# 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?

_{0}cos

^{2}φ where φ is the phase difference and I

_{0}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 = 4I

_{0}cos

^{2}

^{ }2π = 4I

_{0}= 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 = 4I

_{0}cos

^{2}(2π/3) = 4I

_{0}x 1/4 = K/4 ( Using (1)).

This is the required answer . Hope that helps.