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Question #43325

Difficulty: 8/10 | 🟢 Free

The Question

A plane electromagnetic wave propagating in a vacuum has a wave vector k⃗\vec{k} lying in the xyxy-plane. The wave vector makes a small angle α\alpha with the positive xx-axis. The wave is linearly polarized with its electric field vector parallel to the zz-axis, and its root-mean-square (RMS) electric field magnitude is Erms=10π V/mE_{\text{rms}} = 10\pi \text{ V/m}. The wave is incident upon an opaque screen at x=0x = 0 containing a single narrow slit of width bb, extending from y=−b/2y = -b/2 to y=b/2y = b/2. According to Huygens' principle, the slit acts as a continuous array of secondary wavelet sources. Assuming the small angle approximation (sin⁡α≈α\sin\alpha \approx \alpha) is valid and α=λ2b\alpha = \frac{\lambda}{2b} (where λ\lambda is the wavelength), calculate the time-averaged intensity II of the electromagnetic wave at a point PP located on a distant screen at x=Dx = D (D≫bD \gg b) exactly on the xx-axis (y=0y = 0). Let cc be the speed of light and ϵ0\epsilon_0 be the vacuum permittivity. Assume standard Fraunhofer diffraction conditions where the secondary wavelets have uniform amplitude across the slit.

Options

  • A.

    400 c \epsilon_0 \text{ W/m}^2

  • B.

    100π2cϵ0 W/m2100 \pi^2 c \epsilon_0 \text{ W/m}^2

  • C.

    200 c \epsilon_0 \text{ W/m}^2

  • D.

    800 c \epsilon_0 \text{ W/m}^2

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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

Instantaneous vs RMS vs Peak|Phase Constant Determination|Huygens Principle Wavefronts

Why This Question Is Tricky

Common Mistakes

Students incorrectly apply the superposition principle directly to the RMS field values instead of the instantaneous phase-shifted vector components.

By integrating the RMS field across the slit (∫(Erms/b)dy=Erms\int (E_{\text{rms}}/b) dy = E_{\text{rms}}), they ignore the spatial phase variation (k⃗⋅r⃗\vec{k}\cdot\vec{r}) introduced by the off-axis incidence angle α\alpha.

Huygens' principle strictly mandates the summation of instantaneous field vectors; applying RMS values to the wavefront leads to an incorrect intensity profile because intensity is proportional to the time-average of the square of the sum, not the square of the sum of RMS values.

This leads to the trap answer 100π2cϵ0100 \pi^2 c \epsilon_0.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptInstantaneous vs RMS vs Peak
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree