Single Slit Diffraction

Single slit diffraction occurs when a wave passes through a narrow opening and spreads into the region beyond it. Rather than producing a sharp beam with perfectly defined edges, the wave forms a pattern containing a broad central maximum surrounded by alternating dark and weaker bright fringes.

For light, the effect becomes especially noticeable when the width of the slit is comparable to the wavelength.

The pattern forms because waves emerging from different parts of the same slit interfere with one another.

The most important relationship for the dark fringes, or diffraction minima, is:

a sin θ = nλ

where:

• a = slit width
• θ = angle from the center
• n = 1, 2, 3, ...
• λ = wavelength

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What Is Diffraction?

Diffraction is the spreading of waves when they pass through an opening or around an obstacle.

Diffraction can occur with many types of waves, including:

• water waves
• sound waves
• light
• radio waves
• microwaves
• matter waves

The amount of diffraction depends strongly on the relationship between the wavelength and the size of the opening.

A useful general rule is:

wavelength comparable to opening size → significant diffraction

wavelength much smaller than opening → less noticeable diffraction

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Diffraction Through a Wide Opening

Suppose waves approach an opening that is much wider than their wavelength.

Most of the wave continues approximately forward.

Some spreading occurs near the edges, but the overall diffraction is relatively small.

Therefore:

a >> λ → relatively little spreading

where a is the width of the opening.

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Diffraction Through a Narrow Opening

Now reduce the width of the opening.

The spreading becomes more noticeable.

As the slit becomes narrower relative to the wavelength, the wave spreads through a greater range of angles.

Therefore:

smaller slit width → greater diffraction

This relationship is extremely important in single slit diffraction.

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Diffraction and Wavelength

The wavelength also affects the amount of spreading.

For the same slit:

longer wavelength → greater diffraction

shorter wavelength → less diffraction

This explains why long-wavelength radio waves can diffract significantly around large structures while visible light generally produces much less obvious diffraction around everyday objects.

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Why a Single Slit Produces an Interference Pattern

At first, it may seem surprising that interference can occur with only one slit.

The key idea is that the slit is not treated as a single point source.

According to the wave model, different parts of the opening can be considered sources of secondary wavelets.

These wavelets travel slightly different distances to points on the screen.

They therefore arrive with different phases.

At some positions they reinforce each other.

At other positions they cancel.

The result is:

single opening → diffraction → internal interference → bright and dark pattern

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Huygens' Principle

Huygens' principle provides a useful way of understanding diffraction.

It states that every point on a wavefront can be considered a source of secondary wavelets.

As the incoming wave reaches the slit, many points across the opening act as sources.

The secondary wavelets spread beyond the slit.

Their superposition produces the observed diffraction pattern.

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Constructive and Destructive Interference

The waves coming from different parts of the slit can interfere.

When their displacements reinforce one another:

constructive interference → greater intensity

When their displacements cancel:

destructive interference → lower intensity

At certain angles, the contributions from across the slit cancel almost completely.

These locations form the diffraction minima, which appear as dark fringes.

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The Single Slit Pattern

A typical single slit diffraction pattern contains:

• one wide, bright central maximum
• a dark minimum on each side
• weaker secondary maxima
• additional minima
• progressively weaker fringes farther from the center

The pattern is symmetrical about its center.

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The Central Maximum

The large bright region at the center is called the central maximum.

It occurs around:

θ = 0°

At the exact center, waves from across the slit travel approximately equal distances to the screen.

Their contributions therefore reinforce strongly.

This produces the brightest part of the diffraction pattern.

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Why the Central Maximum Is Bright

At the center of the screen:

path differences ≈ 0

The waves emerging from different parts of the slit therefore arrive nearly in phase.

Their amplitudes combine constructively.

Since light intensity is related to the square of amplitude:

I ∝ A²

strong constructive interference produces a high-intensity central region.

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Why the Central Maximum Is Wider

The first dark minimum occurs on either side of the central maximum.

The central maximum therefore extends:

from the first minimum on one side to the first minimum on the other side

The neighbouring secondary maxima lie between successive minima.

For small angles, the central maximum is approximately twice as wide as each secondary bright fringe.

This is one of the most recognizable features of single slit diffraction.

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Central Maximum vs Secondary Maxima

The secondary maxima also become progressively weaker farther from the center.

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Diffraction Minima

The dark regions in the diffraction pattern are called minima.

They occur where waves from different parts of the slit cancel through destructive interference.

The condition for single slit minima is:

a sin θ = nλ

where:

• a = slit width
• θ = angle of the minimum
• λ = wavelength
• n = 1, 2, 3, ...

Notice that:

n = 0 is not a minimum

At θ = 0°, the central maximum occurs instead.

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First Diffraction Minimum

For the first minimum:

n = 1

Therefore:

a sin θ₁ = λ

or:

sin θ₁ = λ/a

This angle determines the edge of the central maximum.

The first minima occur symmetrically at:

+θ₁ and −θ₁

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Second Diffraction Minimum

For the second minimum:

n = 2

Therefore:

a sin θ₂ = 2λ

or:

sin θ₂ = 2λ/a

Additional minima follow for n = 3, 4, and so on, provided the required value of sin θ is physically possible.

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Why the First Minimum Occurs

The first minimum can be understood by dividing the slit into two equal halves.

For the first minimum:

a sin θ = λ

The path difference between light from the very top and very bottom of the slit is one wavelength.

Now pair each point in the upper half with a corresponding point in the lower half.

The path difference within each pair is:

λ/2

A half-wavelength difference corresponds to a phase difference of:

180°

Each pair cancels through destructive interference.

The total intensity therefore reaches a minimum.

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Higher-Order Minima

For higher minima, the slit can be divided into additional sections whose contributions cancel in pairs.

The general result remains:

a sin θ = nλ

This relationship predicts the dark fringes rather than the exact positions of the secondary bright fringes.

That distinction is important.

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Worked Example: First Minimum

Light with wavelength:

λ = 600 nm

passes through a slit of width:

a = 0.20 mm

Convert to metres:

λ = 6.00 × 10⁻⁷ m

a = 2.00 × 10⁻⁴ m

For the first minimum:

a sin θ = λ

sin θ = λ/a

sin θ = (6.00 × 10⁻⁷)/(2.00 × 10⁻⁴)

sin θ = 0.00300

Therefore:

θ ≈ 0.172°

The first minima occur approximately 0.172° on either side of the center.

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Worked Example: Second Minimum

Using the same slit and wavelength:

a = 2.00 × 10⁻⁴ m

λ = 6.00 × 10⁻⁷ m

For:

n = 2

a sin θ = 2λ

sin θ = 2λ/a

sin θ = 0.00600

Therefore:

θ ≈ 0.344°

The second minimum occurs farther from the center.

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Finding Slit Width

The minimum condition can be rearranged:

a = nλ/sin θ

Suppose the first minimum occurs at:

θ = 0.20°

and:

λ = 500 nm

Then:

a = (1)(500 × 10⁻⁹)/sin(0.20°)

Using:

sin(0.20°) ≈ 0.00349

a ≈ 1.43 × 10⁻⁴ m

Therefore:

a ≈ 0.143 mm

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Finding Wavelength

The same equation can be rearranged:

λ = a sin θ/n

Suppose:

a = 0.10 mm

The second minimum occurs at:

θ = 0.60°

Then:

a = 1.0 × 10⁻⁴ m

n = 2

λ = (1.0 × 10⁻⁴)sin(0.60°)/2

Using:

sin(0.60°) ≈ 0.01047

λ ≈ 5.24 × 10⁻⁷ m

Therefore:

λ ≈ 524 nm

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From Angle to Position on a Screen

Experiments often measure the distance of a minimum from the center rather than its angle.

If:

• y = distance from the center
• L = slit-to-screen distance

then:

tan θ = y/L

For small angles:

sin θ ≈ tan θ ≈ y/L

Therefore, the minimum condition becomes approximately:

ay/L = nλ

and:

y_n = nλL/a

This gives the approximate position of the nth dark fringe.

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Worked Example: Position of the First Minimum

Suppose:

λ = 500 nm

a = 0.25 mm

L = 2.0 m

Convert:

λ = 5.00 × 10⁻⁷ m

a = 2.50 × 10⁻⁴ m

For the first minimum:

y₁ = λL/a

y₁ = (5.00 × 10⁻⁷)(2.0)/(2.50 × 10⁻⁴)

y₁ = 0.0040 m

Therefore:

y₁ = 4.0 mm

The first dark fringes appear approximately 4.0 mm on either side of the center.

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Width of the Central Maximum

The first minima occur at:

+y₁ and −y₁

Therefore, the full width of the central maximum is:

W = 2y₁

Using:

y₁ = λL/a

we obtain:

W = 2λL/a

for the small-angle approximation.

This is an important relationship.

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Worked Example: Central Maximum Width

Suppose:

λ = 600 nm

L = 3.0 m

a = 0.30 mm

Convert:

λ = 6.00 × 10⁻⁷ m

a = 3.00 × 10⁻⁴ m

Using:

W = 2λL/a

W = 2(6.00 × 10⁻⁷)(3.0)/(3.00 × 10⁻⁴)

W = 0.012 m

Therefore:

W = 12 mm

The central maximum extends approximately 6 mm to either side of the center.

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Effect of Slit Width

From:

sin θ₁ = λ/a

a smaller slit width produces a larger diffraction angle.

Therefore:

a decreases → θ increases → pattern spreads

Conversely:

a increases → θ decreases → pattern narrows

This inverse relationship is fundamental to diffraction.

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Effect of Wavelength

Again:

sin θ₁ = λ/a

A larger wavelength produces a larger diffraction angle.

Therefore:

λ increases → diffraction increases

For the same slit:

• red light spreads more than blue light
• longer-wavelength waves diffract more than shorter-wavelength waves

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Red Light vs Blue Light

Suppose:

λ_red = 650 nm

λ_blue = 450 nm

Both pass through the same slit.

Because:

650 nm > 450 nm

the red light has the larger diffraction angle.

Therefore:

red light → wider central maximum

blue light → narrower central maximum

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Worked Comparison of Two Wavelengths

Suppose:

a = 0.20 mm

L = 2.0 m

For blue light:

λ = 450 nm

Central width:

W = 2λL/a

W = 2(450 × 10⁻⁹)(2.0)/(2.0 × 10⁻⁴)

W = 9.0 mm

For red light:

λ = 650 nm

W = 2(650 × 10⁻⁹)(2.0)/(2.0 × 10⁻⁴)

W = 13 mm

The longer-wavelength red light produces the broader central maximum.

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Effect of Screen Distance

From:

W = 2λL/a

increasing L increases the physical width of the pattern on the screen.

Therefore:

L increases → central maximum becomes wider on the screen

However, changing L does not change the diffraction angle produced by the slit.

It simply allows the spreading wave to travel farther, producing a larger physical separation on the screen.

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Angular Width of the Central Maximum

The first minima occur at:

−θ₁ and +θ₁

Therefore, the angular width of the central maximum is:

2θ₁

For small angles:

θ₁ ≈ λ/a

when θ is measured in radians.

Therefore:

angular width ≈ 2λ/a

A narrower slit produces a larger angular width.

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Intensity Pattern

The single slit pattern is not a set of equally bright fringes.

The central maximum is the strongest.

The secondary maxima become progressively weaker farther from the center.

The intensity distribution has a characteristic shape:

• large central peak
• zero intensity at first minima
• much smaller secondary peak
• another minimum
• still smaller secondary peak
• continuing outward

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Mathematical Intensity Distribution

For a uniformly illuminated ideal slit, the intensity can be described by:

I = I₀(sin β/β)²

where:

β = πa sin θ/λ

and:

• I = intensity at angle θ
• I₀ = intensity at the center
• a = slit width
• λ = wavelength

The full equation is more advanced than the minimum condition, but it explains why the central maximum is much brighter than the secondary maxima.

At the minima:

sin β = 0

which leads to:

a sin θ = nλ

for non-zero integers n.

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Single Slit vs Double Slit

Single slit and double slit experiments both involve wave interference, but their patterns are different.

In a real double slit experiment, each individual slit also diffracts the light, so the double slit interference pattern is contained inside a broader diffraction envelope.

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Slit Width vs Slit Separation

This distinction is essential.

For single slit diffraction:

a = width of one slit

and:

a sin θ = nλ

For double slit interference:

d = separation between the two slits

and bright fringes satisfy:

d sin θ = nλ

Using d instead of a is a common calculation error.

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Diffraction of Sound

Diffraction is not unique to light.

Sound wavelengths are often comparable to the dimensions of:

• doors
• windows
• walls
• corridors

As a result, sound can diffract noticeably around obstacles and through openings.

This is one reason you can sometimes hear someone speaking from another room even when you cannot see them directly.

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Diffraction of Radio Waves

Radio wavelengths can be much longer than visible-light wavelengths.

Some radio waves can therefore diffract significantly around:

• buildings
• hills
• other obstacles

The exact behaviour depends on wavelength, obstacle size, atmospheric conditions, and other propagation effects.

Diffraction is therefore important in communication engineering.

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Why Light Usually Appears to Travel in Straight Lines

Visible-light wavelengths are extremely small compared with most everyday openings.

For example, visible wavelengths are roughly hundreds of nanometres, while a doorway is about a metre wide.

Therefore:

a >> λ

and diffraction through the doorway is extremely small.

Light consequently appears to travel approximately in straight lines in ordinary situations.

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Diffraction Limits

Diffraction places a fundamental limit on how narrowly waves can be confined and how precisely optical systems can resolve small details.

This affects:

• microscopes
• telescopes
• cameras
• optical instruments

Even a perfect lens cannot eliminate diffraction.

The wave nature of light therefore creates fundamental limits on image resolution.

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Diffraction and Telescopes

Light entering a telescope passes through a finite aperture.

Because of diffraction, light from a distant point does not form a perfectly infinitesimal point in the image.

Instead, it forms a small diffraction pattern.

Larger apertures generally reduce diffraction spreading and allow finer details to be resolved.

This is one reason large telescope apertures are scientifically valuable.

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Diffraction and Microscopes

Microscopes also face diffraction limits.

Very small objects can only be distinguished if the imaging system can resolve their separate diffraction patterns.

Using shorter wavelengths can improve resolution.

This is one reason electron microscopy can achieve much higher resolution than ordinary visible-light microscopy: electrons can have extremely short de Broglie wavelengths.

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Single Slit Diffraction and Matter

Matter particles can also diffract.

According to de Broglie's relationship:

λ = h/p

where:

• λ = matter wavelength
• h = Planck's constant
• p = momentum

Electrons, neutrons, atoms, and other particles can exhibit diffraction when interacting with structures of an appropriate size.

This provides further evidence for the wave nature of matter.

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Experimental Investigation

A basic optical investigation can use:

• monochromatic laser
• narrow single slit
• screen
• ruler
• known slit width

The pattern on the screen can be measured.

Useful quantities include:

• central maximum width
• distance to first minimum
• slit-to-screen distance
• wavelength
• slit width

The relationship:

y₁ = λL/a

can then be tested experimentally.

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Improving Measurements

A good diffraction experiment should:

• measure L carefully
• identify the center of the pattern
• measure between dark minima when possible
• use a sufficiently large screen distance
• repeat measurements
• use consistent units
• keep the slit and screen aligned

The central maximum width can often be measured more reliably than the position of one edge alone.

