What happens to the fringe width in a double-slit experiment if the slit separation is reduced to one-third?
**Interference** occurs when two coherent waves superpose, path difference Δ = d sinθ, bright fringe when Δ = n λ, n integer, seventh bright Δ=7λ, dark fringe Δ = (2n-1)λ/2, second dark Δ=3λ/2, third dark 5λ/2, sixth dark 11λ/2, distance of bright fringe from central y = n λ D/d, D screen distance, d slit separation, for λ=650 nm d=0.5 mm D=1.0 m fifth dark? Actually dark y=(2n-1)λ D/(2d). Fringe width β = (λ D/d) . If d is reduced to (d/3) , β increases to 3β , i.e., triples. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,
Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence