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Question #43117

Difficulty: 8/10 | 🟢 Free

The Question

An RLC circuit with resistance RR and inductance LL drives an antenna, emitting a plane electromagnetic pulse. Due to the circuit's damping, the emitted pulse is not monochromatic but has a Lorentzian energy fluence spectrum (energy per unit area per unit angular frequency) given by: J(ω)=J0πΓ(ω−ω0)2+Γ2J(\omega) = \frac{J_0}{\pi} \frac{\Gamma}{(\omega - \omega_0)^2 + \Gamma^2} where J0J_0 is the total energy fluence, ω0\omega_0 is the central frequency, and Γ=R2L\Gamma = \frac{R}{2L} is the spectral half-width. This pulse is normally incident on a resonant microscopic solar sail. The sail's effective absorption area depends on the frequency of the incident radiation as: A(ω)=A0γ2(ω−ω0)2+γ2A(\omega) = A_0 \frac{\gamma^2}{(\omega - \omega_0)^2 + \gamma^2} where A0A_0 is the peak area and γ\gamma is the resonance half-width of the sail. Assuming the sail absorbs the radiation completely according to this profile and reflects nothing, what is the total momentum imparted to the sail? (Assume ω0≫Γ,γ\omega_0 \gg \Gamma, \gamma so the integration limits can be approximated as −∞-\infty to ∞\infty).

Options

  • A.

    J0A0c(γΓ+γ)\frac{J_0 A_0}{c} \left( \frac{\gamma}{\Gamma + \gamma} \right)

  • B.

    J0A0c\frac{J_0 A_0}{c}

  • C.

    J0A0c(ΓΓ+γ)\frac{J_0 A_0}{c} \left( \frac{\Gamma}{\Gamma + \gamma} \right)

  • D.

    J0A0c(ΓγΓ+γ)\frac{J_0 A_0}{c} \left( \frac{\sqrt{\Gamma \gamma}}{\Gamma + \gamma} \right)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Source of EM Waves|Energy Density Partitioning|Wave-Particle Duality Constraint

Why This Question Is Tricky

Common Mistakes

This question tests the boundary between idealized monochromatic waves and real-world damped oscillators.

Students commonly fall into the trap of assuming the RLC circuit emits a single frequency ω0\omega_0 (ignoring the bandwidth Γ\Gamma caused by the resistor).

If they assume a monochromatic source, they evaluate the absorption area only at the peak, yielding Eabs=J0A0E_{abs} = J_0 A_0 and p=J0A0/cp = J_0 A_0 / c.

However, the Uncertainty Principle embedded in the circuit's Q-factor dictates a Lorentzian spectral spread.

Calculating the true momentum requires convolving the source's emission spectrum with the target's absorption spectrum, which significantly reduces the transferred momentum due to the spectral mismatch.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Waves - Class 12
ConceptSource of EM Waves
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree