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Question #43197
The Question
A plane electromagnetic wave of vacuum wavelength is propagating inside a non-magnetic dielectric block of refractive index . The wave is incident on the flat interface between the dielectric and vacuum at an angle of incidence . The electric field vector is polarized perpendicular to the plane of incidence. Due to total internal reflection, a standing wave pattern is formed along the direction normal to the interface. What is the perpendicular distance from the interface to the first antinode of the electric field inside the dielectric?
Options
- A.
- B.
- C.
- D.
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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
Total Internal Reflection (TIR)|Standing Wave Nodes/Antinodes|Vector Cross-Products
Why This Question Is Tricky
Common Mistakes
This question tests the student's understanding of boundary conditions during Total Internal Reflection (TIR).
A common misconception is treating a TIR interface as a 'hard wall' (perfect conductor), which would incorrectly place a node exactly at the boundary and lead to an antinode at (Option C) or (Option D).
In reality, the boundary condition requires continuity of the electric field and its derivative, leading to an evanescent wave in the vacuum and a specific phase shift (related to the Goos-Hänchen effect) for the reflected wave.
By forcing the calculation of this phase shift for s-polarization and applying it to the standing wave superposition, the problem discriminates between geometric approximations and rigorous wave optics.
Option B serves as a lethal trap for students who incorrectly apply the p-polarization Fresnel phase shift formula.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Ray Optics and Optical Instruments - Class 12 |
| Concept | Total Internal Reflection (TIR) |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |