Young double-slit experiment

Two light paths can make a bright band or a dark one. The dark band is not evidence that one path has been blocked. The arriving waves can cancel there while the overall pattern redistributes energy elsewhere.

We build the double-slit pattern, shift it with optical path, derive single-slit minima and combine the two scales into a diffraction-modulated interference pattern.

What you will learn to do

Explain coherence and polarization requirements for visible fringes.
Calculate resultant intensity for equal or unequal waves.
Derive double-slit angular and small-angle screen positions.
Predict a phase-plate shift with an explicit path-sign convention.
Explain single-slit minima by pairwise cancellation.
Use the sinc-squared profile including its central limit.
Identify missing orders and distinguish slit-width from slit-separation effects.

Useful before this chapterTravelling waves, superposition and energy →Refraction, internal reflection and dispersion →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

Why does adding two wave amplitudes and then squaring differ from adding their separate intensities?

Compare your reasoning

Squaring the sum introduces a cross term that depends on relative phase. That term produces interference. Adding intensities alone misses the coherent relationship.

Revisit the explanation →

01 / Interference, diffraction and the structure of a fringe pattern

A stable pattern requires a stable phase relationship

When two electromagnetic waves overlap, their electric fields add. The observed intensity is proportional to the time average of the square of the total field. Squaring the sum creates a cross term that can enhance or reduce the intensity depending on relative phase.

For a steady visible fringe pattern, the relative phase must remain sufficiently stable over the observation time. Splitting light from one source into two paths can provide this mutual coherence. Two independent ordinary lamps do not generally maintain a stable relative optical phase.

The waves also need compatible polarization components to interfere in the detected intensity. Orthogonal linear polarizations have no cross term in the total intensity unless a further element projects them onto a common direction.

The elementary formulas here assume monochromatic coherent waves with matching polarization and a stable apparatus. Finite spectral width, source size and motion can reduce fringe visibility even when a geometrical path difference can be calculated.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: A stable pattern requires a stable phase relationship
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

02 / Interference, diffraction and the structure of a fringe pattern

Path difference becomes phase, and phase becomes intensity

For two paths through the same uniform medium, a path difference Δ corresponds to phase difference 2πΔ/λ, where λ is the wavelength in that medium. Any initial source phase difference must be added with a consistent sign.

For coherent waves of individual intensities I1 and I2 with the same polarization, the resultant intensity is I1 + I2 + 2√(I1I2) cos φ. The square roots appear because field amplitude, not intensity, is the quantity being added.

At in-phase arrival, Imax = (√I1 + √I2)². At opposite phase, Imin = (√I1 − √I2)². Complete darkness requires equal amplitudes as well as opposite phase; unequal waves leave a nonzero minimum.

For equal individual intensities I0, the result is 4I0 cos²(φ/2). A fourfold local maximum does not create energy: constructive regions are accompanied by reduced intensity elsewhere. The optical system redistributes the energy flow.

I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi,\qquad\phi=\frac{2\pi\Delta}{\lambda}+\phi_0

Coherent monochromatic waves with matching polarization at the observation point. The sign convention for Δ and initial phase φ0 must be consistent.

WORKED EXAMPLE

Opposite phase does not always mean darkness

Two coherent waves with matching polarization have individual intensities 4 and 9 in the same intensity units. Find maximum and minimum resultant intensities.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: Path difference becomes phase, and phase becomes intensity
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi,\qquad\phi=\frac{2\pi\Delta}{\lambda}+\phi_0
Meaning: Coherent monochromatic waves with matching polarization at the observation point. The sign convention for Δ and initial phase φ0 must be consistent.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

03 / Interference, diffraction and the structure of a fringe pattern

Two narrow slits turn a viewing angle into a path difference

Let two narrow slits be separated by distance d and illuminated in phase. In the far field, the rays reaching a given direction are nearly parallel, and their path difference is d sin θ. Here θ is measured from the central perpendicular direction.

Bright interference orders satisfy d sin θ = mλ for integer m. Dark directions for equal slit amplitudes satisfy d sin θ = (m + 1/2)λ. The central m = 0 bright direction has equal paths.

On a flat screen at distance D, the exact angle-to-position relation is y = D tan θ. For small θ, sin θ ≈ tan θ ≈ y/D, giving ym ≈ mλD/d and fringe spacing β = λD/d.

The angular path relation already assumes the far-field slit geometry. The equally spaced screen-fringe formula makes an additional small-angle approximation. At large angles, use the trigonometric mapping rather than extending equal spacing indefinitely.

d\sin\theta_m=m\lambda,\qquad y_m\approx\frac{m\lambda D}{d},\qquad\beta\approx\frac{\lambda D}{d}

Narrow coherent in-phase slits, far-field observation; the screen-position and equal-spacing approximations additionally require small angles.

WORKED EXAMPLE

Keep wavelength, slit separation and screen distance in consistent units

A double-slit setup uses wavelength 600 nm, slit separation 0.30 mm and screen distance 1.5 m. Find the small-angle fringe spacing and the third bright fringe position from the central maximum.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: Two narrow slits turn a viewing angle into a path difference
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
d\sin\theta_m=m\lambda,\qquad y_m\approx\frac{m\lambda D}{d},\qquad\beta\approx\frac{\lambda D}{d}
Meaning: Narrow coherent in-phase slits, far-field observation; the screen-position and equal-spacing approximations additionally require small angles.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

04 / Interference, diffraction and the structure of a fringe pattern

A phase plate shifts the whole interference pattern

For paths through different materials, phase depends on optical path length, the integral of n ds, rather than geometrical length alone. With vacuum wavelength λ0, a fixed optical-path difference produces phase difference 2πΔOPL/λ0.

A thin plate of thickness t and index n inserted normally in one air path replaces air optical path t by nt. The added optical path is η = (n − 1)t, neglecting reflections and geometric beam displacement.

Define the upper slit at positive y and use Δ = lower-path optical length minus upper-path optical length. Near the centre, the geometric contribution is dy/D. If the plate covers the upper slit, the total difference is dy/D − η.

The new central bright direction therefore satisfies y = ηD/d, a shift toward the covered slit. This direction follows from the sign convention and equal optical paths; it should not be memorized independently of the geometry.

WORKED EXAMPLE

Predict the shift direction as well as its magnitude

In the previous setup, place a thin normal plate of index 1.50 and thickness 1.2 μm over the upper slit. Neglect reflections and ray displacement. Find the central-fringe shift.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: A phase plate shifts the whole interference pattern
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

05 / Interference, diffraction and the structure of a fringe pattern

A finite opening contains many interfering contributions

A slit with finite width a is not one point source. Different parts of the illuminated aperture contribute waves with different phases to an off-axis direction. Their superposition gives a diffraction pattern even when there is only one slit.

For the first minimum, divide the slit into two equal halves. At a direction satisfying a sin θ = λ, each point in one half can be paired with a corresponding point half a slit-width away. The paired path difference is λ/2, so equal contributions cancel.

For the mth nonzero minimum, the aperture can be divided into 2m zones and paired similarly. This gives a sin θ = mλ for nonzero integers m. The central direction m = 0 is a maximum, not a minimum, because every contribution is in phase there.

A narrower slit produces a wider angular diffraction pattern. The inverse width scale is a consequence of a smaller aperture needing a larger observation angle to accumulate the same phase difference across it.

a\sin\theta_m=m\lambda,\qquad m=\pm1,\pm2,\ldots\quad\text{(single-slit minima)}

Uniformly illuminated slit, normal incidence and Fraunhofer far-field observation. Only orders with |mλ/a| ≤ 1 can be geometrically reached.

WORKED EXAMPLE

Measure the width between the first two dark directions

A uniformly illuminated slit of width 0.20 mm is observed on a screen 2.0 m away with wavelength 500 nm. Find the small-angle width of its central diffraction maximum between first minima.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: A finite opening contains many interfering contributions
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
a\sin\theta_m=m\lambda,\qquad m=\pm1,\pm2,\ldots\quad\text{(single-slit minima)}
Meaning: Uniformly illuminated slit, normal incidence and Fraunhofer far-field observation. Only orders with |mλ/a| ≤ 1 can be geometrically reached.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

06 / Interference, diffraction and the structure of a fringe pattern

Adding the slit continuously gives a sinc-squared envelope

At angle θ, a point at transverse aperture coordinate x contributes a phase proportional to x sin θ. For uniform illumination, integrate the complex field contributions across x from −a/2 to +a/2. Their sum is proportional to sin α/α, where α = πa sin θ/λ.

