The electric field of an electromagnetic wave travelling through vacuum is given by the equation ` E = E_(0) "sin"(kx - omegat)` The quantity that is independent of wavelength is

Options

A. `komega`
B. `(k)/(omega)`
C. `k^(2)omega`
D. `omega`

Explanation

To determine the quantity that is independent of wavelength from the given electric field equation of an electromagnetic wave, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Equation**: The electric field of the electromagnetic wave is given by:
\[ E = E_0 \sin(kx - \omega t) \]
Here, \(E_0\) is the amplitude, \(k\) is the wave number, and \(\omega\) is the angular frequency.

2. **Identify the Relationships**:
- The wave number \(k\) is related to the wavelength \(\lambda\) by the equation:
\[ k = \frac{2\pi}{\lambda} \]
- The angular frequency \(\omega\) is related to the frequency \(\nu\) by:
\[ \omega = 2\pi\nu \]
- The frequency \(\nu\) is related to the speed of light \(c\) and wavelength \(\lambda\) by:
\[ \nu = \frac{c}{\lambda} \]

3. **Express \(\omega\) in Terms of \(k\)**:
- Substitute the expression for \(\nu\) into the equation for \(\omega\):
\[ \omega = 2\pi \left(\frac{c}{\lambda}\right) \]
- Thus, we can express \(\omega\) in terms of \(k\):
\[ \omega = k \cdot c \]

4. **Calculate the Ratio \(\frac{\omega}{k}\)**:
- Now, we can find the ratio:
\[ \frac{\omega}{k} = \frac{kc}{k} = c \]
- This shows that the ratio \(\frac{\omega}{k}\) is equal to the speed of light \(c\).

5. **Conclusion**:
- The speed of light \(c\) is a constant value, approximately \(3 \times 10^8 \, \text{m/s}\), and is independent of the wavelength of the electromagnetic wave.

### Final Answer:
The quantity that is independent of wavelength is the speed of light \(c\).

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