Physics · Wave Optics · NEET
It depends on how the two waves line up when they meet. If a crest meets a crest (waves in step), they add and you get a bright point — constructive interference. If a crest meets a trough (waves fully out of step), they cancel and you get a dark point — destructive interference. In numbers: constructive needs path difference = n(lambda), destructive needs path difference = (2n minus 1) lambda/2, where n = 0, 1, 2, ... and lambda is the wavelength.
Bright fringe (constructive): path difference Delta = n(lambda), so 0, lambda, 2 lambda, 3 lambda, and so on. Dark fringe (destructive): path difference Delta = (2n minus 1) lambda/2, so lambda/2, 3 lambda/2, 5 lambda/2, and so on. Notice bright fringes sit at whole multiples of the wavelength, and dark fringes sit at odd multiples of half the wavelength.
Constructive: phase difference phi = 2n(pi) — that is 0, 2pi, 4pi, ... (the waves are in step). Destructive: phase difference phi = (2n minus 1) pi — that is pi, 3pi, 5pi, ... (the waves are exactly opposite). Path difference and phase difference are linked by phi = (2pi/lambda) times Delta, so a path difference of lambda equals a phase difference of 2pi.
A path difference of lambda/2 means one wave has traveled an extra half wavelength. In that extra half wavelength a crest turns into a trough. So where one wave has a crest, the other now has a trough. They point in opposite directions and cancel, giving zero intensity — a dark fringe. Any odd number of half wavelengths (lambda/2, 3 lambda/2, ...) does the same thing.
No. For dark fringes use Delta = (2n minus 1) lambda/2 with n = 1, 2, 3, ..., so the smallest path difference is lambda/2 (the first minimum). If you plug n = 0 you get a negative value, which has no meaning here. For bright fringes, though, n = 0 IS allowed and gives the central bright fringe at zero path difference.
No, they are two ways of describing the same lag. Path difference (Delta) is the extra distance one wave travels, measured in metres. Phase difference (phi) is the angular lag, measured in radians. They convert with phi = (2pi/lambda) times Delta. One full wavelength of path (Delta = lambda) equals one full cycle of phase (phi = 2pi).
In a Young's double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference
In Young's double slit experiment, using monochromatic light of wavelength lambda, the intensity of light at a point on the screen where the path difference is lambda is K units. The intensity of light at a point where the path difference is lambda/3 will be
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The two waves must arrive in step. In terms of path difference: Delta = n(lambda) where n = 0, 1, 2, ... In terms of phase difference: phi = 2n(pi). The waves then add to give maximum brightness.
The two waves must arrive exactly out of step. Path difference Delta = (2n minus 1) lambda/2 with n = 1, 2, 3, ..., or phase difference phi = (2n minus 1) pi. The waves cancel and the point is dark.
Use phi = (2pi/lambda) times Delta. So a path difference equal to one wavelength gives a phase difference of 2pi (one full cycle), and a path difference of lambda/2 gives a phase difference of pi (half a cycle).
Yes. The sources must be coherent (constant phase relationship) and usually of the same wavelength. Without coherence the bright and dark points shift randomly and no steady interference pattern is seen.
Yes. Interference only redistributes light energy — it does not create or destroy it. The energy missing from the dark fringes shows up as extra brightness in the bright fringes, so overall energy is conserved.
Zero path difference (Delta = 0, using n = 0 in the bright condition). At the centre of a symmetric setup both waves travel equal distances, arrive in step, and produce the central bright fringe.