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Worked Experimental Problem

A single slit produces a central maximum:

16.0 mm wide

The screen is:

2.50 m

from the slit.

The light wavelength is:

640 nm

Find the slit width.

Using:

W = 2λL/a

Rearrange:

a = 2λL/W

Convert:

λ = 6.40 × 10⁻⁷ m

W = 1.60 × 10⁻² m

Then:

a = 2(6.40 × 10⁻⁷)(2.50)/(1.60 × 10⁻²)

a = 2.00 × 10⁻⁴ m

Therefore:

a = 0.200 mm

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A Complete Diffraction Problem

Light of wavelength:

λ = 550 nm

passes through a slit:

a = 0.10 mm

The screen is:

L = 1.5 m

First find the first minimum.

a sin θ₁ = λ

sin θ₁ = (550 × 10⁻⁹)/(1.0 × 10⁻⁴)

sin θ₁ = 0.0055

Therefore:

θ₁ ≈ 0.315°

Using the small-angle approximation:

y₁ = λL/a

y₁ = (550 × 10⁻⁹)(1.5)/(1.0 × 10⁻⁴)

y₁ = 8.25 mm

Therefore, the central maximum width is:

W = 2y₁

W = 16.5 mm

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Predicting Changes

The equation:

W = 2λL/a

makes several useful predictions.

If λ doubles:

W doubles

If L doubles:

W doubles

If a doubles:

W halves

If a is halved:

W doubles

These proportional relationships can often solve problems without a full calculation.

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Example: Multiple Changes

Suppose an experiment is changed so that:

• wavelength doubles
• slit width is halved
• screen distance remains unchanged

Since:

W ∝ λ/a

doubling λ gives:

×2

Halving a gives another:

×2

Therefore:

W_new = 4W_original

The central maximum becomes four times as wide.

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Limits on Diffraction Orders

The minimum condition is:

a sin θ = nλ

Since:

sin θ ≤ 1

we must have:

nλ ≤ a

Therefore, only a limited number of diffraction minima can exist for a particular slit width and wavelength.

If a proposed value of n gives:

sin θ > 1

that diffraction minimum cannot physically occur.

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Example: Possible Minima

Suppose:

a = 2.5λ

The minimum condition gives:

sin θ = nλ/a

For n = 1:

sin θ = 1/2.5 = 0.4

Possible.

For n = 2:

sin θ = 2/2.5 = 0.8

Possible.

For n = 3:

sin θ = 3/2.5 = 1.2

Impossible.

Therefore, only the first and second minima occur on each side before reaching 90°.

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A Reliable Problem-Solving Strategy

First identify:

• wavelength λ
• slit width a
• screen distance L
• diffraction order n
• angle θ
• screen position y

For angular minima, use:

a sin θ = nλ

For small-angle screen positions:

y_n = nλL/a

For central maximum width:

W = 2λL/a

Always convert units before calculating.

Useful conversions include:

1 mm = 10⁻³ m

1 μm = 10⁻⁶ m

1 nm = 10⁻⁹ m

Finally, check whether the result makes physical sense.

A narrower slit or longer wavelength should produce more, not less, spreading.

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Common Mistakes

Thinking diffraction only occurs when a slit is extremely narrow

Diffraction always occurs, but it becomes more noticeable when the opening size is comparable to the wavelength.

Thinking the single slit pattern requires two separate slits

Different parts of the same opening contribute waves that interfere.

Using n = 0 for a dark fringe

The minimum equation uses n = 1, 2, 3, ...

At θ = 0°, there is a central maximum.

Thinking the central maximum has the same width as the other bright fringes

It is approximately twice as wide as the secondary maxima under the usual small-angle conditions.

Thinking all bright fringes have equal intensity

The secondary maxima become progressively weaker.

Thinking a wider slit causes more diffraction

A narrower slit produces greater angular spreading.

Thinking shorter wavelengths diffract more

For the same slit, longer wavelengths produce greater diffraction.

