Question:hard
The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by
\[ \vec{E}(z,t)=E_0\left[\hat{x}\cos(\omega t-kz)+\hat{y}\sin(\omega t-kz)\right] \]
in a right-handed orthogonal co-ordinate system.
Which of the following is the correct polarization of the electromagnetic wave?
The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by
\[ \vec{E}(z,t)=E_0\left[\hat{x}\cos(\omega t-kz)+\hat{y}\sin(\omega t-kz)\right] \]
in a right-handed orthogonal co-ordinate system.
Which of the following is the correct polarization of the electromagnetic wave?
\[ \vec{E}(z,t)=E_0\left[\hat{x}\cos(\omega t-kz)+\hat{y}\sin(\omega t-kz)\right] \]
in a right-handed orthogonal co-ordinate system.
Which of the following is the correct polarization of the electromagnetic wave?
Show Hint
Since \(E_x=E_0\cos\phi\) and \(E_y=E_0\sin\phi\) have equal amplitude and a 90 degree phase difference, the wave is circularly polarized; use the right-hand rule along the propagation direction to fix the handedness.
Updated On: Jul 20, 2026
- Right-handed circularly polarized
- Left-handed circularly polarized
- Linearly polarized
- Linearly polarized with \(-45^{\circ}\) angle to \(\hat{x}\)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Write the field as the real part of a rotating phasor.
Using $\cos\phi=\text{Re}(e^{j\phi})$ and $\sin\phi=\text{Re}(-je^{j\phi})$, the field becomes
\[ \vec{E}=\text{Re}\Big[E_0(\hat{x}-j\hat{y})e^{j(\omega t-kz)}\Big] \]
Step 2: Recognize the standard phasor form for circular polarization.
A transverse field phasor of the form $E_0(\hat{x}-j\hat{y})$ is the textbook signature of a circularly polarized wave, since $\hat{x}$ and $-j\hat{y}$ have equal magnitude and are in phase quadrature, a 90 degree gap contained in the $-j$ factor. Had the sign been $+j$ instead of $-j$, the handedness would flip.
Step 3: Match the sign convention to handedness for $+z$ propagation.
For a wave travelling in $+z$, the standard convention assigns the phasor form $\hat{x}-j\hat{y}$ to right-handed circular polarization and $\hat{x}+j\hat{y}$ to left-handed circular polarization.
Step 4: Compare with the given field.
Our phasor is exactly $E_0(\hat{x}-j\hat{y})$, matching the right-handed case term for term.
Step 5: Cross check using the real-time snapshot.
At $t=0$, $z=0$: $\vec{E}=E_0\hat{x}$. A short time later the field has swung toward $+\hat{y}$, consistent with the right-handed rotation sense identified from the phasor.
\[ \boxed{\text{Right-handed circularly polarized}} \]
Using $\cos\phi=\text{Re}(e^{j\phi})$ and $\sin\phi=\text{Re}(-je^{j\phi})$, the field becomes
\[ \vec{E}=\text{Re}\Big[E_0(\hat{x}-j\hat{y})e^{j(\omega t-kz)}\Big] \]
Step 2: Recognize the standard phasor form for circular polarization.
A transverse field phasor of the form $E_0(\hat{x}-j\hat{y})$ is the textbook signature of a circularly polarized wave, since $\hat{x}$ and $-j\hat{y}$ have equal magnitude and are in phase quadrature, a 90 degree gap contained in the $-j$ factor. Had the sign been $+j$ instead of $-j$, the handedness would flip.
Step 3: Match the sign convention to handedness for $+z$ propagation.
For a wave travelling in $+z$, the standard convention assigns the phasor form $\hat{x}-j\hat{y}$ to right-handed circular polarization and $\hat{x}+j\hat{y}$ to left-handed circular polarization.
Step 4: Compare with the given field.
Our phasor is exactly $E_0(\hat{x}-j\hat{y})$, matching the right-handed case term for term.
Step 5: Cross check using the real-time snapshot.
At $t=0$, $z=0$: $\vec{E}=E_0\hat{x}$. A short time later the field has swung toward $+\hat{y}$, consistent with the right-handed rotation sense identified from the phasor.
\[ \boxed{\text{Right-handed circularly polarized}} \]
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