Physics · Wave Optics · NEET
This is the most common mix-up. In Young's double slit you compare TWO slits, so path difference nλ means both waves arrive in step and add up (bright). In a SINGLE slit you treat the slit as many tiny sources spread across its width 'a'. When the path difference between the top edge and bottom edge is nλ, you can split the slit into n equal strips. Inside each strip, every point cancels with a matching point half a strip away (they are λ/2 out of step). So the whole slit cancels to zero, giving a dark band. Same formula nλ, opposite result, because the physics of one slit is different from two slits.
If you put n = 0 you get sinθ = 0, which means θ = 0, the exact centre of the screen. But the centre is the CENTRAL MAXIMUM, the brightest point, not a minimum. All the tiny wavelets arrive with zero path difference at θ = 0 and add up fully. So n = 0 is excluded. Minima start from n = 1 (first dark band), then n = 2, n = 3, and so on, on both sides of the centre.
'a' is the WIDTH of the single slit (the size of the opening), not the distance between two slits. 'θ' is the angle measured from the central line (the straight-through direction) to the point on the screen where the dark band appears. λ is the wavelength of light. For small angles, sinθ ≈ θ ≈ y/D, where y is the distance of the dark band from the centre and D is the slit-to-screen distance.
The dark minima are at a sinθ = nλ. The bright secondary maxima fall roughly halfway between them, at a sinθ = (2n+1)λ/2, that is 3λ/2, 5λ/2, 7λ/2... (odd multiples of λ/2). This is the reverse of Young's double slit, where minima are at odd half multiples. NEET loves testing this swap: single slit MINIMA use full nλ, single slit secondary MAXIMA use odd (2n+1)λ/2.
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 × 10⁻⁵ cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For minima (dark bands) in single slit diffraction. The bright secondary maxima are at a sinθ = (2n+1)λ/2.
n = 1, 2, 3... (any non-zero integer). n = 0 is not allowed because it points to the central maximum, which is bright, not dark.
For the first minimum, n = 1, so sinθ = λ/a. For small slits and angles, θ ≈ λ/a radians.
Single slit minima: a sinθ = nλ (full multiples). Double slit (YDSE) minima: path difference = (2n−1)λ/2 (odd half multiples). The conditions are opposite, so do not mix them.
No. The central bright band at θ = 0 is the central maximum. Minima only appear on either side of it, starting at a sinθ = λ.
