GATE 2023 EC Q.53: solution and answer

GATE 2023 · ECQ.53 · 2 marks · MSQ

The electric field of a plane electromagnetic wave is

E = \overset{a}{^}_{x} C_{1 x} cos (ω t - β z) + \overset{a}{^}_{y} C_{1 y} cos (ω t - β z + θ) V/m .

Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?

(A)
$C_{1 x} = 1, C_{1 y} = 1, θ = π /4$
(B)
$C_{1 x} = 2, C_{1 y} = 1, θ = π /2$
(C)
$C_{1 x} = 1, C_{1 y} = 2, θ = 3 π /2$
(D)
$C_{1 x} = 2, C_{1 y} = 1, θ = 3 π /4$

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Answer and solution

ANSWER

Options (A), (B) and (D) give a left-handed elliptically polarized wave. Option (C) is right-handed elliptical.

Official answer: A, B, DSource: Official answer key, GATE 2023 EC (IIT Kanpur)

The phase term is $ω t - β z$ , so the wave travels in the $+ z$ direction. Look at the plane $z = 0$ : $E_{x} = C_{1 x} cos ω t$ and $E_{y} = C_{1 y} cos (ω t + θ)$ , so $E_{y}$ leads $E_{x}$ by $θ$ .
Sense of rotation. The vector $E \times \frac{d E}{d t}$ points along the axis about which $E$ turns, by the right-hand rule. Its $z$ -component is $E_{x} \frac{d E _{y}}{d t} - E_{y} \frac{d E _{x}}{d t} = - ω C_{1 x} C_{1 y} [cos ω t sin (ω t + θ) - cos (ω t + θ) sin ω t] = - ω C_{1 x} C_{1 y} sin θ$ This does not depend on $t$ , so the field never reverses its direction of rotation. If $sin θ > 0$ the $z$ -component is negative, so $E$ turns in the negative sense about $+ z$ (the positive sense about $- z$ ).
Handedness (IEEE convention). Point the thumb along the direction of travel ( $+ z$ ). The wave is left-handed if the fingers of the left hand curl the way $E$ turns, which is the negative sense about $+ z$ : $(E \times \frac{d E}{d t}) \cdot \overset{a}{^}_{z} < 0$ . So we need $sin θ > 0$ , i.e. $0 < θ < π$ ( $E_{y}$ leads $E_{x}$ ). For $π < θ < 2 π$ , $sin θ < 0$ and the wave is right-handed.
Elliptical, not linear or circular. The polarization is linear if $θ = 0$ or $π$ . It is circular only if $C_{1 x} = C_{1 y}$ and $θ = π /2$ or $3 π /2$ . Every other combination with both amplitudes non-zero is elliptical.
(A) $C_{1 x} = C_{1 y} = 1$ , $θ = π /4$ . The amplitudes are equal but $θ \neq = π /2$ , so the wave is elliptical (an ellipse tilted at $4 5^{\circ}$ ). $sin (π /4) = \frac{1}{2} > 0$ , so it is left-handed: LHEP.
(B) $C_{1 x} = 2$ , $C_{1 y} = 1$ , $θ = π /2$ . The amplitudes differ, so the wave is elliptical (an upright ellipse, major axis along $x$ ). $sin (π /2) = 1 > 0$ , so it is left-handed: LHEP.
(C) $C_{1 x} = 1$ , $C_{1 y} = 2$ , $θ = 3 π /2$ . The wave is elliptical, but $sin (3 π /2) = - 1 < 0$ : $E_{y}$ lags $E_{x}$ by $π /2$ and the wave is right-handed elliptical (RHEP). Not correct.
(D) $C_{1 x} = 2$ , $C_{1 y} = 1$ , $θ = 3 π /4$ . The wave is elliptical and $sin (3 π /4) = \frac{1}{2} > 0$ , so it is left-handed: LHEP.
The left-handed elliptical cases are (A), (B) and (D).

KEY IDEA

For $E = \overset{a}{^}_{x} C_{1 x} cos (ω t - β z) + \overset{a}{^}_{y} C_{1 y} cos (ω t - β z + θ)$ , travelling along $+ z$ , $(E \times d E / d t) \cdot \overset{a}{^}_{z} = - ω C_{1 x} C_{1 y} sin θ$ sets the handedness. With $sin θ > 0$ ( $E_{y}$ leads $E_{x}$ ) the wave is left-handed in the IEEE convention, and with $sin θ < 0$ it is right-handed. The polarization is elliptical unless $θ = 0$ or $π$ (linear), or the amplitudes are equal and $θ = π /2$ or $3 π /2$ (circular).

WATCH OUT FOR

One mistake is using the optics convention, where the wave is viewed coming towards the observer. That reverses the handedness, so it would pick (C) alone. Another is rejecting (A) because its amplitudes are equal: equal amplitudes give a circle only when $θ = π /2$ or $3 π /2$ , and with $θ = π /4$ the path is a tilted ellipse. It is also a mistake to call (B) circular just because $θ = π /2$ , since its amplitudes are unequal.

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GATE 2023 EC Q.53: solution and answer

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