Answer

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**Hint:**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{align}

& \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{align}$

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{align}

& \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{align}$

Now put the value of k in equation (4)

$\begin{align}

& \Rightarrow 10\left( \widehat{s} \right)=8\widehat{k}-6\widehat{j} \\

& \Rightarrow \widehat{s}=\dfrac{8\widehat{k}-6\widehat{j}}{10} \\

\end{align}$

$\therefore \widehat{s}=\dfrac{4\widehat{k}-3\widehat{j}}{5}$

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).

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