Full chapter
Superposition
All 9 topics and the revision summary on one page.
01
Add overlapping waves
At a common position and instant, overlapping waves contribute to one resultant disturbance. Add the signed quantities, keeping their physical directions and reference zero.
The principle of superposition
In a linear wave model, the resultant displacement at a point is the algebraic sum of the displacements that the individual waves would produce there at that instant:
For an upward-positive string, an upward displacement is positive and a downward displacement negative. Do not add a value from one position to another, or combine values taken at different times.
When the oscillating quantity has more than one direction, add its vector components. For light, the relevant oscillating quantity is the electric field; a drawn field curve does not represent material particles following that curve.
Worked pulse overlap
Keep the displacement signs
Two pulses contribute +3.0 mm and -2.0 mm at their coincident peaks.
= +1.0 mm
Add signed displacements at the same place and time
Brown long dashes: the +3.0 mm pulse. Green short dashes: the -2.0 mm pulse. Solid blue: their sum. All rows use the same displacement scale; their vertical offsets only separate the views.
The pulses continue after they pass. A smaller resultant during overlap does not mean either pulse has been destroyed.
The individual pulses are a way of analysing the overlap; the string itself has one resultant displacement at each point. In the ideal linear model, the pulses pass through one another and retain their shapes.
Optional check At one position and instant, two overlapping string pulses contribute displacements of +3.0 mm and -2.0 mm. What is the resultant displacement in the linear model?
Interference concerns the combined disturbance
Interference is the combined effect of overlapping waves. Contributions reinforcing one another produce constructive interference; contributions opposing one another reduce the resultant amplitude and produce destructive interference.
Destructive interference at one place does not mean energy has been destroyed. The energy distribution depends on the complete wave pattern and the medium. Add coherent field or displacement contributions before finding an intensity; intensities are not signed displacements.
Two contributions can cancel at one instant without making a permanent node. A node of a standing wave has zero oscillation amplitude at a fixed position throughout the cycle.
02
Build a standing wave
Equal-amplitude waves of the same frequency travelling in opposite directions can form a standing wave. Its fixed pattern of oscillation amplitude differs from the travelling profiles that produce it.
Take two waves in the same region and oscillation direction, with the same wavelength and speed but opposite propagation directions. At every position and instant, add their signed displacements. The result is also called a stationary wave.
Add the profiles through a cycle
Each component in this supplied model has amplitude 2.0 mm, wavelength 0.80 m and period T = 0.20 s. The view covers x = 0 to 0.80 m.
Two travelling profiles form one standing pattern
Component 1 travels right; component 2 travels left. Each has amplitude 2.0 mm. Brown long dashes and green short dashes distinguish the components; solid blue is their point-by-point sum. Every panel shares the same axes.
t = 0
Component 1 = component 2 in this snapshot, so their dashed profiles coincide. The blue resultant is twice either component.
t = T/4 = 0.050 s
The small arrows show particle-motion directions at the antinodes. An instantaneous zero displacement everywhere does not make every point a node.
t = T/2 = 0.100 s
Component 1 = component 2 in this snapshot, so their dashed profiles coincide. The blue resultant is twice either component.
- At t = 0, the resultant displacements at x = 0, 0.20, 0.40, 0.60 and 0.80 m are 0, +4.0, 0, -4.0 and 0 mm.
- At t = T/4 = 0.050 s, the whole resultant profile is at equilibrium. The two component profiles are not both zero; they cancel point by point.
- At t = T/2 = 0.100 s, the nonzero resultant displacements have reversed sign.
Identify amplitude, nodes and antinodes
A node has zero displacement at all times and therefore zero oscillation amplitude. The fixed nodes here are x = 0, 0.40 and 0.80 m. An antinode has the greatest oscillation amplitude: here 4.0 mm at x = 0.20 and 0.60 m.
Amplitude describes the full oscillation, not the displacement in one frame. At T/4, an antinode passes through equilibrium at its greatest speed; it has not become a node.
Why does the zero profile not make every point a node?
