Why Waves Spread Further Through a Narrow Gap

Keep the wave frequency and water depth fixed. Straight water-wave crests approach a barrier with a 12 cm gap. You narrow the gap to 4 cm.

Predict what happens. Does the smaller opening produce a thinner stream of waves, or do the crests spread further sideways?

They spread further sideways. The wavefront emerging from the narrower gap becomes more curved, allowing the waves to travel into a wider range of directions.

The reason is that a narrow opening produces a smaller range of path differences between contributions from different parts of the opening. At a given angle, these contributions have a smaller phase difference, so they cancel less strongly. Waves can therefore reinforce over a wider range of directions.

Three ideas explain this result:

• Boundary effects determine which parts of the incoming wavefront can pass through the opening.
• Path difference and phase difference determine how the contributions from different parts of the opening relate to one another.
• Superposition determines the resulting wave when those contributions overlap.

Understanding how these ideas work together explains why narrowing a gap increases diffraction.

01Investigating what happens when the gap narrows

Before explaining the physics, consider how you would test the effect experimentally.

Setting up a controlled experiment

A ripple tank provides a straightforward way to investigate diffraction.

Arrange the equipment in this order:

1. A straight wave generator sits at one end of a shallow tank of water.
2. The generator produces parallel wavefronts at a constant frequency.
3. A barrier crosses the tank, leaving an adjustable opening of width \(a\).
4. The region beyond the barrier allows you to observe the transmitted wavefronts.
5. A strobe light illuminates the tank from above, making the moving wavefronts easier to observe.

Damping material along the tank boundaries helps reduce unwanted reflections returning into the observation region.

The important part is controlling the experiment.

Keep the following quantities constant:

• Water depth.
• Wave generator frequency.
• Incident wave amplitude.
• Barrier position and orientation.

Change only the gap width.

Why?

Wave speed depends on the properties of the medium. For the same water depth and wave frequency, the wavelength remains approximately constant.

Recall:

\[
v=f\lambda
\]

where:

• \(v\) is wave speed, in metres per second (m/s).
• \(f\) is frequency, in hertz (Hz).
• \(\lambda\) is wavelength, in metres (m).

If you change the water depth or generator frequency while narrowing the gap, you introduce another variable. You can no longer confidently attribute the change in spreading to the opening alone.

What should the experiment show?

Suppose the measured wavelength is \(2.0\text{ cm}\).

Compare three gap widths.

These are illustrative conditions, not recorded experimental measurements.

With the widest gap, the central portion of the wavefront continues almost straight ahead. Some spreading occurs near the edges.

As the gap narrows, the transmitted wavefront becomes increasingly curved. At a gap only twice the wavelength, the spreading is much more obvious.

The observation supports the idea that the opening’s width affects how the transmitted wavefront propagates.

However, the experiment alone does not explain exactly why. To understand that, we need to examine what happens to the different parts of the wave as they pass through the opening.

02Why the contributions from a narrow gap stay more in step

The opening changes the available wavefront

Imagine a straight water-wave crest approaching a barrier.

The barrier blocks the wave from travelling directly through the solid sections. Only the part of the wavefront that reaches the opening can contribute to the transmitted wave.

The Huygens-Fresnel model provides a useful way to understand what happens next.

It treats different points across the opening as sources of secondary wavelets. These wavelets spread into the region beyond the barrier and overlap.

This is a mathematical model of wave propagation. It does not mean that tiny independent generators physically appear across the opening.

Here is the important distinction.

A wide opening allows contributions from points spread across a large distance. A narrow opening restricts those contributions to a smaller distance.

That changes the path differences between the contributions reaching an observation point.

Path difference: how much further does one contribution travel?

Consider a point \(P\) beyond the opening.

If \(P\) lies directly along the central axis, contributions from across the opening travel approximately equal distances to reach it.

They arrive in phase.

Now move \(P\) away from the central axis.

The contributions from different positions across the opening travel slightly different distances. One part of the wave has further to travel than another.

The difference between these distances is called the path difference.

For a single opening of width \(a\), the path difference between contributions from opposite edges is approximately:

\[
\boxed{\delta=a\sin\theta}
\]

where:

• \(\delta\) is the path difference, in metres.
• \(a\) is the gap width, in metres.
• \(\theta\) is the angle from the central axis to the observation direction.

This expression assumes normally incident plane waves and an observation point sufficiently far away that the outgoing rays are approximately parallel.

Notice something important: wavelength does not appear in this equation.

At a fixed observation angle, the path difference depends on the opening’s width.

If you halve the gap width, you halve the path difference between contributions from its opposite edges at that angle.

Phase difference: where are the waves in their cycles?

Path difference and phase difference are related, but they are not the same quantity.

Path difference measures a distance in metres.

Phase difference describes how far apart two waves are in their oscillation cycles. It is measured in radians or degrees.

One complete wavelength of extra travel corresponds to one complete oscillation cycle, or \(360^\circ\).

Half a wavelength corresponds to half a cycle, or \(180^\circ\).

The relationship is:

\[
\boxed{\Delta\phi=\frac{2\pi\delta}{\lambda}}
\]

where:

• \(\Delta\phi\) is the magnitude of the phase difference, in radians.
• \(\delta\) is the path difference, in metres.
• \(\lambda\) is the wavelength, in metres.

For waves that begin in phase and travel through the same medium, this equation converts their path difference into the phase difference caused by their different travel distances.

Think of phase as a position around a circular dial. An extra full wavelength brings a wave back to the same position on the dial. An extra half-wavelength puts it on the opposite side.

The analogy helps with the repeating cycle, although actual waves involve continuous oscillations rather than objects moving around a dial.

The crucial connection: a narrower gap produces less phase variation

Now combine the two equations.

Since:

\[
\delta=a\sin\theta
\]

the phase difference between contributions from opposite edges is:

\[
\boxed{\Delta\phi=
\frac{2\pi a\sin\theta}{\lambda}}
\]

This equation explains the original question.

Imagine observing the wave at an angle of \(20^\circ\).

If the opening is wide, contributions from its opposite edges have a relatively large path difference. Their phases can be quite different.

If the opening is narrow, the path difference is smaller. Their phases are closer together.

As a result, contributions from the narrow opening reinforce one another over a wider range of directions.

The wider opening experiences stronger cancellation at smaller angles away from the central axis.

This is the central physical explanation:

Narrowing the gap reduces the phase variation across the opening at a given observation angle. Consequently, waves can reinforce further away from the central direction, producing greater angular spreading.

03How superposition produces the spreading pattern

Having established the phase difference, we can now explain what the overlapping waves actually do.

The principle of superposition states that when waves overlap, their instantaneous disturbances add.

There are two important possibilities.

Constructive interference

When contributions arrive in phase, their disturbances reinforce one another.

A crest meeting another crest produces a larger resultant crest.

Destructive interference

When equal-amplitude contributions arrive \(180^\circ\) out of phase, their disturbances cancel.

A crest meeting a trough produces zero resultant displacement at that instant.

For a single opening, there are contributions from many points across the gap, not just two.

This matters when determining where the first dark minimum occurs.

Why the first minimum occurs when the path difference is one wavelength

For a single slit of width \(a\), the first dark minimum occurs when:

\[
\boxed{a\sin\theta_1=\lambda}
\]

Here, \(\theta_1\) is the angle between the central axis and the first minimum.

There is a subtle point in this result.

You might expect destructive interference when the path difference between the opposite edges is half a wavelength.

That would be correct if you were comparing just two equal-amplitude waves.

But a slit contains contributions from across its entire width.

At the first minimum, divide the slit into two equal halves. Pair each point in the upper half with a corresponding point in the lower half.

The separation between the two members of each pair is \(a/2\).

When the path difference between the opposite edges is one wavelength, the path difference between each corresponding pair is:

\[
\delta_{\text{pair}}
=\frac{a}{2}\sin\theta_1
\]

Using the first-minimum condition:

\[
\delta_{\text{pair}}
=\frac{1}{2}\lambda
\]

Each pair therefore arrives \(180^\circ\) out of phase.

The contributions cancel pair by pair, producing the first dark minimum.

This is why the first minimum condition uses a full wavelength of path difference across the entire opening, rather than half a wavelength.

What happens as the gap gets narrower?

Rearrange the first-minimum equation:

\[
\sin\theta_1=\frac{\lambda}{a}
\]

For a fixed wavelength, decreasing \(a\) increases the value of \(\sin\theta_1\).

Therefore, the first minimum moves further away from the central axis.

The central region of constructive interference becomes wider.

Consider the following illustrative model predictions for a wavelength of \(2.0\text{ cm}\).

The central maximum extends approximately from \(-\theta_1\) to \(+\theta_1\).

The narrowest opening therefore produces the widest central maximum.

These estimates use the ideal single-slit, far-field model. Real ripple-tank observations can differ because of finite barrier thickness, reflections, and other boundary effects.

There is also a useful limiting case. If \(a\) is smaller than \(\lambda\), the equation cannot produce a first minimum at a physical forward angle. This does not mean the transmitted wave has equal intensity in every direction. It means the simple model predicts no first dark minimum within the forward angular range.

04Boundary effects are not the same as interference

There is one more distinction worth getting right.

The barrier itself affects the wave, but that effect is different from the path differences and superposition that determine the resulting pattern.

What about the reflected wave?

When an incoming wave reaches the barrier, some energy can be reflected.

The reflected wave travels back towards the wave generator. It can overlap with the incoming wave and produce regions of reinforcement and cancellation.

In some circumstances, this produces a standing-wave-like pattern upstream of the barrier.

That is a boundary and superposition effect, but it is not the same as the diffraction of the transmitted wave beyond the opening.

The phase change on reflection depends on the boundary conditions. You should not automatically assume that every reflection produces a \(180^\circ\) phase change.

For the diffraction question, concentrate on the wave contributions passing through the opening.

Does more spreading mean more energy gets through?

No.

A narrower gap can produce greater angular spreading while transmitting less total energy.

The barrier blocks a larger portion of the incoming wavefront. The transmitted wave can also distribute its energy over a wider range of directions.

Consequently, the wave amplitude at a particular observation point might decrease even though diffraction has increased.

This is an important experimental distinction:

• Angular spreading describes the range of directions occupied by the transmitted wave.
• Amplitude describes the size of the disturbance at a particular position.
• Transmitted energy describes how much energy passes through the opening.

They are related, but they are not interchangeable.

Also remember that narrowing the gap does not automatically change the wavelength. In the same medium, the frequency and wave speed determine the wavelength. The opening changes the geometry of the transmitted wavefront.

05Worked examples

06Questions and solutions

Question 1

Plane water waves have a wavelength of \(3.0\text{ cm}\). They approach a barrier normally.

The barrier has two possible opening widths: \(15\text{ cm}\) and \(6.0\text{ cm}\).

Calculate the wavelength-to-gap ratio for each opening and identify which produces stronger diffraction.

Solution 1

The \(6.0\text{ cm}\) opening produces stronger diffraction because its width is smaller relative to the wavelength.

The wavelength-to-gap ratio is:

\[
\frac{\lambda}{a}
\]

For the wider opening:

\[
\frac{\lambda}{a}
=\frac{3.0}{15}
\]

\[
\boxed{\frac{\lambda}{a}=0.20}
\]

For the narrower opening:

\[
\frac{\lambda}{a}
=\frac{3.0}{6.0}
\]

\[
\boxed{\frac{\lambda}{a}=0.50}
\]

The narrower opening has a larger wavelength-to-gap ratio.

This means the path differences across the opening produce a smaller phase variation at any fixed observation angle. Contributions therefore remain reinforced over a wider range of directions.

Remember that the absolute size of the opening is not enough to predict diffraction. You must compare it with the wavelength.

Question 2

A monochromatic laser with a wavelength of \(540\text{ nm}\) passes normally through a single slit.

A screen is positioned \(1.80\text{ m}\) from the slit. The first dark minima are measured \(6.5\text{ mm}\) above and below the central maximum.

(a) Calculate the width of the slit.

(b) The screen is moved to \(2.70\text{ m}\) from the slit. Calculate the new width of the central maximum, assuming the wavelength and slit width remain unchanged.

Solution 2

(a) The slit width is approximately \(0.150\text{ mm}\).

First calculate the diffraction angle using the screen geometry.

\[
\tan\theta=\frac{y}{L}
\]

Substitute:

\[
\tan\theta=
\frac{6.5\times10^{-3}}{1.80}
\]

\[
\tan\theta=0.00361
\]

For this small angle:

\[
\sin\theta\approx0.00361
\]

The first-minimum equation is:

\[
a\sin\theta=\lambda
\]

Rearrange:

\[
a=\frac{\lambda}{\sin\theta}
\]

Substitute:

\[
a=\frac{540\times10^{-9}}{0.00361}
\]

\[
a=1.50\times10^{-4}\text{ m}
\]

Therefore:

\[
\boxed{a=0.150\text{ mm}}
\]

(b) The new central maximum width is \(19.5\text{ mm}\).

The wavelength and slit width have not changed, so the diffraction angle remains constant.

The distance from the centre to the first minimum is proportional to the screen distance for small angles.

Therefore:

\[
\frac{y_2}{y_1}=\frac{L_2}{L_1}
\]

Substitute:

\[
\frac{y_2}{6.5}
=\frac{2.70}{1.80}
\]

\[
y_2=6.5\times1.5
\]

\[
y_2=9.75\text{ mm}
\]

The full central maximum width is:

\[
W=2y_2
\]

\[
W=2(9.75)
\]

\[
\boxed{W=19.5\text{ mm}}
\]

The important distinction is that moving the screen further away increases the measured width of the pattern, but it does not increase the angular diffraction.

Question 3

A student investigates single-slit diffraction and makes the following claim:

“The first dark minimum occurs when the path difference between the top and bottom edges of the slit is half a wavelength. The two edge contributions are then \(180^\circ\) out of phase and cancel.”

Explain why this reasoning is incorrect.

Derive the correct first-minimum condition using the principle of superposition.

Solution 3

The correct first-minimum condition is:

\[
\boxed{a\sin\theta_1=\lambda}
\]

The student’s mistake is treating the entire opening as though it contains only two wave contributions.

A single slit produces contributions from points distributed continuously across its width. Destructive interference at the first minimum involves all these contributions.

To see why, divide the opening into two equal halves.

Pair every point in the upper half with a corresponding point in the lower half.

The distance between the members of each pair is:

\[
\frac{a}{2}
\]

For complete cancellation, each pair must have a path difference of half a wavelength.

Therefore:

\[
\frac{a}{2}\sin\theta_1
=\frac{\lambda}{2}
\]

Multiply both sides by two:

\[
\boxed{a\sin\theta_1=\lambda}
\]

At this angle, every corresponding pair has a phase difference of \(180^\circ\), so their equal-amplitude contributions cancel.

The student’s proposed condition would give:

\[
a\sin\theta=\frac{\lambda}{2}
\]

At this point, the path difference between corresponding points in opposite halves would only be \(\lambda/4\).

Their phase difference would be \(90^\circ\), not \(180^\circ\).

Consequently, the contributions would not cancel pair by pair.

The key distinction is that two-source interference and single-slit diffraction involve different arrangements of wave contributions. You cannot automatically apply the two-source cancellation condition to the opposite edges of a continuous opening.

