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Question #43230

Difficulty: 8/10 | 🟢 Free

The Question

A circular aperture of radius aa is illuminated by a coherent monochromatic plane wave of wavelength λ\lambda. To modify the standard Airy diffraction pattern, a thin, partially reflecting dielectric structure is placed across the aperture. This structure causes the transmitted electric field at the aperture plane to be a coherent superposition of a directly transmitted wave and an internally reflected wave. The directly transmitted wave has a uniform real amplitude E0E_0. The internally reflected wave has an amplitude E0/2E_0/2 and a radially varying relative phase shift Δϕ(r)=π2(ra)2\Delta\phi(r) = \frac{\pi}{2} \left(\frac{r}{a}\right)^2 due to the structure's varying thickness, where rr is the radial distance from the center. If the on-axis intensity of the Fraunhofer diffraction pattern for a completely empty aperture of radius aa (with uniform amplitude E0E_0) is I0I_0, what is the on-axis intensity II of the modified aperture?

Options

  • A.

    I0(1+2π+2π2)I_0 \left( 1 + \frac{2}{\pi} + \frac{2}{\pi^2} \right)

  • B.

    I0(1+2π2)I_0 \left( 1 + \frac{2}{\pi^2} \right)

  • C.

    I0(1−2π+2π2)I_0 \left( 1 - \frac{2}{\pi} + \frac{2}{\pi^2} \right)

  • D.

    94I0\frac{9}{4} I_0

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Resolving Power Limits|Boundary Condition Reflection|Huygens Principle Wavefronts|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental difference between coherent amplitude superposition and incoherent intensity summation in wave optics.

Students often fall into the trap of calculating the on-axis intensity of the directly transmitted wave and the internally reflected wave separately, and then adding them (yielding Option B).

However, because the waves originate from the same coherent source and are superimposed before diffraction, their complex electric field amplitudes must be integrated over the aperture first.

The radially varying phase shift creates a complex integral, generating a crucial interference cross-term 2πI0\frac{2}{\pi}I_0 that is entirely missed if the Huygens-Fresnel principle is bypassed.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptResolving Power Limits
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree