Physics · Wave Optics · NEET
Yes. In YDSE every bright fringe is separated from the next bright fringe by beta = lambda D / d, and every dark fringe is separated from the next dark fringe by the SAME beta. The bright and dark fringes alternate, and the distance from a bright fringe to the next dark fringe is exactly beta/2. So the whole pattern is a set of equally spaced, equal-width bands. This is why you can find lambda by just measuring the width of one band.
Fringe width (beta = lambda D / d) is a LINEAR distance measured on the screen, in metres. Angular fringe width (theta = lambda / d) is an ANGLE, measured from the slits. They are linked by beta = theta times D. Key point: angular width theta = lambda / d does NOT depend on the screen distance D, but linear width beta does. NEET 2018 tested exactly this - they gave angular width and asked how to change d.
From beta = lambda D / d, if you keep D and d fixed, then beta rises and falls with lambda. Physically, a longer wavelength means the two waves fall out of step over a longer distance on the screen, so bright bands sit farther apart. That is why red fringes (about 700 nm) are wider than violet fringes (about 400 nm). NEET 2022 used this: switching 600 nm to 400 nm makes each fringe narrower, so MORE fringes fit in the same region.
No. The position of the n-th bright fringe is y_n = n lambda D / d, which does depend on n. But the WIDTH is the difference y_(n+1) - y_n = lambda D / d = beta, and n cancels out. So the first fringe gap equals the tenth fringe gap - the pattern is uniformly spaced. This is a very common trap: students think outer fringes are wider, but they are not (in the small-angle YDSE approximation).
beta = lambda D / d. If D becomes 2D and d becomes 2d, the factor 2 cancels: beta stays the same. Always plug both changes into the formula together instead of guessing. Only the RATIO D/d matters for fringe width, not their individual values.
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
Try the real previous-year questions from this chapter — each with the answer and a full solution.
beta = lambda D / d, where lambda is the wavelength of light, D is the distance from the slits to the screen, and d is the separation between the two slits. It gives the distance between two consecutive bright (or two consecutive dark) fringes.
The path difference at a point y on the screen is approximately y d / D. For a bright fringe this equals n lambda, giving y_n = n lambda D / d. The n-th and (n+1)-th bright fringes differ by beta = y_(n+1) - y_n = lambda D / d. The same result comes out for dark fringes.
Yes. Linear fringe width beta = lambda D / d increases with D - moving the screen farther makes fringes wider. But the ANGULAR width theta = lambda / d does not depend on D at all. NEET 2023 tested this difference directly.
Angular fringe width is theta = beta / D = lambda / d. It is the angle (from the slits) between two consecutive bright fringes. It depends only on lambda and d, not on the screen distance D.
Inside a medium of refractive index n, the wavelength becomes lambda/n, so the fringe width becomes beta/n - the fringes get narrower. This is covered fully on the YDSE-in-a-medium page.
Yes, in the standard small-angle YDSE the pattern is a set of equally spaced, equal-width bright and dark bands. The width beta = lambda D / d does not depend on the fringe order n.
