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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is
  • a)
    infinite
  • b)
    five
  • c)
    three
  • d)
    zero
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The maximum number of possible interference maxima for slit-separation...
d sin θ = nλ

As sin θ can not exceed 1,
So when n = 0, +-1, +- 2,
sin θ = 0, +- 1/2 , +- 1;
Thus 5 maxima are allowed
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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, isa)infiniteb)fivec)threed)zeroCorrect answer is option 'B'. Can you explain this answer?
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