Interference vs Diffraction: Differences for HSC Physics

Put a laser through one narrow slit and you get a broad central bright band, with weaker bright bands on either side. Put the same laser through two narrow slits and you get a series of alternating bright and dark fringes. Both patterns involve waves spreading out and overlapping. So what is actually different?

**Diffraction describes the spreading of waves as they pass through an opening or around an obstacle. Interference describes what happens when waves superpose, producing reinforcement or cancellation.**

The important detail is that these are not completely separate physical mechanisms. Diffraction patterns are produced by interference between different parts of the same wavefront. The distinction is mainly about what causes the pattern and how we describe it.

For HSC Physics, keep three ideas separate:

- **Boundary effects:** an opening or obstacle changes the directions in which waves can spread.
- **Path and phase difference:** waves travelling along different paths can arrive at different points in their cycles.
- **Superposition:** when the waves overlap, their disturbances add to produce the observed pattern.

Understanding how these ideas connect is much more useful than memorising that diffraction means bending and interference means overlapping.

## 1. The difference: one describes spreading, the other describes superposition

Imagine water waves travelling towards a barrier with a narrow gap.

Before reaching the barrier, the wavefronts are approximately straight. After passing through the gap, the waves spread out into the region behind the barrier. This is diffraction.

[IMAGE: Aerial side-by-side scientific illustration of straight, evenly spaced water wavefronts approaching a solid barrier with a narrow opening, then emerging as curved semicircular wavefronts. Include the barrier, gap, incident direction, and spreading wavefronts. Show the physical geometry clearly.]

Now imagine two sets of water waves arriving at the same location. If a crest from one wave meets a crest from the other, the resulting disturbance is larger. If a crest meets a trough, the disturbances oppose each other.

This is interference, explained by the principle of superposition.

[IMAGE: Two-panel water-wave illustration. Left panel: two coherent circular sources produce overlapping wavefronts, with constructive interference where crest meets crest. Right panel: crest meets trough at a location of destructive interference. Include a resultant-wave representation beneath each.]

The distinction is easiest to remember in a comparison:

| Feature | Diffraction | Interference |
|---|---|---|
| What it describes | Spreading around an obstacle or through an opening | The combined effect of overlapping waves |
| Physical situation | Waves encounter a boundary or opening | Two or more wave contributions reach the same region |
| Main factor affecting the pattern | Wavelength relative to opening or obstacle size | Relative phase, path difference, and amplitude |
| Typical observation | Spreading and a diffraction intensity pattern | Alternating regions of reinforcement and cancellation |
| Does superposition occur? | Yes | Yes |

A useful starting rule is that diffraction is especially noticeable when the opening or obstacle is comparable in size to the wavelength.

For example, water waves passing through a gap roughly one wavelength wide spread out substantially. If the gap is many wavelengths wide, the waves mostly continue forwards, with spreading concentrated near the edges.

This is not a sharp boundary between diffraction and no diffraction. Diffraction occurs to some extent whenever waves encounter finite openings or edges. The wavelength-to-opening ratio determines how noticeable it is.

[IMAGE: Three side-by-side top-down water-wave diagrams showing the same wavelength approaching openings of different widths: a narrow gap comparable to one wavelength, a medium gap of several wavelengths, and a very wide gap. Show strong spreading, moderate spreading, and mostly forward propagation respectively. Keep wavefront spacing identical in all three.]

## 2. Why diffraction and interference are connected

Here is where the simple distinction needs another layer.

When light passes through a slit, it does not behave as though the opening is a tiny window that simply redirects the entire beam. Different parts of the wavefront across the opening contribute to the light arriving at the screen.

These contributions can reinforce or cancel one another.

The Huygens-Fresnel model helps explain this. It treats different points across an advancing wavefront as sources of secondary wavelets. The wavelets spread into the region beyond the opening and superpose.

This gives us a way to understand why diffraction produces bright and dark regions.

[IMAGE: Scientific illustration of a plane light wave approaching a single vertical slit. Show the incident straight wavefronts, the two edges of the slit, several secondary wavelets originating across the open width, and curved emerging wavefronts spreading beyond the barrier. Clearly distinguish the solid barrier from the open slit.]

Think about a point on a distant screen that is slightly off the central axis.

Light travelling through the top of the slit and light travelling through the bottom of the slit cover different distances to reach that point. Their path lengths are not equal.

The difference between those distances is the path difference, usually represented by \(\delta\).

That path difference produces a phase difference:

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

where:

- \(\Delta\phi\) is the phase difference, in radians;
- \(\delta\) is the path difference, in metres;
- \(\lambda\) is the wavelength, in metres.

One complete wavelength of extra travel corresponds to a phase difference of \(2\pi\) radians, or \(360^\circ\). Half a wavelength corresponds to \(\pi\) radians, or \(180^\circ\).

For waves that begin in phase:

- A path difference of \(0,\lambda,2\lambda,\ldots\) produces constructive interference.
- A path difference of \(\frac{\lambda}{2},\frac{3\lambda}{2},\ldots\) produces destructive interference for two equal-amplitude contributions.

But there is a catch with a single slit. There are contributions from across the entire opening, not just two waves travelling along two paths.

For the first dark minimum, the slit can be divided into pairs of corresponding points. Each pair is separated by half the slit width. At the first minimum, the path difference between the two members of each pair is half a wavelength, so their contributions cancel pair by pair.

This is why the first dark minimum occurs when the path difference between light from opposite edges of the slit is one wavelength.

[IMAGE: Annotated single-slit cancellation diagram at the first dark minimum. Show a vertical slit of width a divided into two equal halves. Draw corresponding points in the upper and lower halves, with rays travelling towards the same off-axis observation direction. Label the path difference between opposite edges as lambda, and the path difference within each corresponding pair as lambda/2. Show paired contributions cancelling, without depicting an incorrect two-ray-only model.]

The key insight is that diffraction is the spreading pattern, while interference between the wave contributions across the opening explains its bright and dark regions.

## 3. What the boundaries actually change

A boundary matters because it restricts which parts of the incoming wavefront can continue through an opening or around an obstacle.

Consider a plane wave approaching a single slit of width \(a\).

- The central part of the wavefront passes through the opening.
- The barrier blocks the wave outside the opening.
- The transmitted wave spreads into the space beyond the barrier.
- Contributions from different positions within the slit reach an observation point with different path lengths.

The geometry determines the path differences. The wavelength determines how those path differences translate into phase differences.

This separation is important.

At a fixed observation angle, the path difference across a slit of width \(a\) is:

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

Here, \(\theta\) is the angle between the central axis and the direction towards the observation point.

Notice that wavelength does not appear in this expression. If the slit width and observation angle stay fixed, the geometric path difference stays fixed.

However, the phase difference does depend on wavelength:

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

A shorter wavelength therefore produces a larger phase difference for the same path difference.

This is one reason shorter and longer wavelengths produce different diffraction patterns through the same opening.

### The single-slit diffraction condition

For a single slit, dark minima occur when:

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

where:

- \(a\) is the slit width, in metres;
- \(\theta\) is the angle to the \(m\)-th dark minimum;
- \(m=1,2,3,\ldots\) identifies the dark minimum;
- \(\lambda\) is the wavelength, in metres.

The central bright maximum lies between the first minima on either side.

For small angles, \(\sin\theta\approx\tan\theta\). If the screen is a distance \(L\) from the slit, the distance from the centre of the pattern to the first minimum is approximately:

\[
y_1\approx\frac{L\lambda}{a}
\]

Therefore, the central maximum has an approximate width:

\[
W\approx\frac{2L\lambda}{a}
\]

[IMAGE: Single-slit diffraction pattern on a distant screen, with the slit and screen shown in side view. On the screen, show a broad, bright central maximum and progressively weaker side maxima separated by dark minima. Annotate the slit width a, screen distance L, first-minimum angle theta, first-minimum position y1, and central maximum width W.]

### Worked example: Finding the width of a diffraction maximum

A monochromatic laser with wavelength \(600\text{ nm}\) passes through a slit of width \(0.20\text{ mm}\). A screen is positioned \(2.0\text{ m}\) behind the slit.

Calculate the approximate width of the central bright maximum.

