Question

If the wavelength of the signal to be transmitted by an antenna is increased by \(2\) times, then the effective power radiated by the antenna

• A. Becomes 1/4 times the initial value
• B. Increases by 4 times the initial value
• C. Remains same
• D. Becomes 1/2 times the initial value

Hint:

For an antenna, \[ P\propto \nu^2 \] and since \[ \nu=\frac{c}{\lambda}, \] \[ P\propto \frac{1}{\lambda^2}. \] So if wavelength doubles, \[ P \rightarrow \frac{P}{4}. \]

Answer 1

The Correct Option is A

Concept: The power radiated by an antenna is proportional to the square of its frequency. \[ P \propto \nu^2. \] Since \[ \nu=\frac{c}{\lambda}, \] we get \[ P\propto\left(\frac{1}{\lambda}\right)^2. \] Hence, \[ P\propto\frac{1}{\lambda^2}. \]

Step 1: Relate power and wavelength. If the wavelength is doubled, \[ \lambda' = 2\lambda. \] Therefore, \[ P' = \frac{1}{(2\lambda)^2}. \] \[ P' = \frac{1}{4\lambda^2}. \]

Step 2: Find the ratio of powers. \[ \frac{P'}{P} = \frac{\frac{1}{4\lambda^2}} {\frac{1}{\lambda^2}} = \frac14. \] Thus, \[ P'=\frac14 P. \]

Step 3: State the result. The effective radiated power becomes one-fourth of its original value. \[ \boxed{\frac{P'}{P}=\frac14} \] Therefore, \[ \boxed{\text{Answer = (A)}} \]