The refractive index and the permeability of a medium are respectively 1.5 and `5xx10^-7Hm^-1`. The relative permitivity of the medium is nearly

Options

A. 25
B. 15
C. 10
D. 6

Explanation

To find the relative permittivity (ε_r) of the medium given its refractive index (n) and permeability (μ), we can follow these steps:

### Step 1: Identify the given values
- Refractive index (n) = 1.5
- Permeability (μ) = 5 × 10^(-7) H/m

### Step 2: Calculate the velocity of electromagnetic waves in the medium (V)
The relationship between refractive index (n) and the speed of light (c) in vacuum is given by:
\[ n = \frac{c}{V} \]
From this, we can express the velocity (V) in the medium as:
\[ V = \frac{c}{n} \]
Where \( c \) (the speed of light in vacuum) is approximately \( 3 × 10^8 \) m/s.

Substituting the values:
\[ V = \frac{3 × 10^8 \text{ m/s}}{1.5} = 2 × 10^8 \text{ m/s} \]

### Step 3: Use the relationship between velocity, permeability, and permittivity
The speed of electromagnetic waves in a medium is also related to its permeability (μ) and permittivity (ε) by the formula:
\[ V = \frac{1}{\sqrt{\mu \epsilon}} \]
Squaring both sides gives:
\[ V^2 = \frac{1}{\mu \epsilon} \]
Rearranging this for ε gives:
\[ \epsilon = \frac{1}{V^2 \mu} \]

### Step 4: Substitute the values to find ε
Now, substituting the values we have:
- \( V = 2 × 10^8 \) m/s
- \( μ = 5 × 10^{-7} \) H/m

First, calculate \( V^2 \):
\[ V^2 = (2 × 10^8)^2 = 4 × 10^{16} \text{ m}^2/\text{s}^2 \]

Now substitute into the equation for ε:
\[ \epsilon = \frac{1}{4 × 10^{16} × 5 × 10^{-7}} \]

Calculating the denominator:
\[ 4 × 10^{16} × 5 × 10^{-7} = 20 × 10^{9} = 2 × 10^{10} \]

Thus:
\[ \epsilon = \frac{1}{2 × 10^{10}} = 5 × 10^{-11} \text{ F/m} \]

### Step 5: Calculate the relative permittivity (ε_r)
The relative permittivity (ε_r) is given by:
\[ \epsilon_r = \frac{\epsilon}{\epsilon_0} \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.854 × 10^{-12} \text{ F/m} \).

Now substituting the values:
\[ \epsilon_r = \frac{5 × 10^{-11}}{8.854 × 10^{-12}} \]

Calculating this gives:
\[ \epsilon_r ≈ 5.65 \]

### Conclusion
Thus, the relative permittivity of the medium is approximately 6.