Correct Answer - Option 1 : 10 dB
Radiation pattern \(F\left( \theta \right) = {\cos ^2}\theta ,0 \le \theta \le \;\pi \;/2\)
U(θ) = F2(θ)
Directivity of antenna is given as
\(D = \frac{{4\pi {U_{max}}}}{{{P_{rad}}}}\)
\({P_{rad}} = \mathop \smallint \limits_{\phi = 0}^{2\pi } \mathop \smallint \limits_{\theta = 0}^{\frac{\pi }{2}} \left( {{{\cos }^4}{\rm{\theta }}} \right)\left( {sin\theta } \right)d\theta .d\phi = 2\pi \mathop \smallint \limits_{\theta = 0}^{\frac{\pi }{2}} \left( {{{\cos }^4}{\rm{\theta }}} \right)\left( {\sin {\rm{\theta).(d\theta }}}\right)\)
\(= 2\pi \mathop \smallint \limits_{ - 1}^0 {t^4}dt = 2\pi \left. {\frac{{{t^5}}}{5}} \right|_{ - 1}^0 = \frac{{2\pi }}{5}\)
\({U_{max}} = {\left. {{{\cos }^4}\theta } \right|_{max}} = 1\)
⇒ Directivity \(D = \frac{{4\pi \left( 1 \right)}}{{\left( {\frac{{2\pi }}{5}} \right)}} = 10 \Rightarrow 10\log 10 = 10\;dB\)