in young's double slit experiment,using monochromatic light of wavelength lemda the intensity of light at a point on the screen where path difference is lemda,us k unit.find the intensity of light at a point where path difference is lemda/3?

Intensity of point in YDSE is given by 4I0cos2φ where ​φ is the phase difference and I0 is the intensity of the source.
The relation between path difference δ and phase difference ​φ​ is given by
​φ​ = (2π/λ  )δ.
If path difference is λ , then phase difference is given by ​φ​​ = (2π/λ ) x λ  = 2π.
Intensity corresponding to phase difference 2π = 4I0cos2 2π = 4I0 = K  ...........(1) (given).
Now , we need to find the intensity at a point where  δ = λ/3.
Phase difference corresponding to δ = λ/3
 is given by ​​φ​​ = (2π/λ ) x λ/3  = 2π/3.
So required intensity = 4I0cos2 (2π/3) = 4I0 x 1/4 = K/4  ( Using (1)).
This is the required answer . Hope that helps.
  • 45
Thank you
  • -7

  • 10
Here is the answer👇

  • 27
What are you looking for?