Question 4

A ripple tank produces straight water waves with a frequency of \(12\text{ Hz}\) and a measured wavelength of \(20\text{ mm}\).

The waves approach a barrier normally.

An experiment initially uses an opening \(80\text{ mm}\) wide. The opening is then narrowed to \(20\text{ mm}\).

After narrowing the opening, a student observes:

• Greater spreading of the transmitted waves.
• A lower wave amplitude at a fixed observation point behind the barrier.
• A reflected wave pattern on the incoming side of the barrier.

Answer the following:

(a) Calculate the wave speed.

(b) Use the ideal single-slit model to compare the first-minimum angles for the two opening widths.

(c) Explain why the transmitted amplitude might decrease even though diffraction increases.

(d) Explain why the reflected wave pattern should not be confused with the transmitted diffraction pattern.

Solution 4

(a) The wave speed is \(0.24\text{ m/s}\).

Use the wave equation:

\[
v=f\lambda
\]

Convert the wavelength:

\[
\lambda=20\text{ mm}=0.020\text{ m}
\]

Substitute:

\[
v=12\times0.020
\]

\[
\boxed{v=0.24\text{ m/s}}
\]

(b) The narrower opening produces much greater angular spreading.

The first-minimum equation is:

\[
a\sin\theta_1=\lambda
\]

For the \(80\text{ mm}\) opening:

\[
\sin\theta_1=\frac{20}{80}
\]

\[
\sin\theta_1=0.25
\]

Therefore:

\[
\boxed{\theta_1=14.5^\circ}
\]

For the \(20\text{ mm}\) opening:

\[
\sin\theta_1=\frac{20}{20}
\]

\[
\sin\theta_1=1
\]

Therefore:

\[
\boxed{\theta_1=90^\circ}
\]

In the ideal model, the first minimum has moved to the grazing direction. There is no first dark minimum within the forward angular region.

This indicates very strong spreading. It does not mean that the transmitted wave has equal amplitude in all directions.

(c) The transmitted amplitude can decrease because the narrower barrier opening allows less energy through.

The narrower opening blocks a greater portion of the incident wavefront.

Some incident energy can be reflected, and the transmitted energy can be distributed over a wider angular region.

Consequently, the amplitude measured at one fixed point may decrease.

This does not contradict increased diffraction. The angular distribution of the transmitted wave and the amplitude at a particular location describe different things.

(d) The reflected wave pattern is produced by the overlap of incoming and reflected waves.

The barrier reflects part of the incident wave back towards the generator.

These waves can superpose, producing reinforcement and cancellation on the incoming side.

The transmitted diffraction pattern forms on the opposite side of the barrier. It results from contributions passing through the opening and their subsequent superposition.

The reflected pattern is evidence of a boundary interaction, not a direct measurement of the transmitted diffraction.

You must consider these two regions separately when interpreting the experiment.

Question 5

A monochromatic plane wave of wavelength \(\lambda\) normally illuminates a single slit of width \(6\lambda\).

The apparatus is modified. The single slit is replaced by two identical, narrow slits, each with a width of \(3\lambda\). Their centres are separated by \(9\lambda\).

Both slits are illuminated equally and in phase.

The total open width is still \(6\lambda\), because:

\[
3\lambda+3\lambda=6\lambda
\]

A student argues:

“The total open width has not changed, so the diffraction pattern should remain the same.”

The screen is sufficiently far away for the Fraunhofer diffraction model to apply. Assume uniform illumination across each slit, coherent contributions, and negligible interaction between the two openings.

(a) Calculate the first diffraction-envelope minimum for the original single slit.

(b) Calculate the first diffraction-envelope minimum for the new two-slit arrangement.

(c) Calculate the angle of the first interference minimum and the first non-central interference maximum in the new arrangement.

(d) Determine which interference maximum is missing because it coincides with a diffraction-envelope minimum. Explain why this occurs.

