Physics · Wave Optics · NEET
A bright fringe means the two light waves arrive in step (in phase). This happens when one wave travels an EXTRA distance equal to a whole number of complete wavelengths: 0, lambda, 2 lambda, 3 lambda... Each full wavelength brings the wave back to the same point in its cycle, so the two crests line up and add. This full adding (constructive interference) makes maximum brightness. So the condition is path difference = n lambda, where n = 0, 1, 2, 3...
A dark fringe means the two waves arrive exactly out of step (out of phase). This needs a path difference of half a wavelength, or one and a half, or two and a half, and so on: lambda/2, 3 lambda/2, 5 lambda/2... Here a crest of one wave meets a trough of the other, so they cancel (destructive interference) and give darkness. We write this as path difference = (n - 1/2) lambda with n = 1, 2, 3, or equally as (2n-1) lambda/2. Both mean the same odd multiple of half a wavelength.
The central fringe (right at the centre of the screen, y = 0) is always BRIGHT. At that point light from both slits travels equal distances, so the path difference is zero. Zero counts as n = 0 in the bright condition (path difference = 0 = 0 x lambda), so the two waves are perfectly in step and add fully. This is why the middle of a YDSE pattern is the brightest spot.
Bright fringe position: y_bright = n lambda D / d, where n = 0, 1, 2... (D = slit-to-screen distance, d = slit separation). Dark fringe position: y_dark = (n - 1/2) lambda D / d = (2n-1) lambda D / (2d), where n = 1, 2, 3... The 1st dark fringe sits halfway between the central bright (n=0) and the 1st bright fringe.
Phase difference = (2 pi / lambda) x path difference. For a bright fringe, path difference = n lambda, so phase difference = 2 n pi (a whole number of full turns, in phase). For a dark fringe, path difference = (2n-1) lambda/2, so phase difference = (2n-1) pi (an odd multiple of pi, exactly out of phase). Bright = even multiple of pi, Dark = odd multiple of pi.
Young's double slit experiment is first performed in air and then in a medium other than air. It is found that the 8th bright fringe in the medium lies where the 5th dark fringe lies in air. The refractive index of the medium is nearly
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Path difference between the two waves must equal a whole number of wavelengths: path difference = n lambda, where n = 0, 1, 2, 3... This gives constructive interference (maximum brightness).
Path difference must be an odd multiple of half a wavelength: path difference = (2n-1) lambda/2 = (n - 1/2) lambda, where n = 1, 2, 3... This gives destructive interference (minimum, i.e. darkness).
No. Both bright fringes and dark fringes are equally spaced, and the gap between two next bright fringes equals the fringe width beta = lambda D / d. A dark fringe always lies exactly between two bright fringes.
The wavelength shrinks to lambda/mu inside the medium (mu = refractive index), so all path-difference conditions use lambda/mu instead of lambda. Fringes come closer together, but the bright = n lambda and dark = (2n-1) lambda/2 rules stay the same in form.
