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In a Young’s double slit experiment the intensity, at a point where the path difference is fraction numerator lambda over denominator 6 end fraction(λ being the wavelength of the light used),blank i s blank I. If I subscript 0 end subscript denotes the maximum intensity, I divided by I subscript 0 end subscript is equal to

  1. fraction numerator 1 over denominator square root of 2 end fraction    
  2. fraction numerator square root of 3 over denominator 2 end fraction    
  3. fraction numerator 1 over denominator 2 end fraction    
  4. fraction numerator 3 over denominator 4 end fraction    

The correct answer is: fraction numerator 3 over denominator 4 end fraction


    Phase difference equals fraction numerator 2 pi over denominator lambda end fraction cross times path difference
    ie, ϕ equals fraction numerator 2 pi over denominator lambda end fraction blank cross times fraction numerator lambda over denominator 6 end fraction equals fraction numerator pi over denominator 3 end fraction
    As, I equals I subscript m a x end subscript cos to the power of 2 end exponent invisible function application open parentheses fraction numerator ϕ over denominator 2 end fraction close parentheses
    Or fraction numerator I over denominator I subscript m a x end subscript end fraction equals cos to the power of 2 end exponent invisible function application open parentheses fraction numerator ϕ over denominator 2 end fraction close parentheses
    Or fraction numerator I over denominator I subscript 0 end subscript end fraction equals cos to the power of 2 end exponent invisible function application open parentheses fraction numerator pi over denominator 6 end fraction close parentheses equals fraction numerator 3 over denominator 4 end fraction

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