Fringe width w = λD/d; with D → 2D and d → d/2, w becomes 4 times larger.

Transverse waves, such as light, can undergo diffraction when encountering obstacles or apertures comparable to their wavelength.

With amplitude ratio 3:2, Imax/Imin = (3+2)^2/(3-2)^2 = 25/1, so option B is correct.

Fringe width β = Dλ/d; using D ≈ 80 cm, λ=6563 Å and d=0.3 cm gives β ≈ 0.0182 cm.

A fringe corresponds to loci of points having a constant phase difference, producing constructive or destructive interference.

The block header indicates option A as the answer; option A is presented as the relation matching the constructive interference condition given in the figure.

The angular width of the central maximum is ≈ 2λ/a and is independent of the distance to the screen, though the linear fringe width depends on that distance.

Fringe width β ∝ 1/d, so if slit separation d doubles, fringe width halves (decreased by two times), option B.

Destructive interference occurs when the path difference is an odd multiple of λ/2, producing out-of-phase waves that cancel.

Max intensity occurs with constructive interference: (√I + 2√I)^2 = (3√I)^2 = 9I; min with destructive: (2√I - √I)^2 = I.