Question

What should be the Length of dipole antenna For a carrier wave having frequency $3 \times {10^8}Hz$
A. $0.5m$
B. $0.25m$
C. $1.5m$
D. $1m$

Answer 1

In this question the frequency of carrier wave is it given from this we can calculate the wavelength of the wave by the formula $c = n\lambda$
Where $c \Rightarrow$ speed of carrier wave
$n \Rightarrow$ frequency of carrier wave
$\lambda \Rightarrow$ wavelength of carrier wave
To transmit this carrier wave by dipole antenna. The length of the dipole antenna must be half of the wavelength of carrier wave
$L = \dfrac{\lambda }{2}$
$L \Rightarrow$ length of antenna

Complete step by step answer:
Dipole antenna is the one of the simple and widely used antenna This antenna consist of Two wires the size of each wire is approximately 1/4 of the wavelength off the desired frequency of operation
So the total length of this antenna is about half of the wavelength so it is also known as half wave dipole antenna.
Step 1
In our question the frequency of carrier wave is given
$n = 3 \times {10^8}Hz$
First we have to calculate the wavelength of this career wave by applying formula $c = n\lambda$
The carrier wave is electromagnetic wave so it’s speed is equal to the speed of light we know that speed of light is
$c = 3 \times {10^8}m/s$
Put these value in about the question equation
$\Rightarrow c = n\lambda$
$\Rightarrow 3 \times {10^8} = 3 \times {10^8} \times \lambda$
$\Rightarrow \lambda = \dfrac{{3 \times {{10}^8}}}{{3 \times {{10}^8}}}$
By solving it
$\Rightarrow \lambda = 1m$
In this app we get the wavelength of carrier wave is $1m$
Step 2
As we know to transmit this carrier wave from a dipole antenna the length of dipole antenna must be equal to the $\dfrac{\lambda }{2}$
Put the value of $\lambda$ in formula $L = \dfrac{\lambda }{2}$
$\Rightarrow L = \dfrac{1}{2}m$
$\therefore$ the length of antenna is $0.5m$

Hence in this question the option A is correct .

Note: By these simple formulas we can find the length of the dipole antenna for a given frequency carrier wave.