Question:
For the electric field of an electromagnetic wave given below, which of the following statements is correct?
\[ \vec{E} = \hat{x}\, E_0 \cos(\omega t) + \hat{y}\, 2E_0 \cos\left(\omega t + \frac{\pi}{2}\right) \]
For the electric field of an electromagnetic wave given below, which of the following statements is correct?
\[ \vec{E} = \hat{x}\, E_0 \cos(\omega t) + \hat{y}\, 2E_0 \cos\left(\omega t + \frac{\pi}{2}\right) \]
\[ \vec{E} = \hat{x}\, E_0 \cos(\omega t) + \hat{y}\, 2E_0 \cos\left(\omega t + \frac{\pi}{2}\right) \]
Show Hint
Write \(E_y=2E_0\cos(\omega t+\pi/2)=-2E_0\sin\omega t\) and eliminate \(t\) between \(E_x\) and \(E_y\).
The result is an ellipse equation with semi-axes \(E_0\) and \(2E_0\).
The result is an ellipse equation with semi-axes \(E_0\) and \(2E_0\).
Updated On: Jul 28, 2026
- The electric field is linearly polarised with slope 2.
- The electric field is circularly polarised with radius \(E_0\).
- The electric field is elliptically polarised with a ratio of major to minor axis being 2.
- The electric field is unpolarised with the two components being phase shifted by \(\pi/2\).
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The polarisation state of a wave is fixed by how its two transverse field components compare in amplitude and in phase. Two components in phase (or exactly out of phase by \(\pi\)) give linear polarisation. Two components with equal amplitude and a \(\pi/2\) phase difference give circular polarisation. Two components with unequal amplitude and a \(\pi/2\) phase difference give elliptical polarisation, with the ellipse's axes lined up with the coordinate axes.
Step 2: Key Formula or Approach:
Write out both components explicitly as functions of time and use \(\cos(\omega t + \pi/2) = -\sin(\omega t)\):
\[ E_x = E_0\cos(\omega t), \qquad E_y = 2E_0\cos\left(\omega t + \frac{\pi}{2}\right) = -2E_0\sin(\omega t) \]
Step 3: Detailed Explanation:
From these, \(E_x/E_0 = \cos\omega t\) and \(E_y/(2E_0) = -\sin\omega t\). Squaring both and adding removes the time dependence:
\[ \left(\frac{E_x}{E_0}\right)^2 + \left(\frac{E_y}{2E_0}\right)^2 = \cos^2\omega t + \sin^2\omega t = 1 \]
This is the equation of an ellipse in the \(E_x\)-\(E_y\) plane, with semi-axis \(E_0\) along \(x\) and semi-axis \(2E_0\) along \(y\). As time goes on, the tip of the electric field vector traces this ellipse once every period, so the wave is elliptically polarised.
Step 4: Get the axis ratio and rule out the other options.
The ratio of the major axis to the minor axis is \(2E_0 : E_0 = 2\), matching option (C). Option (A) is wrong because the phase difference here is \(\pi/2\), not \(0\) or \(\pi\), so the field does not oscillate along a single fixed line. Option (B) needs the two amplitudes to be equal so the ellipse becomes a circle, but here they are \(E_0\) and \(2E_0\), not equal, so it cannot be circular. Option (D) is wrong because unpolarised describes an incoherent mix of many independent waves with randomly varying phase, not a single coherent wave like this one, which has one fixed, well defined amplitude and phase relationship at all times.
Final Answer:
The wave traces an ellipse with major to minor axis ratio 2, so the field is elliptically polarised. \[ \boxed{\text{Option (C): axis ratio } = 2} \]
The polarisation state of a wave is fixed by how its two transverse field components compare in amplitude and in phase. Two components in phase (or exactly out of phase by \(\pi\)) give linear polarisation. Two components with equal amplitude and a \(\pi/2\) phase difference give circular polarisation. Two components with unequal amplitude and a \(\pi/2\) phase difference give elliptical polarisation, with the ellipse's axes lined up with the coordinate axes.
Step 2: Key Formula or Approach:
Write out both components explicitly as functions of time and use \(\cos(\omega t + \pi/2) = -\sin(\omega t)\):
\[ E_x = E_0\cos(\omega t), \qquad E_y = 2E_0\cos\left(\omega t + \frac{\pi}{2}\right) = -2E_0\sin(\omega t) \]
Step 3: Detailed Explanation:
From these, \(E_x/E_0 = \cos\omega t\) and \(E_y/(2E_0) = -\sin\omega t\). Squaring both and adding removes the time dependence:
\[ \left(\frac{E_x}{E_0}\right)^2 + \left(\frac{E_y}{2E_0}\right)^2 = \cos^2\omega t + \sin^2\omega t = 1 \]
This is the equation of an ellipse in the \(E_x\)-\(E_y\) plane, with semi-axis \(E_0\) along \(x\) and semi-axis \(2E_0\) along \(y\). As time goes on, the tip of the electric field vector traces this ellipse once every period, so the wave is elliptically polarised.
Step 4: Get the axis ratio and rule out the other options.
The ratio of the major axis to the minor axis is \(2E_0 : E_0 = 2\), matching option (C). Option (A) is wrong because the phase difference here is \(\pi/2\), not \(0\) or \(\pi\), so the field does not oscillate along a single fixed line. Option (B) needs the two amplitudes to be equal so the ellipse becomes a circle, but here they are \(E_0\) and \(2E_0\), not equal, so it cannot be circular. Option (D) is wrong because unpolarised describes an incoherent mix of many independent waves with randomly varying phase, not a single coherent wave like this one, which has one fixed, well defined amplitude and phase relationship at all times.
Final Answer:
The wave traces an ellipse with major to minor axis ratio 2, so the field is elliptically polarised. \[ \boxed{\text{Option (C): axis ratio } = 2} \]
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