Confusing slit width with slit separation

Single slit diffraction uses slit width a. Double slit interference commonly uses slit separation d.

Using W = λL/a for the full central width

λL/a gives the approximate distance from the center to the first minimum. The full central maximum width is 2λL/a.

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Key Terms

Diffraction — spreading of a wave as it passes through an opening or around an obstacle.

Single slit diffraction — diffraction pattern produced when waves pass through one narrow opening.

Huygens' principle — model in which points on a wavefront act as sources of secondary wavelets.

Superposition — addition of overlapping wave displacements.

Interference — combination of overlapping waves.

Constructive interference — interference that increases resultant amplitude.

Destructive interference — interference that decreases resultant amplitude.

Central maximum — broad, bright central region of a single slit diffraction pattern.

Secondary maximum — weaker bright region on either side of the central maximum.

Diffraction minimum — dark region produced by destructive interference.

Slit width — width a of the opening through which the wave passes.

Diffraction angle — angle θ between the central direction and a feature in the diffraction pattern.

Wavelength — distance between corresponding points on successive wave cycles.

Intensity — wave power transferred per unit area; for light it determines the brightness of the pattern.

Angular width — angle between the two sides of a diffraction feature such as the central maximum.

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Key Takeaways

• Diffraction is the spreading of waves through openings and around obstacles.
• Single slit diffraction occurs because contributions from different parts of one slit interfere.
• A single slit pattern contains a broad bright central maximum surrounded by dark minima and weaker secondary maxima.
• The central maximum is the brightest part of the pattern.
• The central maximum is approximately twice as wide as neighbouring secondary maxima.
• Secondary maxima become progressively weaker farther from the center.
• Single slit minima satisfy a sin θ = nλ, where n = 1, 2, 3, ...
• The first minimum defines one edge of the central maximum.
• Narrower slits produce greater diffraction.
• Longer wavelengths produce greater diffraction.
• For small angles, the position of a minimum is approximately y_n = nλL/a.
• The central maximum width is approximately W = 2λL/a.
• Increasing screen distance increases the physical width of the pattern but does not change its angular diffraction pattern.
• Red light produces greater diffraction than blue light through the same slit.
• Single slit diffraction is different from double slit interference, although both involve superposition.
• Slit width a must not be confused with double slit separation d.
• Diffraction places fundamental limits on the resolving ability of optical instruments.
• Sound, radio waves, light, and matter waves can all exhibit diffraction.
• The diffraction pattern provides strong evidence for the wave nature of light.

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Check Your Understanding

1. Define diffraction and explain when it becomes particularly noticeable.
2. Explain how interference can occur even though a single slit experiment contains only one opening.
3. Describe the appearance of a typical single slit diffraction pattern.
4. Why is the central maximum brighter than the surrounding secondary maxima?
5. Explain why the central maximum is approximately twice as wide as the neighbouring bright fringes.
6. State the condition for a single slit diffraction minimum and explain each symbol.
7. Light of wavelength 500 nm passes through a slit 0.20 mm wide. Calculate the angle of the first minimum.
8. Using the same values from Question 7, calculate the angle of the second minimum.
9. A 600 nm laser passes through a 0.30 mm slit onto a screen 2.0 m away. Calculate the approximate distance from the center to the first minimum.
10. Using your result from Question 9, calculate the full width of the central maximum.
11. Predict what happens to the diffraction pattern if the slit width is reduced by half. Explain using the diffraction equation.
12. Red light and blue light pass separately through the same slit. Which produces the wider central maximum? Explain why.
13. A central maximum is 20 mm wide on a screen 2.5 m from a slit. The wavelength is 500 nm. Calculate the slit width.
14. Explain the difference between the roles of slit width in single slit diffraction and slit separation in double slit interference.
15. A student says, "Making a slit narrower should make the light beam narrower." Evaluate this statement using diffraction and the relationship between slit width, wavelength, and the angular width of the central maximum.