Physics · Wave Optics · NEET
It decreases. Fringe width beta = lambda D / d. Inside water the wavelength becomes lambda/n (n = 4/3 for water), so beta_water = beta_air / n = (3/4) beta_air. The fringes come closer together, so the pattern looks more crowded. Only the wavelength term changes; D and d stay the same.
Frequency is fixed by the source and cannot change when light enters a new medium (that would break energy at the boundary). Speed drops to v = c/n inside the medium. Since v = f x lambda and f is fixed, wavelength must drop: lambda_medium = lambda/n. Smaller lambda means smaller fringe width.
If the whole apparatus (both slits and screen) is dipped in one uniform medium, there is no sideways shift - the central bright fringe stays at the centre, and every fringe simply moves closer to the centre (shrinks). A sideways SHIFT happens only when a thin transparent sheet is placed over ONE slit, which is a different problem (path difference type).
No. d and D are physical distances between the apparatus parts and do not depend on the medium. Only the wavelength changes. So in beta = lambda D / d, you replace only lambda by lambda/n. This is the single most common exam catch.
The conditions use the path difference in terms of the medium wavelength. Bright: path difference = m(lambda/n). Dark: path difference = (m + 1/2)(lambda/n). Because the effective wavelength is smaller, more fringes fit in the same screen space, which is why an 8th fringe in the medium can line up with a lower-numbered fringe in air.
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
In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away was found to be 0.2 degree. What will be the angular width of the first minima if the entire experimental apparatus is immersed in water? (n_water = 4/3)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
beta_medium = lambda D / (n d) = beta_air / n, where n is the refractive index of the medium. For water, n = 4/3, so the fringe width becomes 3/4 of the air value.
Fringe width depends on wavelength. Water slows the light and shrinks its wavelength to lambda/n, so each fringe occupies less space on the screen and the whole pattern becomes tighter.
Yes. Because each fringe is narrower (beta/n), more fringes fit within the same screen width. This is why a higher-order fringe in the medium can coincide with a lower-order fringe measured in air.
No. Immersing the whole apparatus makes fringes narrower with no sideways shift. A thin sheet over one slit introduces an extra path (n-1)t and shifts the whole pattern sideways without changing fringe width. Do not mix the two.
The central bright fringe stays exactly at the centre because the path difference there is still zero for both slits. Only the spacing of the surrounding fringes decreases.
