If both the length of an antenna and the wavelength of the signal to be transmitted are doubled, the power radiated by the antenna

Options

A. is doubled
B. is halved
C. remains constant
D. is quadrupled

Explanation

To solve the problem, we need to analyze how the power radiated by an antenna changes when both the length of the antenna (L) and the wavelength of the signal (λ) are doubled.

### Step-by-Step Solution:

1. **Understand the relationship between power, length, and wavelength**:
The power radiated by an antenna is given by the formula:
\[ P \propto \frac{L}{\lambda^2} \]
where \(P\) is the power radiated, \(L\) is the length of the antenna, and \(\lambda\) is the wavelength of the signal.

2. **Define the initial conditions**:
Let the initial length of the antenna be \(L\) and the initial wavelength be \(\lambda\). Thus, the initial power can be expressed as:
\[ P = k \frac{L}{\lambda^2} \]
where \(k\) is a proportionality constant.

3. **Apply the changes to the length and wavelength**:
According to the problem, both the length of the antenna and the wavelength are doubled:
- New length \(L' = 2L\)
- New wavelength \(\lambda' = 2\lambda\)

4. **Calculate the new power**:
Substitute the new values into the power formula:
\[ P' = k \frac{L'}{\lambda'^2} = k \frac{2L}{(2\lambda)^2} \]

5. **Simplify the expression**:
Now simplify the equation:
\[ P' = k \frac{2L}{4\lambda^2} = k \frac{L}{2\lambda^2} \]

6. **Compare the new power with the initial power**:
Notice that:
\[ P' = \frac{1}{2} \left( k \frac{L}{\lambda^2} \right) = \frac{1}{2} P \]
This indicates that the new power \(P'\) is half of the initial power \(P\).

### Conclusion:
The power radiated by the antenna is halved when both the length of the antenna and the wavelength of the signal are doubled.

### Final Answer:
The power radiated by the antenna will be half of the original power.

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