GATE 2020 EC Q.43: solution and answer

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

1. Image theory. A perfectly conducting plane can be replaced by the image of the source. For a vertical electric current element, the image points the same way (current $I\delta l$ along $+{\hat{a}}_z$) and sits at $z=-h$. The problem becomes a two-element array in free space, with the elements at $z=\pm h$ and in phase.

2. Path difference. In the far field, looking along direction $\theta$, the real dipole is $h\cos \theta$ closer to the observer than the origin is, and the image is $h\cos \theta$ farther away:
   $$E_{\theta }\propto \frac{e^{-jkr}}{r}\sin \theta \left[e^{jkh\cos \theta }+e^{-jkh\cos \theta }\right]=\frac{e^{-jkr}}{r}2\sin \theta \cos (kh\cos \theta )$$

3. Pattern multiplication. The pattern is the element factor $\sin \theta$ times the array factor $AF=\cos (kh\cos \theta )$. It holds only in the upper half-space, $0\le \theta \le 90^{\circ }$. The array factor has maxima, $|AF|=1$, where
   $$kh\cos \theta =n\pi ,n=0,1,2,\dots$$

4. $n=0$ does not help. $n=0$ gives $\cos \theta =0$, so $\theta =90^{\circ }$ (along the plane). That lobe is there for every $h$. A maximum at $60^{\circ }$ therefore needs $n\ge 1$, and the smallest $h$ comes from $n=1$.

5. Solve for $h$. With $\cos 60^{\circ }=\frac{1}{2}$ and $k=2\pi /\lambda$:
   $$\frac{2\pi }{\lambda }h\cdot \frac{1}{2}=\pi \Rightarrow h=\lambda$$
   At $h=\lambda$ the array-factor maxima fall at $\cos \theta =0,\frac{1}{2},1$, that is $\theta =90^{\circ },60^{\circ },0^{\circ }$. The one at $0^{\circ }$ is wiped out by the null of $\sin \theta$.

6. Rule out the other options. At $\theta =60^{\circ }$, $kh\cos \theta =\pi h/\lambda$. For $h=0.25\lambda$ this is $\pi /4$, and for $h=0.75\lambda$ it is $3\pi /4$. Neither is a multiple of $\pi$. For $h=0.5\lambda$ it is $\pi /2$, and $\cos (\pi /2)=0$, so $60^{\circ }$ is a null, not a maximum.

7. Check on the full pattern. The $\sin \theta$ element factor shifts the peak of $|\sin \theta \cos (kh\cos \theta )|$ slightly. For $h=\lambda$ the lobe between the nulls at $41.4^{\circ }$ and $75.5^{\circ }$ peaks at about $61^{\circ }$. Differentiating,
   $$\frac{d}{d\theta }\left[\sin \theta \cos (kh\cos \theta )\right]=\cos \theta \cos (kh\cos \theta )+kh{\sin }^2\theta \sin (kh\cos \theta )$$
   Setting this to zero at $\theta =60^{\circ }$ gives $\frac{1}{2}\cos \frac{kh}{2}+\frac{3}{4}kh\sin \frac{kh}{2}=0$, that is $\tan \left(\frac{kh}{2}\right)=-\frac{2}{3kh}$, so $h\approx 0.97\lambda$. That is not one of the options. The nearest option is still $\lambda$, and $\lambda$ is exactly what the standard array-factor condition gives. Hence the answer is (A), $h=\lambda$.