Physics · Wave Optics · NEET
Both are correct but they mean different things. The first minimum sits at angle theta = lambda/a from the centre. That lambda/a is the HALF angular width (centre to one edge). The FULL angular width is the angle from the first minimum on the left to the first minimum on the right, so you double it: full angular width = 2*lambda/a. In NEET, 'angular width of central maximum' almost always means the full value 2*lambda/a. Read the question: if it says 'width of central maximum' use 2*lambda/a; if it asks 'position of first minimum' use lambda/a.
Angular width is an angle measured in radians (2*lambda/a). Linear width is an actual length on the screen measured in metres or cm. You convert angle to length by multiplying by the screen distance D. So linear width = D * (angular width) = D * 2*lambda/a = 2*D*lambda/a. Angular width does not depend on D; linear width does. Move the screen back and the patch of light gets physically wider, but the angle stays the same.
The central maximum stretches from the first minimum on one side to the first minimum on the other side, so it covers two 'units' of width. Every other bright band sits between two neighbouring minima that are only one unit apart. Spacing between successive minima on the screen is D*lambda/a, so a side maximum has width D*lambda/a, but the central one has width 2*D*lambda/a - exactly twice. This is why the central maximum is the brightest and the widest.
Angular width = 2*lambda/a, so a and width are inversely related. Make the slit wider (larger a) and the central maximum gets narrower and sharper. Make the slit very narrow (small a) and it spreads out wide. In the limit of a very wide slit, the pattern shrinks to a single sharp point - that is why ray optics (a straight-line beam) works when the slit is large compared to the wavelength.
Inside a medium of refractive index n the wavelength shrinks to lambda/n. Since width is proportional to lambda, the angular width becomes 2*lambda/(n*a) - it decreases by the factor n. For water (n = 4/3) the width becomes 3/4 of its air value. Frequency and slit width do not change, only the wavelength does, so the pattern gets narrower.
The angular width of the central maximum in the Fraunhofer diffraction for lambda = 6000 A is theta0. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 x 10^-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
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.
Angular width (full) = 2*lambda/a, where lambda is the wavelength of light and a is the slit width. The half angular width (centre to first minimum) is lambda/a.
Linear width = 2*D*lambda/a, where D is the distance from the slit to the screen. It is the angular width multiplied by D, giving an actual length on the screen.
No. Angular width = 2*lambda/a contains no D, so moving the screen does not change the angle. Only the linear width (2*D*lambda/a) changes with D because the same angle spreads over a longer distance.
All waves from the slit arrive nearly in phase at the centre, giving maximum intensity. It is also twice as wide (2*D*lambda/a) because it spans the first minima on both sides, while each side maximum is only D*lambda/a wide.
Multiply the angular width by the screen distance D. Since the angle is small, width on screen = D * angle = D * 2*lambda/a = 2*D*lambda/a.
