Phasor Addition and the Mathematics of Interference
Phasors represent coherent waves as arrows. Their vector sum gives the resultant amplitude, and intensity is proportional to its square. This explains the two-slit intensity profile and the narrowing principal maxima of equally spaced multiple slits.
What a learner can do afterwards
- Adds two or more waves of equal frequency with a phasor diagram and reads off the resultant amplitude
- Derives the intensity pattern of a two-slit interference figure
- Explains how the pattern sharpens as the number of slits increases
1 · Read
Recall two slits: whole-wavelength path differences preserve an in-phase start. For coherent waves of equal frequency and matching polarisation, a phasor is an arrow of length A (amplitude) and angle phi (phase). Its components are A cos(phi) and A sin(phi). Sum horizontal components to X and vertical ones to Y; the resultant amplitude is R = √(X² + Y²). Head to tail, the resultant joins the first tail to the last head. Use one angle unit throughout.
For two equal phasors separated by phase phi, choose an axis halfway between them. Components perpendicular to that axis cancel. Components along it add to 2A cos(phi/2), giving resultant length |2A cos(phi/2)|. In the same medium, light intensity is proportional to amplitude squared. If either wave alone gives intensity I1, their sum gives I = 4I1 cos²(phi/2). Doubling amplitude multiplies intensity by four.
For initially in-phase slits separated by d, the far-screen path difference is approximately d sin(theta). Theta is measured from the central direction; lambda is wavelength in the medium. The phase gap is phi = 2pi d sin(theta)/lambda radians. For equal contributions, each giving intensity I1 alone, I(theta) = 4I1 cos²(pi d sin(theta)/lambda). Maxima have d sin(theta) = m lambda for integer m. Equal amplitudes cancel at odd half-wavelength path differences.
To find intensity from a path difference delta, first convert distance to phase: phi = 2pi delta/lambda radians. For initially in-phase waves, this is their arriving phase gap. With equal contributions, each giving intensity I1 alone, halve phi and use I = 4I1 cos²(phi/2). Substitute I1 and the cosine-squared value, keeping the intensity units. Adding the two separate intensities alone misses the phase-dependent interference term.
N equal phasors with successive phase steps of 360/N degrees close a regular polygon and sum to zero. Four phasors at 0°, 90°, 180° and 270° cancel in opposite pairs. If those four waves instead arrive in phase, their resultant amplitude is 4A and intensity is 16I1. More generally, N equal in-phase waves give amplitude NA and intensity N²I1.
The cells show components to add. They do not draw the directed phasor polygon.
For N equally spaced slits with equal coherent contributions, all phasors align at principal maxima. The nearest minima occur at adjacent phase gaps ±360/N degrees from alignment. Increasing N reduces these gaps. At fixed wavelength and slit spacing, those minima move closer to each principal maximum, narrowing it. Secondary maxima can remain between principal maxima; the intensity is not zero everywhere between them.
Interference changes where energy arrives. In a lossless arrangement, a dark fringe does not destroy energy. Reduced intensity in some regions is balanced by enhanced intensity elsewhere when the full energy flow is counted. A finite screen need not collect that whole flow. Opening more slits can also increase transmitted power, so comparisons must state what illumination is held fixed.
Add coherent phasors to find amplitude, then square it for intensity. Two equal waves give I = 4I1 cos²(phi/2). More equally spaced slits narrow principal maxima.
The lesson, practice questions and flashcards were written for this dataset.
The readings and flashcards were written for this dataset. Linked videos credited to Khan Academy.
More to read
Following phase from the slits
Suppose two narrow slits emit coherent light with matching polarisation and zero starting phase difference. An extra travel distance of one wavelength adds one full phase cycle. The fields therefore arrive in phase again whenever their path difference is an integer number of wavelengths.
On a distant screen, the path difference is approximately d sin(theta). Here d is slit separation, theta is measured from the central direction and lambda is wavelength in the propagation medium. Bright interference directions satisfy d sin(theta) = m lambda for integer m. An odd half-wavelength difference makes the fields oppose. Their arriving amplitudes must also be equal for complete darkness.
For small angles near the screen centre, sin(theta) is approximately y/D. Substituting gives y_m ≈ m lambda D/d. Consecutive integer values of m differ by one, so neighbouring fringes have spacing w = lambda D/d. Rearranging gives lambda = wd/D. All lengths must use consistent units.
From two arrows to an intensity
Phasors keep track of phase without drawing every rapid oscillation. An arrow at phase phi has components A cos(phi) and A sin(phi), where A is amplitude. Coherent contributions with matching polarisation can be added component by component.
Take two equal arrows with phase gap phi. Choose a reference axis halfway between them. Their perpendicular components have opposite signs and cancel. Their components along the reference axis each equal A cos(phi/2). The sum has length |2A cos(phi/2)|.
The intensity of light in a fixed medium is proportional to the square of its field amplitude. If one contribution alone gives intensity I1, the combined result is I = 4I1 cos²(phi/2). At zero phase gap, the amplitude doubles and intensity is 4I1. At a half-cycle gap, the equal fields cancel.
These statements concern intensity at a particular location. They do not say that light energy has been created at a maximum or destroyed at a minimum. Conservation applies to the complete energy flow through the arrangement.
Locating the minima beside a principal peak
Consider N equally spaced slits that contribute equal coherent fields with matching polarisation. At a principal maximum, all their phasors point together. The amplitude is NA and intensity is N²I1, where I1 is the intensity from one contribution at that angle.
Now move away from that direction. Adjacent phasors turn through a phase gap phi. When the gap reaches 360/N degrees, the N equal arrows close a polygon. The resultant is zero. The same happens on the other side with the opposite phase gap. These are the nearest minima beside the principal maximum.
For fixed wavelength and slit spacing, a larger N requires a smaller change of angle to reach those minima. That is why the principal maximum narrows. Between principal maxima, weaker secondary maxima can remain. The field does not cancel at every intermediate angle.
The N² peak comparison holds when each slit contributes the same amplitude as before. Opening additional slits can admit more light, so this is not automatically a comparison at fixed total transmitted power.
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