An EM wave from air enters a medium. The electric fields are
`overset(vec)E_(1)=E_(01)hatx cos[2piv((z)/(c )-t)]` in air and
`overset(vec)E=E_(02)hatx cos [k(2z-ct)]` in medium, where the wave number k and frequency v refer to their values in air. The medium is non-magnetic. If `in_(r_(1))` and `in_(r_(2))` refer to relative permittivities of air and medium respectively, which of the following options is correct ?

Options

A. Option1 `(in_(r_(1)))/(in_(r_(2)))=(1)/(2)`
B. Option2 `(in_(r_(1)))/(in_(r_(2)))=4`
C. Option3 `(in_(r_(1)))/(in_(r_(2)))=2`
D. Option4 `(in_(r_(1)))/(in_(r_(2)))=(1)/(4)`

Answer

To solve the problem, we need to analyze the given electric fields in air and in the medium, and then find the ratio of the relative permittivities of air and the medium.

### Step-by-Step Solution:

1. **Identify the Electric Fields**:
- In air, the electric field is given by:
\[ \vec{E_1} = E_{01} \hat{x} \cos\left[2\pi v\left(\frac{z}{c} - t\right)\right] \]
- In the medium, the electric field is:
\[ \vec{E} = E_{02} \hat{x} \cos\left[k(2z - ct)\right] \]

2. **Understand the Wave Properties**:
- The wave number \( k \) and frequency \( v \) refer to their values in air. The speed of light in air is \( c \).

3. **Determine the Speed of Light in the Medium**:
- The speed of light in a medium is given by:
\[ v = \frac{c}{n} \]
- The refractive index \( n \) can be expressed in terms of the relative permittivity \( \epsilon_r \) and relative permeability \( \mu_r \):
\[ n = \sqrt{\epsilon_r \mu_r} \]
- Since the medium is non-magnetic, \( \mu_r = 1 \). Thus:
\[ n = \sqrt{\epsilon_r} \]

4. **Relate the Speeds**:
- The speed of light in the medium can also be expressed as:
\[ v = \frac{c}{\sqrt{\epsilon_r}} \]
- Therefore, the square of the speed of light in the medium is:
\[ v^2 = \frac{c^2}{\epsilon_r} \]

5. **Find the Ratio of Permittivities**:
- The ratio of the relative permittivities of air (\( \epsilon_{r1} \)) and the medium (\( \epsilon_{r2} \)) can be expressed as:
\[ \frac{\epsilon_{r1}}{\epsilon_{r2}} = \frac{v^2}{c^2} \]
- We know \( v^2 = \frac{c^2}{\epsilon_{r2}} \), thus:
\[ \frac{\epsilon_{r1}}{\epsilon_{r2}} = \frac{c^2 / \epsilon_{r2}}{c^2} = \frac{1}{\epsilon_{r2}} \]

6. **Calculate the Value of \( \epsilon_{r2} \)**:
- From the electric field in the medium, we can determine the coefficient of \( z \) in the cosine function, which gives us the wave speed:
\[ k = \frac{2\pi}{\lambda} \quad \text{and} \quad \text{velocity} = \frac{\omega}{k} \]
- The coefficient of \( z \) in the medium's electric field is \( 2k \), leading to:
\[ v = \frac{\omega}{2k} \]

7. **Final Calculation**:
- Substituting \( v = \frac{c}{2} \):
\[ \epsilon_{r2} = \left(\frac{c}{v}\right)^2 = \left(\frac{c}{c/2}\right)^2 = 4 \]
- Thus, the ratio becomes:
\[ \frac{\epsilon_{r1}}{\epsilon_{r2}} = \frac{1}{4} \]

### Conclusion:
The correct answer is:
\[ \frac{\epsilon_{r1}}{\epsilon_{r2}} = \frac{1}{4} \]