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What is the length of a wave dipole at $15\;MHz$?
A. $15\;cm$
B. $12\;cm$
C. $10\;cm$
D. None of these.
Answer
488.4k+ views
Hint: In a half wave dipole antenna, the length of the half wave dipole is equal to the half of the wavelength of the wave. Hence, with the value of the wave length, the length of the half wave can be calculated. The wavelength can be calculated from the relationship between the speed, frequency and wavelength of the wave.
Useful formula:
The length of the half wave dipole,
$l = \dfrac{\lambda }{2}$
Where, $l$ is the length of the half wave dipole and $\lambda $ is the wavelength of the wave.
The relationship between speed, wavelength and frequency of the wave,
$c = \dfrac{\lambda }{\upsilon }$
Where, $c$ is the speed of light, $\lambda $ is the wavelength of the wave and $\upsilon $ is the frequency of the wave.
Given data:
The frequency of the wave, $\upsilon = 15\;MHz = 15 \times {10^6}\;Hz$
Complete step by step solution:
The relationship between speed, wavelength and frequency of the wave,
$\lambda = \dfrac{c}{\upsilon }\;.........................................\left( 1 \right)$
Since, the normal wave moves with the speed of light. So, $c = 3 \times {10^8}\;m{s^{ - 1}}$
Substitute the values of $c$ and $\upsilon $ in equation (1), we get
$
\lambda = \dfrac{{3 \times {{10}^8}\;m{s^{ - 1}}}}{{15 \times {{10}^6}\;Hz}}\; \\
\lambda = 0.2 \times {10^2}\;m \\
\lambda = 20\;m \\
$
The length of the half wave dipole,
$l = \dfrac{\lambda }{2}$
Substitute the value of $\lambda $ in above equation, we get
$
l = \dfrac{{20\;m}}{2} \\
l = 10\;m \\
l = 1000\;cm \\
$
Thus, the answer is not given in the option.
Hence, the option (D) is correct.
Note: While solving the relation between the speed, wavelength and frequency of the wave, the speed of the wave is not given in the question. So, it is assumed to be equal to the speed of the light in the medium. And the SI unit of the frequency $\left( \upsilon \right)$ is $Hz$. It is also denoted as ${s^{ - 1}}$.
Useful formula:
The length of the half wave dipole,
$l = \dfrac{\lambda }{2}$
Where, $l$ is the length of the half wave dipole and $\lambda $ is the wavelength of the wave.
The relationship between speed, wavelength and frequency of the wave,
$c = \dfrac{\lambda }{\upsilon }$
Where, $c$ is the speed of light, $\lambda $ is the wavelength of the wave and $\upsilon $ is the frequency of the wave.
Given data:
The frequency of the wave, $\upsilon = 15\;MHz = 15 \times {10^6}\;Hz$
Complete step by step solution:
The relationship between speed, wavelength and frequency of the wave,
$\lambda = \dfrac{c}{\upsilon }\;.........................................\left( 1 \right)$
Since, the normal wave moves with the speed of light. So, $c = 3 \times {10^8}\;m{s^{ - 1}}$
Substitute the values of $c$ and $\upsilon $ in equation (1), we get
$
\lambda = \dfrac{{3 \times {{10}^8}\;m{s^{ - 1}}}}{{15 \times {{10}^6}\;Hz}}\; \\
\lambda = 0.2 \times {10^2}\;m \\
\lambda = 20\;m \\
$
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The length of the half wave dipole,
$l = \dfrac{\lambda }{2}$
Substitute the value of $\lambda $ in above equation, we get
$
l = \dfrac{{20\;m}}{2} \\
l = 10\;m \\
l = 1000\;cm \\
$
Thus, the answer is not given in the option.
Hence, the option (D) is correct.
Note: While solving the relation between the speed, wavelength and frequency of the wave, the speed of the wave is not given in the question. So, it is assumed to be equal to the speed of the light in the medium. And the SI unit of the frequency $\left( \upsilon \right)$ is $Hz$. It is also denoted as ${s^{ - 1}}$.
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