An EM wave from air enters a medium .The electric fields are ` vecE _1 = vecE_(01 ) hat x cos [ 2 pi v ((z)/(c )-t)] ` in air and ` vecE_2 = E_(02) hat x cos [ k ( 2z-c t ) ]` in medium , where the wave number k and frequency v refer to their values in air . the medium is non - magnetic . if ` epsi_(r_1) and epsi_(r_2)` refer to relative permittivities of air and medium respectively , which of the following options is correct ?

Options

A. `(epsi_(r_1))/( epsi_(r_2)) = 1/4`
B. `(epsi_(r_1))/( epsi_(r_2))=1/2`
C. `(epsi_(r_1))/(epsi_(r_2))=4`
D. `(epsi_(r_1))/( epsi_(r_2))=2`

Explanation

To solve the problem, we need to analyze the given electric fields of the electromagnetic waves in air and in the medium, and then find the relationship between 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:
\[ \vec{E}_1 = E_{01} \hat{x} \cos\left[2\pi \nu \left(\frac{z}{c} - t\right)\right] \]
- In the medium, the electric field is given by:
\[ \vec{E}_2 = E_{02} \hat{x} \cos\left[k(2z - ct)\right] \]

2. **Relate the Wave Parameters**:
- The wave number \( k \) and frequency \( \nu \) refer to their values in air. The speed of light in air is \( c \).
- The relationship between speed, frequency, and wave number is given by:
\[ v = \frac{\omega}{k} \]
- Here, \( \omega = 2\pi\nu \).

3. **Find the Speed of Light in the Medium**:
- For air:
\[ V_1 = \frac{c}{\sqrt{\epsilon_{r1}}} \]
- For the medium, since it is non-magnetic, we can write:
\[ V_2 = \frac{c}{\sqrt{\epsilon_{r2}}} \]

4. **Using the relationship between the speeds**:
- From the given electric field in the medium, we can find the speed of light in the medium. Since the electric field is given in terms of \( k \):
\[ k = \frac{2\pi \nu}{c} \]
- The speed of light in the medium can also be expressed as:
\[ V_2 = \frac{\omega}{k} = \frac{2\pi\nu}{k} \]

5. **Setting up the relationship**:
- Since the medium is non-magnetic, we have:
\[ \epsilon_{r1} = 1 \quad (\text{for air}) \]
- Therefore, we can relate the permittivities:
\[ \frac{V_1}{V_2} = \sqrt{\frac{\epsilon_{r2}}{\epsilon_{r1}}} \]

6. **Substituting the values**:
- We know that:
\[ V_1 = c \quad \text{and} \quad V_2 = \frac{c}{\sqrt{\epsilon_{r2}}} \]
- Thus, we can write:
\[ \frac{c}{\frac{c}{\sqrt{\epsilon_{r2}}}} = \sqrt{\frac{\epsilon_{r2}}{\epsilon_{r1}}} \]
- Simplifying gives:
\[ \sqrt{\epsilon_{r2}} = \frac{c}{V_2} \]

7. **Finding the ratio**:
- From the above relationships, we can derive:
\[ \frac{\epsilon_{r1}}{\epsilon_{r2}} = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{4}} = 4 \]
- Thus:
\[ \frac{\epsilon_{r1}}{\epsilon_{r2}} = \frac{1}{4} \]

### Conclusion:
The ratio of the relative permittivities is:
\[ \frac{\epsilon_{r1}}{\epsilon_{r2}} = \frac{1}{4} \]