If a Young's double slit experiment with light of wavelength λ, the separation of slits is d and distance of screen is D such that D>>d>>λ. If the Fringe width is β, the distance from the point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is

The intensity is given by
Iresultant=I1+I2+2√(I1I2)cosθ.. (i)
and θ is phase difference which is given as
θ=2πλΔx
and Δx is path difference.
at center the path difference is zero i.e.maximum intensity:
Imax+Imax+2Imaxcos0=4Imax.. (ii)
The phase difference for intensity to be half of maximum value
2Imax=Imax+Imax+2Imaxcosθ
⇒θ=π/2
Δx=λ/4
in YDSE Δx is given as
Δx=y d/D
y=λD/4d.. (iii)
The fringe width is given as
β=λD/d
substituting value of λD/d in equation (iii)
y=β/4
correct answer is option A.