GATE EC 2020 Set 1 — Question 53

When an electric dipole is placed above an infinite ideal conducting plane, an image dipole is formed below the plane. For a vertical electric dipole, the image dipole will also be vertical and in the same direction (due to the boundary condition that the tangential electric field is zero on the conducting plane, which means the normal component of the magnetic field is zero, and for a vertical dipole, the image current is in the same direction).The total electric field in the far field is the superposition of the field from the original dipole and its image. The phase difference between the direct ray and the reflected ray (from the image) depends on the path difference and the reflection coefficient.For a vertical electric dipole above a perfect electric conductor (PEC) plane, the image current is in the same direction as the original current. The total field is given by:
$E_\theta^{total} \propto \sin\theta \cdot 2 \sin(kh \cos\theta)$For a maximum in the radiation pattern, the term $2 \sin(kh \cos\theta)$ must be maximum. This occurs when $kh \cos\theta = (2n+1)\frac{\pi}{2}$ for $n = 0, 1, 2, \dots$ (for constructive interference, where $2n+1$ is an odd integer, as the reflection from a PEC introduces a $180^\circ$ phase shift for the tangential E-field, but here we are considering the total field due to the dipole and its image, which is equivalent to two dipoles separated by $2h$).Alternatively, consider the path difference. The path difference between the direct ray and the ray from the image is $2h \cos\theta$. For constructive interference (a maximum), this path difference should be an odd multiple of $\lambda/2$ (since the reflection from a PEC introduces a $180^\circ$ phase shift for the tangential E-field, which is equivalent to an additional $\lambda/2$ path difference).
So, $2h \cos\theta = (2n+1)\frac{\lambda}{2}$, where $n = 0, 1, 2, \dots$We are looking for the minimum value of $h$ for which a maximum occurs at $\theta = 60^\circ$. This corresponds to $n=0$.
$2h \cos(60^\circ) = (2(0)+1)\frac{\lambda}{2}$
$2h \cdot \frac{1}{2} = \frac{\lambda}{2}$
$h = \frac{\lambda}{2}$Let's recheck the formula for total field. For a vertical dipole above a ground plane, the total field is $E_\theta = E_0 \sin\theta [1 - e^{-j2kh\cos\theta}]$. This is for a horizontal dipole. For a vertical dipole, the image current is in the same direction, and the total field is $E_\theta = E_0 \sin\theta [1 + e^{-j2kh\cos\theta}]$.Let's use the image theory for a vertical dipole. The original dipole is at $z=h$ and the image dipole is at $z=-h$. The total field is the sum of the fields from these two dipoles. The phase difference between the two fields at a far-field point is $k(r_1 - r_2) = k(2h \cos\theta)$.For a vertical dipole, the image current is in the same direction. So, the fields add constructively when the phase difference is an even multiple of $\pi$, or $2kh \cos\theta = 2n\pi$, which means $2h \cos\theta = n\lambda$. This would imply maxima at $h = n\lambda / (2\cos\theta)$.However, the standard result for a vertical dipole above a ground plane is that the total field is proportional to $\sin\theta \sin(kh \cos\theta)$.
Let's derive this carefully. The field from a vertical dipole at $z=h$ is $E_1 \propto \sin\theta e^{-jkr_1}$. The field from its image at $z=-h$ is $E_2 \propto \sin\theta e^{-jkr_2}$.
In the far field, $r_1 \approx r - h\cos\theta$ and $r_2 \approx r + h\cos\theta$.
So, $E_{total} \propto \sin\theta (e^{-jk(r-h\cos\theta)} + e^{-jk(r+h\cos\theta)})$
$E_{total} \propto \sin\theta e^{-jkr} (e^{jkh\cos\theta} + e^{-jkh\cos\theta})$
$E_{total} \propto \sin\theta e^{-jkr} (2 \cos(kh\cos\theta))$For a maximum, $|\cos(kh\cos\theta)|$ must be maximum, i.e., $|\cos(kh\cos\theta)| = 1$. This means $kh\cos\theta = n\pi$ for $n = 0, 1, 2, \dots$.We are looking for the minimum value of $h$ for a maximum. This corresponds to $n=1$ (since $n=0$ would imply $h=0$ or $\cos\theta=0$, which is not a general maximum for $h>0$).
So, $kh\cos\theta = \pi$.
Substitute $k = 2\pi/\lambda$:
$(2\pi/\lambda) h \cos\theta = \pi$
$2h \cos\theta = \lambda$Given $\theta = 60^\circ$, so $\cos(60^\circ) = 1/2$.
$2h (1/2) = \lambda$
$h = \lambda$Let's re-evaluate the image current direction. For a vertical electric dipole (current element $I dL \hat{z}$) above a PEC, the image current is also $I dL \hat{z}$ at $z=-h$. This means the fields add constructively. The phase difference is $k(r_1 - r_2) = k(2h \cos\theta)$. For constructive interference, this phase difference should be $2n\pi$.
$2kh \cos\theta = 2n\pi \implies 2h \cos\theta = n\lambda$.
For the first maximum (excluding $n=0$ which means $h=0$), we take $n=1$.
$2h \cos\theta = \lambda$
Given $\theta = 60^\circ$, $\cos\theta = 1/2$.
$2h (1/2) = \lambda \implies h = \lambda$.This matches option (A).Final check: The problem states $E_\theta \propto (e^{-jkr}/r) \sin\theta$. This is the field of a single dipole. When placed above a ground plane, the total field is the sum of the direct field and the reflected field. The reflected field can be considered as coming from an image dipole. For a vertical electric dipole, the image dipole is also vertical and has the same current direction. The path difference is $2h\cos\theta$. The phase difference is $k(2h\cos\theta)$. Since the image is in phase, for a maximum, the phase difference should be $2n\pi$.
$k(2h\cos\theta) = 2n\pi$
$(2\pi/\lambda)(2h\cos\theta) = 2n\pi$
$2h\cos\theta = n\lambda$
For the minimum $h$ (and $h>0$), we take $n=1$.
$2h\cos\theta = \lambda$
$2h\cos(60^\circ) = \lambda$
$2h(1/2) = \lambda$
$h = \lambda$The final answer is $\boxed{\lambda}$.