Solution 5

The student’s conclusion is incorrect. The new arrangement produces a broader diffraction envelope containing additional interference fringes.

The reason is that the two-slit pattern depends on two different distances:

• The width of each individual slit determines the diffraction envelope.
• The separation between the slit centres determines the interference fringes.

Keeping the total open width constant does not keep either of these quantities unchanged.

(a) The original single slit has its first minimum at approximately \(9.6^\circ\).

For the original slit:

\[
a=6\lambda
\]

Use the first-minimum equation:

\[
a\sin\theta_1=\lambda
\]

Substitute:

\[
6\lambda\sin\theta_1=\lambda
\]

Cancel \(\lambda\):

\[
\sin\theta_1=\frac{1}{6}
\]

Therefore:

\[
\boxed{\theta_1=9.6^\circ}
\]

The original central diffraction maximum extends approximately from \(-9.6^\circ\) to \(+9.6^\circ\).

(b) The new diffraction envelope has its first minimum at approximately \(19.5^\circ\).

Each individual slit has width:

\[
b=3\lambda
\]

The first minimum of the individual-slit diffraction envelope occurs when:

\[
b\sin\theta_1=\lambda
\]

Substitute:

\[
3\lambda\sin\theta_1=\lambda
\]

Therefore:

\[
\sin\theta_1=\frac{1}{3}
\]

Hence:

\[
\boxed{\theta_1=19.5^\circ}
\]

The new diffraction envelope is twice as wide in angular terms as the original single-slit central maximum.

However, this does not mean that the screen displays one uninterrupted bright region.

The two slits also produce interference fringes.

(c) The first interference minimum occurs at approximately \(3.2^\circ\), and the first non-central maximum occurs at approximately \(6.4^\circ\).

The slit centres are separated by:

\[
d=9\lambda
\]

For two coherent sources that begin in phase, destructive interference occurs when:

\[
d\sin\theta=
\left(m+\frac12\right)\lambda
\]

where \(m=0,1,2,\ldots\).

For the first minimum, \(m=0\):

\[
9\lambda\sin\theta=\frac{\lambda}{2}
\]

Therefore:

\[
\sin\theta=\frac{1}{18}
\]

\[
\boxed{\theta=3.2^\circ}
\]

Now consider constructive interference.

Its condition is:

\[
d\sin\theta=m\lambda
\]

For the first non-central maximum, \(m=1\):

\[
9\lambda\sin\theta=\lambda
\]

Therefore:

\[
\sin\theta=\frac{1}{9}
\]

\[
\boxed{\theta=6.4^\circ}
\]

The screen therefore contains interference fringes within the broader diffraction envelope.

(d) The third interference maximum is missing because it coincides with the first diffraction-envelope minimum.

For the third interference maximum:

\[
m=3
\]

The constructive-interference condition gives:

\[
9\lambda\sin\theta=3\lambda
\]

Therefore:

\[
\sin\theta=\frac{1}{3}
\]

\[
\boxed{\theta=19.5^\circ}
\]

But this is exactly the angle of the first diffraction-envelope minimum found in part (b).

At that angle, the contributions from across each individual slit cancel.

Therefore, even though the separation between the two slit centres would normally produce constructive interference, each slit contributes zero resultant amplitude.

There is consequently no bright fringe at that position.

This is called a missing interference order.

There is another useful check. At the original single-slit first-minimum angle of \(9.6^\circ\), the new arrangement has an interference minimum because:

\[
d\sin\theta
=9\lambda\times\frac16
\]

\[
=1.5\lambda
\]

This is a destructive-interference condition.

The same angular position is therefore dark in both arrangements, but for different physical reasons.

The key lesson is that the total open width does not determine the entire diffraction pattern. Splitting one opening into two changes the spatial distribution of the transmitting regions and introduces an additional interference condition.

This is the bridge to understanding double-slit experiments: slit width controls the diffraction envelope, while slit separation controls the spacing of the interference fringes.