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NEET PHYSICSWAVE OPTICSMedium

Question

Two points are located at a distance of 10 m10 \text{ m} and 15 m15 \text{ m} from the source of oscillation. The period of oscillation is 0.05 s0.05 \text{ s} and the velocity of the wave is 300 m/s300 \text{ m/s}. What is the phase difference between the oscillations of two points?

A

π3\frac{\pi}{3}

B

2π3\frac{2\pi}{3}

C

π\pi

D

π6\frac{\pi}{6}

Step-by-Step Solution

  1. Identify the Given Data: Path difference, Δx=x2x1=15 m10 m=5 m\Delta x = x_2 - x_1 = 15 \text{ m} - 10 \text{ m} = 5 \text{ m}. Time period, T=0.05 sT = 0.05 \text{ s}. Wave velocity, v=300 m/sv = 300 \text{ m/s}.
  2. Calculate Wavelength (λ\lambda): Wavelength is the distance travelled by the wave in one time period. λ=v×T=300×0.05=15 m\lambda = v \times T = 300 \times 0.05 = 15 \text{ m}.
  3. Calculate Phase Difference (Δϕ\Delta\phi): The relationship between phase difference and path difference is given by Δϕ=2πλΔx\Delta\phi = \frac{2\pi}{\lambda} \Delta x . Substitute the values: Δϕ=2π15×5=10π15=2π3 rad\Delta\phi = \frac{2\pi}{15} \times 5 = \frac{10\pi}{15} = \frac{2\pi}{3} \text{ rad}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSpointslocateddistancesourceoscillation

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