Intensity is the squared amplitude, so I/Icentral = (sin α/α)². At α = 0 the expression has a removable limit of one; it does not predict an undefined physical intensity at the centre.

Nonzero integer multiples of π give the minima. The secondary maxima lie between those minima but are not exactly halfway: differentiating the profile gives tan α = α for noncentral extrema. Half-integer approximations should be labelled as approximations.

The profile assumes a uniformly illuminated rectangular slit in the far field. Illumination variation, near-field propagation or a different aperture shape changes the pattern. A circular aperture, for example, has an Airy pattern rather than this one-dimensional sinc-squared profile.

\frac{I(\theta)}{I(0)}=\left(\frac{\sin\alpha}{\alpha}\right)^2,\qquad\alpha=\frac{\pi a\sin\theta}{\lambda},\qquad\lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}=1

Fraunhofer diffraction by a uniformly illuminated single slit. Central normalization avoids confusion with incident irradiance or total transmitted power.

WORKED EXAMPLE

Evaluate a point inside the central maximum

For a single slit, evaluate the normalized intensity at a direction where a sin θ/λ = 1/2.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: Adding the slit continuously gives a sinc-squared envelope
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
\frac{I(\theta)}{I(0)}=\left(\frac{\sin\alpha}{\alpha}\right)^2,\qquad\alpha=\frac{\pi a\sin\theta}{\lambda},\qquad\lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}=1
Meaning: Fraunhofer diffraction by a uniformly illuminated single slit. Central normalization avoids confusion with incident irradiance or total transmitted power.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Interference, diffraction and the structure of a fringe pattern

Interference fringes sit inside a diffraction envelope

Two finite identical slits have two geometrical scales. Their centre separation d controls the relative phase between slit contributions, while their width a controls the phase spread within each slit.

For equal coherent illumination, the normalized far-field intensity is [sin α/α]² cos² β, with α = πa sin θ/λ and β = πd sin θ/λ. The narrow fringes are modulated by the broader single-slit diffraction envelope when d is appreciably larger than a.

An interference-order direction d sin θ = mλ is missing if it also falls at a diffraction minimum a sin θ = jλ with nonzero integer j. Thus m = j d/a. When d/a is an integer, every corresponding multiple of that order is suppressed.

The cosine maxima identify the underlying interference-order directions. Away from the envelope centre, the exact maxima of the product can shift slightly because the envelope itself varies with angle. Treating the envelope as locally constant is an approximation.

\frac{I(\theta)}{I(0)}=\left(\frac{\sin\alpha}{\alpha}\right)^2\cos^2\beta,\qquad\alpha=\frac{\pi a\sin\theta}{\lambda},\quad\beta=\frac{\pi d\sin\theta}{\lambda}

Two equal nonoverlapping rectangular slits with identical coherent illumination, in the Fraunhofer regime. I(0) is the two-slit central intensity.

Separate the slit-width envelope from the slit-spacing fringes

Inspect a plotted point: q = a sin θ / λ = -2

Diffraction envelope: 0
Two finite slits: 0

Here d/a = 3. The broad envelope vanishes at nonzero integer q; the underlying interference orders at q = ±1 are suppressed because those positions are also diffraction minima. Lines connect the plotted samples.

WORKED EXAMPLE

Use the two length scales to locate a missing order

Two identical slits have centre separation d = 0.60 mm and individual width a = 0.20 mm. Which first nonzero interference orders are suppressed by a single-slit minimum?

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: Interference fringes sit inside a diffraction envelope
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
\frac{I(\theta)}{I(0)}=\left(\frac{\sin\alpha}{\alpha}\right)^2\cos^2\beta,\qquad\alpha=\frac{\pi a\sin\theta}{\lambda},\quad\beta=\frac{\pi d\sin\theta}{\lambda}
Meaning: Two equal nonoverlapping rectangular slits with identical coherent illumination, in the Fraunhofer regime. I(0) is the two-slit central intensity.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

08 / Interference, diffraction and the structure of a fringe pattern

Use scaling and limiting cases to identify the physical cause

If slit separation d increases at fixed width, wavelength and screen distance, the interference fringes become closer while the diffraction envelope width stays unchanged. If slit width a decreases at fixed separation, the envelope broadens while the underlying interference spacing stays unchanged.

