Question:

In the Young's double slit experiment the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wavelength of light used). The intensity at a point where the path difference is λ/4, will be?

K/2

zero

K/4

Imax=4I0=K
at the other point, path difference=λ/4
so phase difference θ=kΔx=2π/λ × λ/4=π/2
I1=I0+I0+2√I0√I0cos(π/2)
I1=2I0=K/2