Superposition And Interference
Two identical loudspeakers driven in phase emit sound of wavelength 0.5 m, and a listener stands so that the path difference between the two sound paths is exactly 0.75 m. What does the listener experience at this point?
Select the correct option:
Solution
Destructive interference and minimum loudness
When two coherent waves overlap, the type of interference is decided by the path difference relative to the wavelength. Constructive interference, giving maximum loudness, requires a path difference of a whole number of wavelengths, while destructive interference, giving minimum loudness, requires an odd number of half-wavelengths. Here the path difference is 0.75 m and the wavelength is 0.5 m, so the path difference equals 0.75/0.5=1.5 wavelengths, which is three half-wavelengths, an odd multiple of λ/2. This is exactly the condition for fully destructive interference, where the two coherent in-phase signals arrive perfectly out of step. The option of constructive interference is wrong because 1.5λ is not a whole number of wavelengths. The option claiming no interference is incorrect, since identical in-phase coherent sources interfere strongly. The option of a beat pattern is wrong because beats arise from different frequencies, not from a fixed path difference at one frequency. This applies the NCERT superposition principle for coherent sources. As a plausibility check, an odd number of half-wavelengths places one wave's crest on the other's trough, cancelling the sound, consistent with the minimum loudness conclusion.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- oscillations and waves
- Topic
- superposition and interference
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Destructive interference and minimum loudness
When two coherent waves overlap, the type of interference is decided by the path difference relative to the wavelength. Constructive interference, giving maximum loudness, requires a path difference of a whole number of wavelengths, while destructive interference, giving minimum loudness, requires an odd number of half-wavelengths. Here the path difference is 0.75 m and the wavelength is 0.5 m, so the path difference equals 0.75/0.5=1.5 wavelengths, which is three half-wavelengths, an odd multiple of λ/2. This is exactly the condition for fully destructive interference, where the two coherent in-phase signals arrive perfectly out of step. The option of constructive interference is wrong because 1.5λ is not a whole number of wavelengths. The option claiming no interference is incorrect, since identical in-phase coherent sources interfere strongly. The option of a beat pattern is wrong because beats arise from different frequencies, not from a fixed path difference at one frequency. This applies the NCERT superposition principle for coherent sources. As a plausibility check, an odd number of half-wavelengths places one wave's crest on the other's trough, cancelling the sound, consistent with the minimum loudness conclusion.
This medium difficulty physics question is from the chapter oscillations and waves, covering the topic of superposition and interference. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse oscillations and waves questions on RankGuru.