The amplitude of the electric field component of an electromagnetic wave is increased from E to 4 E. What is the corresponding change in the intensity of the wave ?

Options

A. The intensity is unchanged by the increase in E .
B. The intensity increases by a factor of sixteen .
C. The intensity increases by a factor of four.
D. The intensity decreases by a factor of four.

Answer

To find the change in intensity of an electromagnetic wave when the amplitude of the electric field component is increased from \( E \) to \( 4E \), we can follow these steps:

### Step 1: Understand the relationship between intensity and electric field amplitude
The intensity \( I \) of an electromagnetic wave is related to the amplitude \( E \) of its electric field by the formula:
\[ I = \frac{1}{2} \epsilon_0 c E^2 \]
where \( \epsilon_0 \) is the permittivity of free space and \( c \) is the speed of light in vacuum.

### Step 2: Calculate the initial intensity
Let the initial amplitude of the electric field be \( E \). The initial intensity \( I_1 \) can be expressed as:
\[ I_1 = \frac{1}{2} \epsilon_0 c E^2 \]

### Step 3: Calculate the new intensity after increasing the amplitude
Now, if the amplitude is increased to \( 4E \), the new intensity \( I_2 \) can be calculated as:
\[ I_2 = \frac{1}{2} \epsilon_0 c (4E)^2 \]
Calculating \( (4E)^2 \):
\[ (4E)^2 = 16E^2 \]
Thus, the new intensity becomes:
\[ I_2 = \frac{1}{2} \epsilon_0 c \cdot 16E^2 = 8 \epsilon_0 c E^2 \]

### Step 4: Find the ratio of the new intensity to the initial intensity
To find the change in intensity, we take the ratio of the new intensity to the initial intensity:
\[ \frac{I_2}{I_1} = \frac{8 \epsilon_0 c E^2}{\frac{1}{2} \epsilon_0 c E^2} \]
This simplifies to:
\[ \frac{I_2}{I_1} = \frac{8}{\frac{1}{2}} = 16 \]
Thus, we find that:
\[ I_2 = 16 I_1 \]

### Conclusion
The intensity of the wave increases by a factor of 16 when the amplitude of the electric field component is increased from \( E \) to \( 4E \).