Intensity in Single-Slit and Double-Slit Diffraction
Continuous phasor addition gives the far-field intensity of a uniformly illuminated slit. For two identical slits under coherent monochromatic normal illumination with common polarisation, the single-slit envelope multiplies the interference factor. Their overlap can suppress an otherwise allowed interference order.
What a learner can do afterwards
- Derives the single-slit intensity pattern by adding contributions across the slit
- Explains missing orders in a two-slit pattern using the diffraction envelope
- Predicts how the pattern changes when the slit is widened
1 · Read
Add phasors head to tail. Radian angle = arc length/radius. For a uniformly illuminated slit with monochromatic normal illumination in the far field, beta = pi a sin(theta)/lambda; a is width and theta is from the beam direction. Signed phase spread is 2beta radians. For beta ≠ 0, the arc has radius r > 0 in field-amplitude units, traced length E0 = 2r|beta| and chord magnitude E = 2r|sin(beta)|. Thus E/E0 = |sin(beta)/beta| and I/I0 = [sin(beta)/beta]^2.
The arc turns through 2beta radians. Half-angle circle geometry gives chord magnitude 2r|sin(beta)|. E0 and I0 are the single slit's central field-amplitude magnitude and intensity. At beta = 0, direct substitution gives 0/0, not zero. The ratio |sin(beta)/beta| approaches 1 as beta approaches zero, so central intensity is I0. At the centre, aligned phasors add in a straight line without an arc radius.
A single-slit minimum needs sin(beta) = 0 with beta nonzero. Thus beta = p pi and a sin(theta) = p lambda for p = ±1, ±2, and so on. Zero is excluded because it is the central maximum. A stated minimum also needs |p lambda/a| ≤ 1 so that its angle exists.
When a exceeds lambda, the central boundaries are theta = ±arcsin(lambda/a). At small angles their full separation is approximately 2lambda/a radians. Increasing a at fixed wavelength narrows this peak. For 550 nm light and a = 2.00 micrometres, the first minima are about ±16.0 degrees. The second minima are about ±33.4 degrees, outside the central peak.
For identical slits of width a and centre spacing d, use uniform coherent monochromatic normal illumination, common polarisation and far-field observation. Define I2,0 as the combined central intensity: I2,0 = 4I0, where I0 is one slit's central intensity. Then I/I2,0 = [sin(beta)/beta]^2 cos²(alpha), with alpha = pi d sin(theta)/lambda. Missing orders require integer m = p d/a for nonzero integer p, with |m lambda/d| ≤ 1.
For the same two-slit model with a > lambda, the first envelope minima have sin(theta) = ±lambda/a. Integer interference orders strictly inside this central envelope satisfy |m| < d/a. Count negative orders, zero once, and positive orders. If d/a is an integer R, there are 2R − 1 such orders; boundary orders ±R are missing. Check physical angles as well.
At beta = 3pi/2 radians, the phasor arc makes one and a half turns. The chord is a diameter, not a nearly closed endpoint gap. Its ratio to arc length is 2/(3pi). Squaring gives I/I0 = 4/(9pi²), about 0.045. This is a value near the first side-lobe maximum, not its exact maximum.
The bars compare two relative intensities, not an intensity curve. The 0.045 value is the estimate at beta = 3pi/2.
For each nonzero integer envelope order p, calculate the candidate m = p d/a. It is a missing interference order only if m is an integer and |m lambda/d| ≤ 1. A noninteger candidate is not an interference order. An integer candidate with required sine above 1 has no physical angle, rather than an envelope-suppressed direction. Apply both checks in the stated identical-slit far-field model.
Divide the chord magnitude by the traced phasor length and square. Nonzero envelope minima can suppress integer, accessible interference orders in the stated two-slit model.
Derivation, diffraction-envelope teaching and the 550 nm example adapted from OpenStax's University Physics Volume 3. Revised questions by the dataset authors.
Readings written for this dataset. Videos credited to Khan Academy.
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