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Question #43734
The Question
A single slit of width is placed parallel to a perfectly reflecting flat optical mirror. The center of the slit is at a height above the mirror. The slit is illuminated normally by monochromatic light of wavelength . A screen is placed at a large distance from the slit, perpendicular to the mirror. The intensity distribution on the screen for (where is the intersection of the mirror plane with the screen) is a combined effect of single-slit diffraction and interference between the direct wave and the reflected wave. Assume the small angle approximation is valid. How many interference maxima are observed on the screen strictly within the central diffraction envelope (i.e., between and the first diffraction minimum)?
Options
- A.
2
- B.
3
- C.
5
- D.
6
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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
Single Slit Diffraction Minima|Boundary Condition Reflection|Wave Propagation Direction
Why This Question Is Tricky
Common Mistakes
Students often incorrectly treat the slit and its virtual image as a standard double-slit setup, forgetting that reflection at a denser optical boundary introduces a phase shift.
This phase shift alters the interference condition, changing the path difference requirement for maxima from to .
Consequently, if the phase shift is ignored, students will calculate the maxima positions as and find only 5 maxima within the central diffraction envelope (since the 6th would perfectly coincide with the diffraction minimum and have zero intensity).
This question rigorously tests the synthesis of wave optics principles with boundary condition effects.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Wave Optics - Class 12 |
| Concept | Single Slit Diffraction Minima |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |