A signal is to be transmitted through a wave of wavelength λ, using a linear antenna. The length l of the antenna and effective power radiated Peff will be given respectively as : (K is a constant of proportionality)

  • Option 1)

    \lambda ,P_{eff}= K\left ( \frac{1}{\lambda } \right )^{2}

  • Option 2)

    \frac{\lambda }{8},P_{eff}= K\left ( \frac{1}{\lambda } \right )

  • Option 3)

    \frac{\lambda }{16},P_{eff}= K\left ( \frac{1}{\lambda } \right )^{3}

  • Option 4)

    \frac{\lambda }{5},P_{eff}= K\left ( \frac{1}{\lambda } \right )^{\frac{1}{2}}

 

Answers (1)

As we learnt in

Range of transmitting antenna -

d_{T}= \sqrt{2h_{T}R}


 

- wherein

h_{T}= height of antenna

R= Radius of earth

 

 Length of antenna = Comparable to \lambda

Power radiated by linear antenna inversly depends on the square wavelength and directly on the length of antenna.

Hence, Power=\mu \left(\frac{1}{\lambda} \right )^{2}

\mu = k

Correct option is 1.

 


Option 1)

\lambda ,P_{eff}= K\left ( \frac{1}{\lambda } \right )^{2}

This is the correct option.

Option 2)

\frac{\lambda }{8},P_{eff}= K\left ( \frac{1}{\lambda } \right )

This is an incorrect option.

Option 3)

\frac{\lambda }{16},P_{eff}= K\left ( \frac{1}{\lambda } \right )^{3}

This is an incorrect option.

Option 4)

\frac{\lambda }{5},P_{eff}= K\left ( \frac{1}{\lambda } \right )^{\frac{1}{2}}

This is an incorrect option.

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