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ExamsJEE MainPhysics

Two coherent waves travelling in the same direction are given by E1 = E0 sin(2*pi*x1/lambda - 2*pi*f*t + pi/6) and E2 = E0 sin(2*pi*x2/lambda - 2*pi*f*t + pi/8). For the superposition of these two waves to give constructive interference, the smallest positive value of the path difference (x2 - x1) must be:

  1. lambda/48
  2. lambda/24
  3. lambda/12
  4. lambda/6

Correct answer: lambda/48

Solution

The phase of wave 1 is phi1 = 2*pi*x1/lambda + pi/6 and of wave 2 is phi2 = 2*pi*x2/lambda + pi/8 (the -2*pi*f*t term is common and cancels). The phase difference is delta = phi1 - phi2 = 2*pi*(x1 - x2)/lambda + (pi/6 - pi/8) = 2*pi*(x1 - x2)/lambda + pi/24. For constructive interference delta = 2*n*pi. Solving for the spatial separation gives x2 - x1 = lambda/48 for the smallest positive value (n = 0).

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