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In the Young’s double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
In the Young’s double slit experiment using a monochromatic ligh...
Imax/2=Im cos^2 (phi/2)
= Cos^2 (phi/2)=1/root 2
= phi/2=n/4
phi=n/2(2n+1)

delta x= wavelength /2n×phi

= wavelength /2n×n/2 (2n+1)

=wavelength /4 (2n+1)

Option D is correct
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Community Answer
In the Young’s double slit experiment using a monochromatic ligh...
The intensity I is given as
  where  Io is the peak intensity
For a phase difference of 2π the path difference is λ
∴ For a phase difference of  the path difference 
option (b) is correct.
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In the Young’s double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity isa)b)c)d)Correct answer is option 'B'. Can you explain this answer?
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