A node remains at equilibrium throughout the cycle. Most points in the T/4 frame have nonzero displacements before and after it. Their zero displacement is temporary; only the fixed node positions have zero oscillation amplitude.
Adjacent antinodes: separation = λ/2
A node to its neighbouring antinode: λ/4
Points within one loop oscillate in phase. Points in neighbouring loops oscillate in antiphase: when one loop moves up, the next moves down. The nodes separating them do not oscillate.
The ideal standing pattern has no net time-averaged energy transport along it. Energy is still present and exchanges locally. Real driven demonstrations need an energy input to replace losses; this does not turn the fixed node pattern into travelling crests.
The graphical model explains formation. The string and microwave experiments and air-column experiment show how the pattern can be observed and measured.
03
Measure string and microwave patterns
A stable standing-wave pattern turns node or minimum positions into a wavelength measurement. Identify what the apparatus detects and count the intervals between matching positions.
In the ideal pattern, neighbouring nodes are λ/2 apart. Repeated detector minima can locate the corresponding fixed pattern for microwaves. A span covering several intervals usually makes a position-reading uncertainty a smaller fraction of the total distance.
A driven stretched string
Stretch a string between effectively fixed supports. Drive it weakly near an end and vary the driving frequency until stable loops appear. The wave and its reflection overlap; stationary low-amplitude positions are displacement nodes.
The supports and driver have different roles. An attachment that moves appreciably is not an exact fixed node. In the ideal fixed-end model, the support separation L contains an integer n of half-wavelength loops:
Keep the support separation, tension and string unchanged when comparing the patterns at different frequencies. After selecting a steady pattern, locate nodes against a fixed ruler rather than reading a moving antinode as a fixed endpoint.
A reflected microwave pattern
Point a microwave transmitter towards a reflecting metal sheet. The incident and reflected waves overlap. Move a detector along the propagation direction, keeping its orientation fixed so that it samples the same electric-field component.
The detector shows repeated signal maxima and minima at fixed positions. It measures signal strength, not signed electric-field displacement. Its minima are not sound-pressure nodes. For the stationary pattern, successive minima or successive maxima are λ/2 apart.
Locate several corresponding nodes or minima
A weakly driven stretched string
A moving attachment is not an exact fixed node. The weak near-end drive approximates the fixed-boundary model. Subtract ruler readings before counting the two node intervals.
Microwaves overlap with their reflection
Keep the source, reflector and detector orientation fixed while finding positions. Real minima can remain above zero; a raw detector reading is not a signed electric-field displacement.
Estimate before calculating
String nodes separated by roughly 0.4 m suggest a wavelength of about 0.8 m. Microwave minima separated by about 1.5 cm suggest a wavelength of a few centimetres. The factor of two comes from the standing pattern.
Worked string readings
Three successive displacement nodes have supplied ruler positions 13.0, 53.0 and 93.0 cm. Their span contains two node intervals:
Node interval = 0.800/2 = 0.400 m
λ = 2 × 0.400 = 0.800 m
The ruler origin is not a node. Subtract the position readings before assigning a separation. The 0.80 m support separation contains two loops, so L = 2λ/2 = λ; it does not contain two complete wavelengths.
Worked microwave readings
Four successive detector minima are at 8.0, 9.5, 11.0 and 12.5 cm. There are three intervals:
λ = 2 × 0.045/3 = 0.030 m
The readings are supplied positions in a model experiment. An absolute field or intensity calibration is not needed to locate minima, but the position scale and the identity of successive minima must be reliable.
Keep the measurement conditions controlled
For the microwave scan, keep source frequency, reflector position and detector alignment fixed. Record position units and the ruler or position readout's resolution. Repeat the scan to judge broad minima; changing detector orientation can change its response without representing a new interference minimum.
Real minima may remain above zero because of background signals or unequal incident and reflected amplitudes. Repeated measurements can reveal variation and improve minimum location; they do not correct a wrong scale or an uncontrolled setup.
For either experiment, label the first and last matching positions and count the intervals between them. Use the full measured span in the wavelength calculation, rather than assuming the apparatus starts at the ruler's zero.
04
Standing sound and air columns
A resonating air column has displacement and pressure patterns with different node positions. Identify which quantity is shown before using a boundary condition or a resonant length.
Separate particle displacement from pressure variation
Sound is longitudinal: an air particle's displacement is along the tube. Pressure variation, Δp, is the change from ambient pressure, measured in Pa. A graph of either quantity uses vertical height to show its value; it does not show particles moving transversely across the tube.
- At an approximately closed end, air cannot move through the boundary: it is a displacement node and a pressure-variation antinode.
- At an approximately open end, pressure stays close to ambient: it is a pressure-variation node and a displacement antinode.
A pressure node means nearly zero pressure variation, not zero absolute pressure or a vacuum. A pressure antinode has the greatest amplitude of pressure variation; it is not permanently at the highest pressure.
The displacement and pressure nodes alternate in space. A displacement node and its nearest pressure node are λ/4 apart. In a tube open at both ends, both ends have the open-end condition: displacement antinodes and pressure nodes. These are spatial relationships; a pair of amplitude envelopes does not establish a time delay between pressure and displacement.
Find resonances with a water boundary
Hold a vibrating tuning fork near an open tube mouth and change the water level to vary the air length. The water surface acts approximately as a closed end. At suitable lengths, the sound grows loud because a standing-wave mode is resonantly driven.
Measure the air length from the water surface to the fixed tube mouth. Do not use the tube's full length or the depth of water. Keep the fork frequency and tube geometry unchanged, and repeat the judgement of the loudness peak.
Separate the apparatus from the air-column model
The tube mouth stays fixed while the water level changes
Air-length model 1: physical length 0.160 m
These model comparisons align the closed end at x = 0. They do not show the apparatus mouth moving. Grey dashed: physical mouth. Brown dotted: effective open end, 0.015 m farther out.
A pressure node means no oscillating pressure variation there, not a vacuum. The two physical lengths differ by 0.350 m = λ/2; their common end correction cancels.
Air-length model 2: physical length 0.510 m
These model comparisons align the closed end at x = 0. They do not show the apparatus mouth moving. Grey dashed: physical mouth. Brown dotted: effective open end, 0.015 m farther out.
A pressure node means no oscillating pressure variation there, not a vacuum. The two physical lengths differ by 0.350 m = λ/2; their common end correction cancels.
In the closed/open model, the first two effective resonant lengths are λ/4 and 3λ/4. The second pattern contains an additional internal displacement node and antinode. Increasing the effective length by λ/2 gives the next resonance.
Estimate the wavelength and speed
A successive resonant-length difference of roughly 0.35 m suggests λ about 0.7 m. With a fork frequency of order 500 Hz, v = fλ is of order 350 m/s. This assumes the same tube and successive resonances; it is a scale estimate, not an independent sound-speed measurement.
Worked resonant lengths
Use the difference to remove a common end correction
At a supplied frequency of 480 Hz, successive physical air lengths are L1 = 0.160 m and L2 = 0.510 m. Their effective open ends lie slightly beyond the tube mouth. Model this with the same correction e at both resonances.
λ = 2(0.510 - 0.160) = 0.700 m
v = fλ = 480 × 0.700 = 336 m/s
The common end correction cancels in the difference. If these are the first two resonances, their effective lengths are:
3λ/4 = 0.525 m
e = 0.175 - 0.160 = 0.015 m
The second length gives the same e = 0.525 - 0.510 = 0.015 m. Using λ = 4 × 0.160 m would treat the physical mouth as the exact effective endpoint and underestimate the wavelength.
Optional check A 480 Hz fork gives successive resonances at physical air lengths 0.160 m and 0.510 m in the same closed/open tube. Assume a common end correction. What are the wavelength and wave speed?
Interpret the measurement limitations
A broad loudness maximum limits how precisely the resonant length can be located. Approach it from both sides and repeat the reading, while keeping the fork and tube arrangement consistent. Read the position scale with the line of sight perpendicular to it to reduce parallax.