Increasing wavelength broadens both scales. Increasing screen distance broadens their small-angle spatial sizes while leaving the angular pattern unchanged. These comparisons identify which parameter a proposed change can affect.

Blocking one slit removes the two-slit cross term but leaves the single-slit diffraction pattern of the remaining aperture. A claim that every bright and dark feature must disappear confuses interference between slits with interference across one slit.

The far-field condition can be achieved at sufficiently large distance or in the focal plane of an appropriate lens. For free propagation a characteristic requirement is D much larger than the relevant aperture scale squared divided by wavelength. A near-field pattern requires a different propagation calculation.

STOP & THINK

One slit in a two-finite-slit setup is blocked. What remains from the open slit?

No pattern at all
Its single-slit diffraction pattern
The same two-slit fringe pattern with identical visibility

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Interference, diffraction and the structure of a fringe pattern
Focus: Use scaling and limiting cases to identify the physical cause
Learning goals:
- Explain coherence and polarization requirements for visible fringes.
- Calculate resultant intensity for equal or unequal waves.
- Derive double-slit angular and small-angle screen positions.
- Predict a phase-plate shift with an explicit path-sign convention.
- Explain single-slit minima by pairwise cancellation.
- Use the sinc-squared profile including its central limit.
- Identify missing orders and distinguish slit-width from slit-separation effects.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

Coherent fields add before the intensity is calculated.
Opposite phase gives complete darkness only for equal compatible amplitudes.
Double-slit fringe spacing λD/d is a far-field small-angle result.
Optical path, not geometry alone, determines phase through different media.
Single-slit minima satisfy a sin θ = mλ with m nonzero.
A narrower slit broadens its diffraction pattern.
Finite double slits combine a diffraction envelope and an interference factor.
A missing interference order coincides with a diffraction zero.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

With wavelength 500 nm, screen distance 2.0 m and slit separation 0.50 mm, what is the small-angle fringe spacing in millimetres?

Answer in mm.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 2

Two coherent same-polarization waves have individual intensities 1 and 16 in common units. What is their minimum resultant intensity?

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 3

For λ = 600 nm, D = 1.0 m and slit width a = 0.30 mm, what is the central diffraction width between first minima in millimetres?

Answer in mm.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 4

If slit centre separation is four times the individual slit width, what is the first positive missing interference order?

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 5

Why is m = 0 excluded from the single-slit minimum condition?

The central contributions arrive in phase and reinforce
Zero-order light has no wavelength
The centre is outside the aperture

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 6

Slit width decreases while slit separation and wavelength stay fixed. What changes in the far-field pattern?

The diffraction envelope broadens while underlying fringe spacing stays fixed
The envelope narrows and spacing doubles
Both scales must stay fixed

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 7

A local maximum of two equal coherent waves can have four times one-wave intensity because:

Energy is created by overlap
Energy flow is redistributed, with reduced intensity elsewhere
The wave frequency necessarily doubles

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

The whole interference setup enters a liquidIntermediate · Free sample · Hints and guided reasoning Count the maxima that survive the slit envelopeAdvanced · Hints and guided reasoning How far apart are neighbouring bright fringes?Intermediate · Hints and guided reasoning Width of the broad central bandIntermediate · Hints and guided reasoning Interference inside a diffraction envelopeAdvanced · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

A transparent plate is inserted into one path of a two-path interference setup. Why can the fringes move without changing the slit spacing?

Compare your reasoning

The plate changes that path’s optical length and relative phase. A different screen position is then needed for the same constructive condition, so the pattern shifts.

Revisit the explanation →

A few days later: change the situation

How could you distinguish the broad single-slit envelope from the narrow two-slit fringes in a finite-slit pattern?

Compare your reasoning

They respond to different geometry: slit width sets the diffraction envelope, while slit separation sets the fine interference spacing in the far-field approximation. Changing one dimension tests which feature it controls.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

Back to chapter contents ↑