The ratio of contributions made by the eletric field and magnetic field components to the intensity of an `EM` wave is.

Options

A. `c : 1`
B. `c^(2) :1`
C. `1 : 1`
D. `sqrt(c) :1`

Explanation

To find the ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic (EM) wave, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Electric and Magnetic Fields**:
The electric field amplitude \( E_0 \) and magnetic field amplitude \( B_0 \) in an electromagnetic wave are related by the equation:
\[ E_0 = c \cdot B_0 \]
where \( c \) is the speed of light.

2. **Calculate Average Energy Density**:
The average energy density contributed by the electric field \( U_E \) is given by:
\[ U_E = \frac{1}{2} \epsilon_0 E_0^2 \]
and the average energy density contributed by the magnetic field \( U_B \) is given by:
\[ U_B = \frac{1}{2} \frac{B_0^2}{\mu_0} \]

3. **Substituting the Relationship**:
We can substitute \( E_0 = c \cdot B_0 \) into the equation for \( U_E \):
\[ U_E = \frac{1}{2} \epsilon_0 (c \cdot B_0)^2 = \frac{1}{2} \epsilon_0 c^2 B_0^2 \]

4. **Using the Speed of Light Relation**:
Recall that the speed of light \( c \) can be expressed in terms of \( \epsilon_0 \) and \( \mu_0 \):
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \implies c^2 = \frac{1}{\mu_0 \epsilon_0} \]
Substitute \( c^2 \) into the energy density equation:
\[ U_E = \frac{1}{2} \epsilon_0 \left(\frac{1}{\mu_0 \epsilon_0}\right) B_0^2 = \frac{1}{2 \mu_0} B_0^2 \]

5. **Finding the Ratio of Average Energy Densities**:
Now we have:
- \( U_E = \frac{1}{2 \mu_0} B_0^2 \)
- \( U_B = \frac{1}{2} \frac{B_0^2}{\mu_0} \)

The ratio of the average energy densities is:
\[ \frac{U_E}{U_B} = \frac{\frac{1}{2 \mu_0} B_0^2}{\frac{1}{2} \frac{B_0^2}{\mu_0}} = 1 \]

6. **Conclusion**:
Therefore, the ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is:
\[ 1 : 1 \]

### Final Answer:
The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is \( 1 : 1 \).