The difference method assumes genuinely successive resonances and a common effective-end correction. Skipping a resonance changes the number of half-wavelength intervals. A changed tube opening can change the correction, so repeating readings alone would not repair that change.
The supplied lengths illustrate the inference. In an experiment, retain the original position readings, their units and the spread of repeated peak locations before reporting a wavelength.
05
Coherence, phase and two sources
A stable interference pattern depends on the phase difference where the waves meet. That difference includes both the source relationship and the distances travelled.
Name the conditions and quantities
Interference is the combined effect of overlapping waves. Coherent waves have the same frequency and a constant phase difference. They can be in phase or have a fixed nonzero phase difference.
Phase difference compares stages of oscillation. Path difference is the difference between distances travelled from the sources to the observation point. An extra wavelength adds one cycle of propagation delay; an extra half-wavelength adds half a cycle.
Diffraction is spreading through an opening or around an edge. It can allow waves from separate slits to overlap. Its opening-width and wavelength dependence is developed in diffraction and a single slit.
Use a path rule only with its source-phase condition
For sources in phase at emission and the same propagation wavelength:
Destructive: path difference = (n + 1/2)λ
n = 0, 1, 2, ... for a path-difference magnitude
Complete cancellation additionally requires equal amplitudes at the observation point. Unequal amplitudes can still interfere, but leave a nonzero minimum. Comparable amplitudes improve the contrast between large and small resultant oscillations.
Worked source and path phase
Two half-cycle differences
Suppose two equal-amplitude sound sources are in antiphase at emission, and their paths to P differ by λ/2.
The source relationship contributes half a cycle. The propagation difference contributes another half-cycle difference. Together they are equivalent to a whole cycle, so the arrivals are in phase and interfere constructively.
Applying the in-phase-source path rule without checking the source condition would give the wrong answer.
Optional check Two coherent sound sources of equal amplitude are in antiphase at emission. Their paths to P differ by half a wavelength. What happens at P?
Recognise the four demonstrations
Water: two dippers driven from a common source generate overlapping ripples. Stable regions of large and small oscillation amplitude can be observed. Keep frequency and water depth fixed. A snapshot showing crests and troughs is a phase map at an instant; it is not itself a map of oscillation amplitude.
Sound: two loudspeakers driven by the same signal generator produce alternating louder and quieter regions as a microphone or listener moves through the overlap. State the relative phase, including any reversed connection. Room reflections and background sound can obscure minima.
Microwaves: one transmitter illuminates two apertures whose outputs overlap. Move an aligned detector through the pattern. Keep polarisation and detector orientation fixed, so a response change caused by rotating the detector is not misidentified as interference.
Identify the common source and the overlap region
Water: two dippers driven in phase
This is an instantaneous phase-line map of the component waves. To locate persistent small- and large-amplitude regions, observe the resultant over time.
Sound: source phase and path phase both matter
One source is phase-reversed. The extra half-wavelength of path supplies another half-cycle difference, giving constructive interference at P. Source phase cannot be omitted from the reasoning.
Microwaves: one transmitter illuminates two apertures
Hold source frequency and detector orientation fixed. Rotating the probe would change its response for a different reason. The light demonstration uses the separate double-slit apparatus below.
Light: illuminate two slits with one coherent source and let their diffracted outputs overlap on a screen. A stable relative phase and compatible polarisation components allow visible fringes. The double-slit apparatus and screen pattern show this arrangement and its wavelength measurement. Ordinary independent lamps generally have rapidly changing relative phase, so their illumination does not normally produce a stable visible fringe pattern.
Separate a stable pattern from a perfectly dark minimum
Observable sustained fringes need overlapping waves of matching frequency with a sufficiently stable relative phase over the observation. For light, the interfering components must have compatible polarisation. The detector or screen must resolve the spatial variation, with enough contrast over background.
These conditions establish a stable pattern. Equal amplitudes and opposite arrival phase are the additional conditions for complete cancellation at a particular minimum. Coherence alone neither guarantees equal amplitudes nor means that every point is constructive.
06
Measure double-slit fringes
A double-slit pattern turns the difference between two light paths into a measurable fringe spacing. Keep slit separation, fringe separation and screen position distinct.
One coherent source illuminates two slits. Their outputs overlap on a screen and form stable bright and dark bands when the relative phase remains fixed. Here the slit outputs are in phase, and the relevant polarisation components are compatible.
Label the geometry
- a: centre-to-centre slit separation.
- D: perpendicular distance from the slit plane to the screen.
- x: separation of neighbouring bright centres.
- y: position on the screen measured from the central maximum.
For a distant screen and small observation angle θ, the path difference is approximately a sin θ, with sin θ ≈ tan θ ≈ y/D. Thus:
Bright centres: ay/D ≈ nλ
Fringe separation x = λD/a
λ = ax/D
The approximation requires D large compared with the slit separation and small y/D in the region used. It does not make sine, tangent and angle identical at arbitrary angles.
Worked fringe spacing
Use wavelength 600 nm, slit separation 0.30 mm and screen distance 2.0 m.
= 0.0040 m = 4.0 mm
Separate apparatus distances from the enlarged fringe scale
One coherent source, two slits, one screen
Enlarged calibrated screen region
The centre-to-centre span is divided by eight intervals, not nine bright centres. Fringe width and slit separation are measured in different places.
Nine consecutive bright centres from -16 mm to +16 mm span 32 mm. They contain eight intervals:
λ = (0.30 × 10-3)(4.0 × 10-3)/2.0
= 6.0 × 10-7 m = 600 nm
Optional check Nine consecutive bright centres span 32 mm in a double-slit pattern. Slit separation is 0.30 mm and perpendicular screen distance is 2.0 m. Under the small-angle approximation, what is the wavelength?
Classify a screen position
At y = 10 mm = 0.010 m in the same model:
= 1.50 × 10-6 m = 2.5λ
This is a minimum for the stated in-phase sources. The angle is approximately y/D = 0.005 rad, consistent with the small-angle approximation. The result uses the distant-screen model, not an exact finite-distance construction.
Make a wavelength measurement
Record the known slit separation with its units. Measure perpendicular D, keep the screen normal to the central axis and identify the central maximum. Locate several corresponding bright centres, measure their total span and divide by the number of intervals.
Use a band's centre rather than an arbitrary bright edge. A supplied image needs a calibrated spatial scale. Background light and broad fringes limit centre-location precision; a larger image alone does not improve the original measurement resolution. Repeat position readings while retaining the original values and their units.
07
Diffraction gratings and wavelength
A grating's regularly spaced slits reinforce light in particular directions. Use the spacing between slits, the order and the actual angle from the normal.
A transmission grating has many parallel slits. With monochromatic light normally incident, adjacent slits start in phase. Towards an outgoing angle θ, their path difference is a sin θ, where a is the slit spacing.
Locate principal maxima
When adjacent contributions differ by an integer number of wavelengths, they reinforce to form a principal maximum:
The central direction has n = 0. Matching positive and negative orders lie on opposite sides of the normal. Use the angle from that normal, not from the grating surface. The spacing a is a length; it is not the quoted number of lines per millimetre.
Estimate the optical scales
A few hundred lines per millimetre means a slit spacing of order micrometres. Visible wavelength is of order half a micrometre, so a/λ is only a few. Expect a few possible orders and first-order angles of order a few tenths of a radian. The exact sine bound, not this estimate, decides the highest integer order.
Worked orders and angles
Convert line density before using the equation
A grating has 600 lines/mm and is illuminated normally by light of wavelength 500 nm.
a = 1/600 000 m
≈ 1.67 × 10-6 m
sin θ = nλ/a = 0.300n
| Order n | sin θ | θ / ° |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 0.300 | 17.46 |
| 2 | 0.600 | 36.87 |
| 3 | 0.900 | 64.16 |
Keep the unrounded spacing in the calculation. Since |sin θ| cannot exceed 1:
Highest integer order = 3
The allowed directions are 0, ±1, ±2 and ±3: seven in total. Order 4 would require sin θ = 1.2 and has no real direction. These are geometrically allowed maxima, not a promise that every spot has equal brightness or can be detected.
The third-order angle is about 64°. Using sin θ ≈ θ there would be inappropriate; use the inverse sine.
Use path difference for maxima and exact angles for large orders
Adjacent-slit path difference
Seven geometrically allowed directions
Measure a first-order screen displacement
Use θ = arctan(0.3145/1.000), then λ = a sin θ for first order. Measurements on both sides help locate the centre and reveal alignment problems.
Optional check Normally incident 500 nm light falls on a grating with 600 lines/mm. What is the highest geometrically allowed principal order and the total number of allowed order directions?
Determine wavelength from a screen measurement
Use a monochromatic source and a grating of known line density. Direct the beam normally onto the grating. Establish the zero-order direction, then measure a chosen order's angle or obtain it from a calibrated perpendicular screen.
For a screen distance D = 1.000 m and a first-order displacement y = 0.3145 m from the zero-order centre:
= arctan(0.3145/1.000) ≈ 17.46°
λ = a sin θ/n
≈ 5.00 × 10-7 m = 500 nm
Here y/D is the tangent, not the exact sine. First find the angle, then use its sine in the grating equation.
Record the order, line density, screen geometry and spot-centre positions with units. Measure matching spots on both sides to help establish the centre and check alignment. Repeated centre readings assess variation, but averaging cannot automatically correct a tilted grating. In a school laser experiment, use the approved equipment and beam-control procedure.
08
Diffraction and a single slit
Waves spread through openings and around edges. For a single slit, the first dark minima bound a broad central intensity maximum.
Compare wavelength with the opening
Diffraction is spreading through an aperture or around an edge. Its angular spread depends on wavelength relative to the opening or obstacle dimensions.
In a ripple tank, keep water depth and source frequency fixed so the wavelength stays the same. Compare a wide gap with one whose width is comparable to that wavelength. The narrower gap gives greater angular spread. The wide gap still has spreading at its edges; diffraction has not disappeared.
Compare spreading at the same wavelength
The ripple-tank views share one spatial scale and the same wavelength. Each blue line represents a wavefront of the same phase. Line thickness does not represent amplitude or transmitted power.
Wide gap: width = 5 wavelengths
Narrow gap: width = 1 wavelength
Sound can spread around an edge
A narrow opening gives more angular spreading, not automatically greater transmitted power. A wide opening still shows diffraction at its edges.
A narrow opening does not automatically transmit more power. Spreading describes the directions occupied by the transmitted wave, not how much energy passes through.
Sound can spread around a corner into a region outside the direct geometrical path. Its wavelength compared with the obstacle size helps determine the spreading. Sound from a loudspeaker also diffracts at its emitting aperture. In a real room, reflections may contribute as well, so sound heard around an obstacle need not have only one cause.
Locate first minima for a uniformly illuminated slit
For a single slit of width b in the far-field model, the first minima satisfy:
There is one first minimum on each side of the central direction. The equation locates minima, unlike the grating equation a sin θ = nλ for principal maxima. The first minima enclose the central maximum; weaker side maxima lie beyond them.
Worked central width
Use slit width b = 0.15 mm, wavelength λ = 600 nm and screen distance D = 2.0 m.
= (600 × 10-9)/(0.15 × 10-3)
= 0.0040
This is a small angle, so θ ≈ 0.0040 rad and tan θ ≈ θ. The distance from the centre to one first minimum is:
= 0.0080 m = 8.0 mm
Full central width = 2y1 ≈ 16 mm
The first minima bound the broad central maximum
For b = 0.15 mm, λ = 600 nm and D = 2.0 m, the first minima are approximately 8.0 mm from the centre. This graph is intensity divided by the central intensity, not displacement.
The 8.0 mm distance is one-sided. The central width is 16 mm between the first two minima. Unlike a grating condition, this slit equation locates minima.
The centre-to-first-minimum distance is 8.0 mm on each side; it is not the full width. Using the inverse sine and then the tangent gives the same 8.0 mm value at the stated precision.
Optional check A single slit of width 0.15 mm is illuminated by 600 nm light. A screen is 2.0 m away. The first minimum is about 8.0 mm from the centre on each side. What is the central maximum's full width?
Predict the effect of changing the slit
At fixed wavelength and screen distance, halving b approximately doubles the small-angle first-minimum distances and central width. The pattern spreads more. This statement about width does not imply that the absolute peak intensity stays fixed as the slit narrows.
When reading a pattern, identify the central direction and the first minima on both sides. Use an intensity-position graph or calibrated image; a sine displacement curve does not represent this light-intensity pattern.
09
Resolve two sources
An aperture spreads each point source into a diffraction pattern. Two nearby source images can overlap, so their separation must be compared with the pattern width.
State the resolution criterion
Under the Rayleigh criterion, two equal sources are just resolved when one individual pattern's central maximum coincides with the other's first minimum. For the aperture-width model used here:
Use wavelength λ and aperture width b in matching length units; the resulting small angle is in radians. The actual source angular separation and the limiting diffraction angle are different quantities.
This comparison uses equal, mutually incoherent sources. Their time-averaged image intensities add. That differs from adding coherent field contributions to produce an interference pattern.
Estimate the angular scale
Visible wavelength of order half a micrometre divided by an aperture width of a few millimetres gives a resolution angle of order 10-4 rad. This is a rough scale check; it is not an angle of order one degree.
Worked aperture comparison
Keep the source separation fixed
Two sources emit light of wavelength 550 nm and have angular separation 0.20 mrad. For aperture width 2.0 mm:
= 2.75 × 10-4 rad = 0.275 mrad
The actual 0.20 mrad separation is smaller, so the sources fall below this resolution criterion. Widen the aperture to 4.0 mm:
= 0.1375 mrad
The unchanged 0.20 mrad separation now exceeds the limit, so the sources are resolved by the criterion. At the boundary between these cases:
= (550 × 10-9)/(0.20 × 10-3)
= 2.75 × 10-3 m = 2.75 mm
Keep the source separation fixed while changing aperture width
Two equal, mutually incoherent sources remain at -0.10 and +0.10 mrad, separated by 0.20 mrad. Brown long dashes and green short dashes show the individual intensity profiles; solid blue is their intensity sum. All panels have the same axes.
Each individual profile is normalised to unit peak to compare widths. This does not claim that different apertures pass the same absolute power or give the same physical peak intensity.
b = 2.0 mm: Below the Rayleigh criterion
First-minimum offset from either source centre: 0.2750 mrad. The source separation remains 0.20 mrad.
b = 2.75 mm: Just resolved by the criterion
First-minimum offset from either source centre: 0.2000 mrad. The brown square marks one profile's first minimum directly below the other profile's central peak.
b = 4.0 mm: Resolved by the criterion
First-minimum offset from either source centre: 0.1375 mrad. The source separation remains 0.20 mrad.
Optional check Two equal incoherent sources of wavelength 550 nm have angular separation 0.20 mrad. Using the stated Rayleigh criterion theta approximately lambda/b, what aperture width just resolves them?
Explain changes in resolving power
A larger aperture or shorter wavelength reduces λ/b and narrows the diffraction patterns. It can improve resolution without changing the actual angular separation of the sources.
Use the stated criterion consistently. It is a model for judging optical resolution, not a universal claim about every observer or image-processing method. Do not substitute a different aperture-shape formula when the width model θR ≈ λ/b is specified.
10
Revision summary
Match the equation to the pattern and its conditions. A standing-wave interval, an interference fringe and a diffraction minimum measure different things.
Superposition and standing patterns
In the linear model, add signed disturbances at the same position and instant. For vector quantities, add components. Coherent light requires electric-field addition before intensity is found; a pair of equal incoherent source images instead has intensities that add.
