JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 4)
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 direction of propagation $$\widehat s$$, is :
$$\widehat s = {{3\widehat i - 4\widehat j} \over 5}$$
$$\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}$$
$$\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$$
$$\widehat s = {{4\widehat j - 3\widehat k} \over 5}$$
Explanation
$$\overrightarrow E = {E_0}\widehat n\sin \left( {\omega t + \left( {6y - 8z} \right)} \right)$$
$$ = {E_0}\widehat n\sin \left( {\omega t + \overrightarrow k .\overrightarrow r } \right)$$
where $$\overrightarrow r = x\widehat i + y\widehat j + z\widehat k$$ and $$\overrightarrow k .\overrightarrow r = 6y - 8z$$
$$ \Rightarrow \overrightarrow k = 6\widehat j - 8\widehat k$$
direction of propagation
$$\widehat s = - \widehat k = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$$
$$ = {E_0}\widehat n\sin \left( {\omega t + \overrightarrow k .\overrightarrow r } \right)$$
where $$\overrightarrow r = x\widehat i + y\widehat j + z\widehat k$$ and $$\overrightarrow k .\overrightarrow r = 6y - 8z$$
$$ \Rightarrow \overrightarrow k = 6\widehat j - 8\widehat k$$
direction of propagation
$$\widehat s = - \widehat k = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$$
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