Question

The minimum length of the dipole antenna for a carrier wave frequency of 200 MHz is nearly

• A. 1.75m
• B. 0.52m
• C. 0.25m
• D. 0.38m
• E. 0.75m

Answer 1

Answer: (E)

The minimum length of a dipole antenna is typically half of the wavelength of the carrier wave. The wavelength \( \lambda \) of a wave is related to the frequency \( f \) by the equation: \[ \lambda = \frac{c}{f} \] where \( c \) is the speed of light (\( c = 3 \times 10^8 \) m/s) and \( f \) is the frequency of the wave. Given that the frequency of the carrier wave is 200 MHz \( = 200 \times 10^6 \) Hz, we can calculate the wavelength: \[ \lambda = \frac{3 \times 10^8}{200 \times 10^6} = 1.5 \, \text{m} \] The minimum length of the dipole antenna is half the wavelength: \[ \text{Minimum length} = \frac{\lambda}{2} = \frac{1.5}{2} = 0.75 \, \text{m} \]

The correct option is (E) : \(0.75\ m\)

Answer 2

To find the minimum length of a dipole antenna, we use the relation: 

Length of dipole antenna = \( \frac{\lambda}{2} \)

Where:
- \( \lambda \) = wavelength of the wave
- \( \lambda = \frac{c}{f} \)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( f = 200 \times 10^6 \, \text{Hz} \)

So,
\( \lambda = \frac{3 \times 10^8}{200 \times 10^6} = 1.5 \, \text{m} \)

Therefore, length of dipole antenna = \( \frac{1.5}{2} = 0.75 \, \text{m} \)

✅ Correct Answer: 0.75 m