The electric field in space is given by `overset(rarr)E=E_(0)sin (omegat+6y-8z)hatn` then the direction of propagation of light wave is

Options

A. `hats=(3hati-4hatj)/(5)`
B. `hats=(-4hatk-3hatj)/(5)`
C. `hats=((-3hatj+4hatk)/(5))`
D. `hats=(4hatj-3hatk)/(5)`

Answer

To find the direction of propagation of the light wave given the electric field \(\overset{rarr}{E} = E_0 \sin(\omega t + 6y - 8z) \hat{n}\), we can follow these steps:

### Step 1: Identify the form of the electric field
The electric field is given in the form:
\[ \overset{rarr}{E} = E_0 \sin(\omega t + 6y - 8z) \hat{n} \]
This can be compared to the general form of a plane wave:
\[ \overset{rarr}{E} = E_0 \sin(\omega t + \overset{rarr}{k} \cdot \overset{rarr}{r}) \]
where \(\overset{rarr}{k}\) is the wave vector.

### Step 2: Rewrite the electric field equation
We can rewrite the electric field equation as:
\[ \overset{rarr}{E} = E_0 \sin(\omega t - ( -6y + 8z) ) \hat{n} \]
This indicates that the wave is propagating in the direction opposite to the vector \((0, 6, -8)\).

### Step 3: Identify the wave vector
From the term \(6y - 8z\), we can identify the wave vector \(\overset{rarr}{k}\):
\[ \overset{rarr}{k} = (0, 6, -8) \]

### Step 4: Determine the direction of propagation
The direction of propagation of the wave is given by the negative of the wave vector:
\[ \overset{rarr}{S} = -\overset{rarr}{k} = (0, -6, 8) \]

### Step 5: Normalize the direction vector
To find the unit vector in the direction of propagation, we calculate its magnitude:
\[ |\overset{rarr}{S}| = \sqrt{0^2 + (-6)^2 + 8^2} = \sqrt{0 + 36 + 64} = \sqrt{100} = 10 \]
Now, we can find the unit vector:
\[ \hat{S} = \frac{\overset{rarr}{S}}{|\overset{rarr}{S}|} = \frac{(0, -6, 8)}{10} = (0, -0.6, 0.8) \]

### Conclusion
The direction of propagation of the light wave is given by the unit vector:
\[ \hat{S} = (0, -0.6, 0.8) \]