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Minima in a single-slit pattern occur when wave contributions from different parts of the slit cancel destructively.
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Formula
a and λ in the same length unit; θ in radians or degrees; m is a nonzero integer.
| Symbol | Meaning | Unit |
|---|---|---|
| Width of the single slit | m | |
| Angle of the m-th intensity minimum from the central axis | rad or ° | |
| Order of the minimum, m≠0 | integer | |
| Wavelength of the light | m |
a\sin\theta_m=m\lambda,\quad m=\pm1,\pm2,\ldotsFor a slit of width a, wave contributions cancel pairwise when the path difference across the slit is an integer multiple of the wavelength. This gives a sinθ=mλ for minima, not maxima. In the small-angle limit on a screen at distance L, ym≈mλL/a. The relation assumes coherent illumination and a far-field (Fraunhofer) geometry; the central maximum lies between m=−1 and m=1 and is twice as wide as the spacing of side minima.
The condition a sinθ=mλ describes far-field minima under coherent monochromatic illumination. The small-angle form y≈mλL/a requires θ≪1 and screen distance L much larger than slit width.
At a minimum, equal regions of the slit cancel in pairs.
Wavelength from the first minimum
The first-order minimum has |m|=1.
First minimum on a screen
Assumes the small-angle approximation and far-field geometry.
For λ=600 nm, a=0.30 mm, and L=2.0 m, how far is the first minimum from the axis?
y1≈λL/a=(6.0·10−7·2.0)/(3.0·10−4)=4.0·10−3 m=4.0 mm.
Check: halving a doubles the distance in the small-angle limit.
Confusing a grating-maximum condition a sinθ=mλ with a single-slit minimum.
For a single slit, this equation with m≠0 marks a minimum; a grating maximum is constructive interference.
Formula cluster
The single-slit pattern complements double-slit maxima and the resolution criterion.
With λ=600 nm, a=0.30 mm and L=2.0 m, the first minimum is at y1≈λL/a=(6.0·10−7·2)/(3.0·10−4)=4.0·10−3 m=4.0 mm. Check: a wider slit makes the angular pattern narrower.
Analyze single-slit patterns, determine wavelength or slit width, and distinguish single slits from double slits and gratings.
Here is how to work through a typical Minima of single-slit diffraction (a sin θₘ = mλ) task step by step:
Task
For λ=600 nm, a=0.30 mm, and L=2.0 m, how far is the first minimum from the axis?
Solution path
y1≈λL/a=(6.0·10−7·2.0)/(3.0·10−4)=4.0·10−3 m=4.0 mm.
Check: halving a doubles the distance in the small-angle limit.
The relation gives the angles of intensity minima where light nearly cancels. Here a is slit width, λ is wavelength, and m is a nonzero integer. Cancellation comes from pairwise destructive interference of wave contributions from different parts of the slit. At m=0 there is no minimum; there is the broad central maximum. This condition gives minima, not principal maxima. Use the small-angle form only for small diffraction angles, with compatible units for wavelength, slit width, and screen distance. The minima occur symmetrically at positive and negative orders, while m=0 marks the bright central maximum.
In a single slit, light from different points within one opening cancels, with minima at a sinθ=mλ. In a double slit, two coherent openings reinforce at d sinθ=mλ. A real double-slit pattern is also modulated by the single-slit envelope. Slit width a and separation d are different quantities that shape different parts of the pattern. This condition gives minima, not principal maxima. Use the small-angle form only for small diffraction angles, with compatible units for wavelength, slit width, and screen distance.
For the first minimum, sinθ1=λ/a. At fixed wavelength, increasing a decreases the angle. On a distant screen, y1≈λL/a also decreases, so the central bright region becomes narrower. This differs from the intuition that a larger opening must always produce a broader light pattern. This condition gives minima, not principal maxima. Use the small-angle form only for small diffraction angles, with compatible units for wavelength, slit width, and screen distance. The angular width between the first minima is about 2λ/a, so the inverse dependence also applies to the full central lobe.
The screen form follows from sinθ≈tanθ≈y/L at small angles. It is appropriate when the first minimum lies near the central axis and screen distance is much larger than slit width. At larger angles, use a sinθ=mλ directly with tanθ=y/L for screen geometry. The pattern must also be observed in the far field or with suitable optics. This condition gives minima, not principal maxima. Use the small-angle form only for small diffraction angles, with compatible units for wavelength, slit width, and screen distance.
Measure slit width a, screen distance L, and distance y1 from the central axis to the first dark fringe. At small angles, λ≈a y1/L. Use one length unit throughout and measure the dark fringe on both sides to average the result and reduce reading error. For larger angles, obtain θ from geometry and use λ=a sinθ. This condition gives minima, not principal maxima. Use the small-angle form only for small diffraction angles, with compatible units for wavelength, slit width, and screen distance.
Quanta exam set
Curated exam set for "Minima of single-slit diffraction":
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Which formula describes Minima of single-slit diffraction?
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How do you rearrange Minima of single-slit diffraction for Wavelength from the first minimum?
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Which common mistake happens with Minima of single-slit diffraction?
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