Adjacent nodes or antinodes: λ/2
Neighbouring node and antinode: λ/4
A standing wave can form from equal-amplitude, same-frequency waves travelling oppositely. Its nodes have zero amplitude at fixed positions. An instantaneous zero profile does not make every point a node. The ideal pattern has no net time-averaged energy transport along it.
For sound, a closed end is approximately a displacement node and pressure-variation antinode; an open end has the opposite conditions. A pressure node is not a vacuum. Successive closed/open resonant lengths differ by λ/2 when the effective-end correction is unchanged:
v = fλ
Revisit the stretched-string and microwave measurements for their distinct driver, reflector and detector roles, and the air-column experiment for length endpoints and end correction. Count intervals between matching positions, not the number of marked positions.
Coherence, phase and double slits
Coherent waves share a frequency and maintain a constant phase difference; they need not be in phase. Observable fringes also need overlap, sufficient stability and contrast, and compatible light polarisation components. Equal amplitude is needed for complete cancellation, rather than for every interference pattern.
Combine source phase with path difference. For in-phase sources, path differences nλ are constructive and (n + 1/2)λ are destructive. An initial antiphase relationship reverses those conditions. The water, sound and microwave demonstrations and light double-slit demonstration make the source and detector roles explicit.
Wavelength: λ = ax/D
Here a is slit separation, D the perpendicular screen distance and x the spacing between adjacent bright centres. The distant-screen, small-angle model gives path difference approximately ay/D at screen position y. Nine centres contain eight fringe intervals.
Distinguish maxima, minima and resolution
- Grating principal maxima: a sin θ = nλ for normal incidence. Convert line density to a spacing first. Require |nλ/a| ≤ 1, and count zero plus both signs of the allowed nonzero orders.
- Single-slit first minima: b sin θ = λ. With a small angle, the full central width is approximately 2λD/b, bounded by one first minimum on each side.
- Rayleigh criterion: θR ≈ λ/b in radians for the stated aperture-width model. Compare this limiting angle with the actual source separation.
For a grating screen measurement, y/D gives tan θ. Use the actual angle and its sine when the angle is not small. A wider aperture narrows its diffraction pattern and can improve resolution without moving the source centres.
a versus b: a separates neighbouring slits; b is the width of a single slit or aperture. x versus y: x is one fringe interval; y is a position from the central maximum. A grating equation locates maxima; a single-slit first-minimum equation locates minima.
Quantities and units
| Quantity | Symbol | Unit or meaning |
|---|---|---|
| Particle displacement | y or ξ | m; signed from equilibrium |
| Displacement amplitude | A | m; maximum magnitude |
| Pressure variation; its amplitude | Δp; Δp0 | Pa; relative to ambient pressure |
| Electric-field amplitude | E0 | N/C |
| Period | T | s |
| Frequency | f | Hz = s-1 |
| Wavelength | λ | m |
| Propagation speed | v | m/s |
| Phase difference | Δφ | rad or explicitly stated degrees |
| Path difference | Δr | m |
| Slit separation | a | m |
| Slit or aperture width | b | m |
| Fringe separation | x | m |
| Screen position from centre | y | m |
| Perpendicular screen distance | D | m |
| Grating line density | 1/a | m-1 |
| Order | n | Integer; dimensionless |
| Diffraction or separation angle | θ | rad; or degrees when stated for trigonometry |
| Intensity | I | W/m2; a normalised ratio is dimensionless |
Symbols can be reused in different models: y is particle displacement in a string graph but screen position in optical geometry. Name the plotted quantity locally. A pressure amplitude is not measured in metres, and a normalised intensity profile does not provide an absolute power.
Read and measure the right feature
Keep source frequency, geometry, alignment and the relevant medium controlled. Locate corresponding nodes, minima, resonances or bright centres, and retain their original position readings with units. Use several intervals where appropriate and state the count.
Use a supplied calibrated scale rather than measuring a schematic drawing. Repeated readings assess variability; they do not fix parallax, an incorrect source model, changed detector orientation or an unaccounted end correction. For a grating, identify the order and zero direction before extracting the wavelength.
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