}^{2}}} \\\ & \Rightarrow k=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 6\widehat{j} \right)}^{2}}+{{\left( -8\widehat{k} \right)}^{2}}} \\\ & \Rightarrow k=\sqrt{36+64} \\\ & \Rightarrow k=\sqrt{100} \\\ & \therefore k=10......\left( 5 \right) \\\ \end{aligned}$
Now put the value of k in equation (4)
$\begin{aligned} & \Rightarrow 10\left( \widehat{s} \right)=8\widehat{k}-6\widehat{j} \\\ & \Rightarrow \widehat{s}=\dfrac{8\widehat{k}-6\widehat{j}}{10} \\\ \end{aligned}$
$\therefore \widehat{s}=\dfrac{4\widehat{k}-3\widehat{j}}{5}$
Here $\widehat{s}$is direction of propagation of the light.

So hence the correct option is (C) .

Note:
So when we compare both the equations then we have to see the sign of the equation for example in equation (3) (I) will take negative (-ve) sign common so that the ( I) can relate the equation and can match the negative (-ve) sign with the other equation. So the correct option is (C).