An electromagnetic wave of frequency `v=3.0 MHz` passes from vacuum into a dielectric medium with permittivity `epsilon=4.0`. Then

Options

A. wavelength and frequency both become half.
B. wavelength is doubled and frequency remains unchanged.
C. wavelength and frequency both remain unchanged.
D. wavelength is halved and frequency remains unchanged.

Explanation

To solve the problem of an electromagnetic wave transitioning from vacuum into a dielectric medium with a given permittivity, we can follow these steps:

### Step 1: Understand the relationship between frequency, wavelength, and velocity of electromagnetic waves.
The fundamental relationship for electromagnetic waves is given by the equation:
\[ v = f \cdot \lambda \]
where:
- \( v \) is the velocity of the wave,
- \( f \) is the frequency,
- \( \lambda \) is the wavelength.

### Step 2: Identify the frequency of the wave.
The frequency of the electromagnetic wave is given as:
\[ f = 3.0 \, \text{MHz} = 3.0 \times 10^6 \, \text{Hz} \]

### Step 3: Determine the velocity of the wave in vacuum.
The speed of light in vacuum is:
\[ c = 3 \times 10^8 \, \text{m/s} \]

### Step 4: Calculate the wavelength in vacuum.
Using the relationship \( v = f \cdot \lambda \), we can find the wavelength in vacuum (\( \lambda_0 \)):
\[ \lambda_0 = \frac{c}{f} = \frac{3 \times 10^8 \, \text{m/s}}{3.0 \times 10^6 \, \text{Hz}} = 100 \, \text{m} \]

### Step 5: Understand the effect of entering a dielectric medium.
When the electromagnetic wave enters a dielectric medium, the frequency remains unchanged, but the velocity and wavelength will change. The velocity of the wave in a dielectric medium is given by:
\[ v = \frac{c}{\sqrt{\epsilon_r}} \]
where \( \epsilon_r \) is the relative permittivity of the medium.

### Step 6: Calculate the velocity in the dielectric medium.
Given that the permittivity \( \epsilon_r = 4.0 \):
\[ v_{\text{medium}} = \frac{c}{\sqrt{4}} = \frac{3 \times 10^8 \, \text{m/s}}{2} = 1.5 \times 10^8 \, \text{m/s} \]

### Step 7: Calculate the new wavelength in the dielectric medium.
Now, we can find the new wavelength (\( \lambda_{\text{medium}} \)) in the dielectric medium using the unchanged frequency:
\[ \lambda_{\text{medium}} = \frac{v_{\text{medium}}}{f} = \frac{1.5 \times 10^8 \, \text{m/s}}{3.0 \times 10^6 \, \text{Hz}} = 50 \, \text{m} \]

### Step 8: Summarize the results.
- The frequency of the wave remains unchanged at \( 3.0 \, \text{MHz} \).
- The wavelength in vacuum is \( 100 \, \text{m} \), and in the dielectric medium, it is \( 50 \, \text{m} \).

### Final Conclusion:
- The wavelength is halved when the wave enters the dielectric medium, while the frequency remains unchanged.