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.
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.
