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.
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