Physics · Wave Optics · NEET
Wider. Fringe width beta = lambda D / d, and lambda is on the top, so beta is directly proportional to lambda (beta ∝ lambda). Red light (about 700 nm) makes wider fringes than blue light (about 400 nm) for the same setup. This is why in NEET 2022 (Q below) changing 600 nm to 400 nm made the fringes narrower, so MORE fringes fit in the same region.
The fringes get NARROWER. In beta = lambda D / d, the term d is in the bottom (denominator), so beta ∝ 1/d. Double the slit separation and you halve the fringe width. This is the trap students miss: d is the only quantity that has an INVERSE effect.
Yes, the linear fringe width beta = lambda D / d increases because beta ∝ D. But the ANGULAR width theta = lambda / d stays the same, because theta does not contain D. So the fringes spread apart on the screen, yet the angle each fringe makes at the slits is unchanged. NEET 2023 tested exactly this.
Fringe width beta = lambda D / d is the actual spacing between two neighbouring bright fringes measured in metres on the screen — it depends on D. Angular width (angular fringe separation) theta = beta / D = lambda / d is the angle subtended at the slits — it does NOT depend on D. Change D and beta changes but theta is fixed.
The region has a fixed length L. Number of fringes n = L / beta, and beta ∝ lambda, so n ∝ 1/lambda. Smaller wavelength means smaller fringe width, so more fringes squeeze into the same length. That gives n1 lambda1 = n2 lambda2.
Inside a medium of refractive index mu, the wavelength becomes lambda/mu, so fringe width becomes beta_medium = beta_air / mu. Water (mu = 1.33) makes the fringes about 1.33 times narrower. Frequency and colour do not change, only wavelength inside the medium shrinks.
In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength lambda of the light used is 5896 A and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20 degree. To increase the fringe angular width to 0.21 degree (with same lambda and D), the separation between the slits needs to be changed to
In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is
Statement I: If the screen is moved away from the plane of the slits, the angular separation of the fringes remains constant. Statement II: If the monochromatic source is replaced by another of larger wavelength, the angular separation of fringes decreases. Choose the correct answer.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Fringe width beta = lambda D / d, where lambda is the wavelength of light, D is the slit-to-screen distance, and d is the separation between the two slits. It is the distance between two neighbouring bright (or dark) fringes.
Increasing the slit separation d, or decreasing D or lambda, makes fringes narrower. Also doing the experiment in a denser medium (like water) shrinks the wavelength and narrows the fringes.
No. Angular width theta = lambda / d has no D in it. Moving the screen changes the linear fringe width beta but never the angular width.
In a medium of refractive index mu, wavelength becomes lambda/mu, so fringe width becomes beta/mu. In water (mu about 1.33) the fringes become about 1.33 times narrower.
Fringe width doubles, because beta is directly proportional to lambda (beta ∝ lambda) when D and d are kept the same.
