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JEE Main 2004
Exam
JEE Main
Year
2004
Subject
Physics
Chapter
Wave Optics

JEE Main 2004 · Physics · Wave Optics

The maximum number of possible interference maxima for...

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

Options

  1. Ainfinite
  2. Bfive
  3. Cthree
  4. Dzero

Correct answer

B. five

Step-by-step solution

For interference maxima, dsin⁡θ=nλd \sin \theta=n \lambda Here d=2λd=2 \lambda ∴sin⁡θ=n/2\therefore \sin \theta=\mathrm{n} / 2 and is satisfied by 5 integral values of n(−2,−1,0,1,2)\mathrm{n}(-2,-1,0,1,2), as the maximum value of sin⁡θ\sin \theta can only be 1 .
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