An electromagnetic wave going through vacuum is described by
` E= E_0sin(kx- omegat).` Which of the following is/are independent of the wavelength?

Options

A. k
B. `omega`
C. `k/omega `
D. `k omega`

Answer

To solve the question regarding which of the given options are independent of the wavelength in the electromagnetic wave described by the equation \( E = E_0 \sin(kx - \omega t) \), we will analyze each option step by step.

### Step 1: Understand the parameters of the wave equation
The equation of the electromagnetic wave is given as:
\[ E = E_0 \sin(kx - \omega t) \]
Here, \( k \) is the wave number, \( \omega \) is the angular frequency, and \( E_0 \) is the amplitude of the wave.

### Step 2: Define the wave number \( k \)
The wave number \( k \) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
where \( \lambda \) is the wavelength. Since \( k \) is directly dependent on \( \lambda \), this option is **not independent of the wavelength**.

### Step 3: Define the angular frequency \( \omega \)
The angular frequency \( \omega \) is defined as:
\[ \omega = 2\pi \nu \]
where \( \nu \) is the frequency. We also know that the speed of light \( c \) is given by:
\[ c = \nu \lambda \]
From this, we can express frequency \( \nu \) as:
\[ \nu = \frac{c}{\lambda} \]
Substituting this into the expression for \( \omega \):
\[ \omega = 2\pi \left(\frac{c}{\lambda}\right) \]
This shows that \( \omega \) also depends on \( \lambda \), making this option **not independent of the wavelength**.

### Step 4: Analyze \( \frac{k}{\omega} \)
Now, we will analyze the expression \( \frac{k}{\omega} \):
\[ \frac{k}{\omega} = \frac{\frac{2\pi}{\lambda}}{2\pi \left(\frac{c}{\lambda}\right)} \]
Simplifying this gives:
\[ \frac{k}{\omega} = \frac{1}{c} \]
This expression is independent of \( \lambda \), hence this option is **independent of the wavelength**.

### Step 5: Analyze \( k \cdot \omega \)
Next, we will analyze the expression \( k \cdot \omega \):
\[ k \cdot \omega = \left(\frac{2\pi}{\lambda}\right) \cdot \left(2\pi \left(\frac{c}{\lambda}\right)\right) \]
This simplifies to:
\[ k \cdot \omega = \frac{4\pi^2 c}{\lambda^2} \]
Since this expression contains \( \lambda \) in the denominator, it is **dependent on the wavelength**.

### Conclusion
From the analysis:
- **Option 1 (k)**: Dependent on \( \lambda \)
- **Option 2 (ω)**: Dependent on \( \lambda \)
- **Option 3 (k/ω)**: Independent of \( \lambda \) (Correct answer)
- **Option 4 (k·ω)**: Dependent on \( \lambda \)

Thus, the only option that is independent of the wavelength is **Option 3: \( \frac{k}{\omega} \)**.