**Step 1**: Identify the relevant quantities.

Convert the measurements into SI units:

\[
\lambda=600\times10^{-9}\text{ m}
\]

\[
a=0.20\times10^{-3}\text{ m}
\]

\[
L=2.0\text{ m}
\]

**Step 2**: Use the small-angle expression for central maximum width.

\[
W\approx\frac{2L\lambda}{a}
\]

**Step 3**: Substitute the values.

\[
W=\frac{2(2.0)(600\times10^{-9})}
{0.20\times10^{-3}}
\]

\[
W=0.012\text{ m}
\]

**Step 4**: State the result.

\[
\boxed{W=12\text{ mm}}
\]

The central bright maximum is approximately \(12\text{ mm}\) wide. This is the distance between the first dark minima on opposite sides of the centre.

The calculation assumes far-field diffraction and a sufficiently small angle for the approximation \(\sin\theta\approx\tan\theta\) to be valid.

04What changes when you use two slits?

Now consider a version of Young’s double-slit experiment.

A monochromatic laser shines towards an opaque barrier containing two narrow, parallel slits. A screen sits some distance behind the barrier. Both slits receive light from the same incoming beam.

Because the same laser illuminates both openings, the waves emerging from the slits have the same frequency and maintain a constant phase relationship. They are coherent sources.

At a point on the screen, light from the two slits has travelled slightly different distances.

For slits separated by distance \(d\), the path difference at an observation angle \(\theta\) is:

\[ \delta=d\sin\theta \]

Here, \(d\) is the centre-to-centre separation of the slits, not the width of either opening.

Assuming the sources are initially in phase, constructive interference occurs when:

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

Destructive interference occurs when:

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

where \(n=0,1,2,3,\ldots\).

The screen therefore displays alternating bright and dark fringes.

For a distant screen and small angles, the distance between adjacent bright fringes is approximately:

\[ \Delta y=\frac{L\lambda}{d} \]

where \(\Delta y\) is the fringe spacing and \(L\) is the distance from the slits to the screen.

Notice that this equation uses the separation between the slits.

But we haven’t finished. Each slit also has a finite width, so light diffracts as it passes through each opening.

The resulting pattern has two features:

1. Closely spaced interference fringes caused by light from the two slits overlapping.
2. A broader diffraction envelope caused by the finite width of each individual slit.

This is an important HSC distinction:

The separation \(d\) controls the spacing of the fine interference fringes. The width \(a\) controls the broad diffraction envelope.

05A controlled experiment: separating the two effects

How could you test the distinction without changing the wavelength or the screen arrangement?

Use a monochromatic laser, a slit mask, and a screen.

Place the laser on one side of the mask and the screen a fixed distance \(L\) behind it. The mask can be changed between a single-slit opening and a double-slit arrangement.

First arrangement: one slit

Start with a single slit of width \(a\).

The incoming wavefront reaches the opening. Different parts of the wavefront contribute to the transmitted wave, which spreads beyond the barrier.

The prediction is a broad central maximum with dark minima and weaker side maxima.

Measure the distance between the first dark minima. This gives the width of the central diffraction maximum.

Second arrangement: two slits

Replace the single opening with two slits. Keep the laser, wavelength, screen distance, and alignment unchanged. Make each slit the same width as the original single slit.

The prediction is now more detailed.

Each opening still produces diffraction, so the broad envelope remains. However, the light from the two openings also interferes, creating fine bright and dark fringes.

The expected screen pattern is therefore a combination of both effects.

What does the comparison establish?

The single-slit arrangement demonstrates diffraction without needing two separate slits.

The double-slit arrangement demonstrates how two coherent openings produce additional interference fringes, while the finite width of each opening still determines the diffraction envelope.

The experiment also shows why saying that diffraction and interference are completely separate phenomena is misleading.

Both patterns depend on superposition. What changes is the arrangement of the wave contributions and the feature being described.

There are practical limitations to this comparison:

• The laser should be sufficiently monochromatic.
• Both slits should receive light from the same coherent source.
• The screen should be sufficiently distant for the small-angle approximations to be useful.
• The slit widths and separation must be known accurately.
• The two-slit arrangement should be aligned so that both openings are illuminated.

These controls help ensure that a change in the observed pattern comes from the aperture arrangement rather than an unrelated change to the experiment.

06How to identify which effect a question is testing

In an HSC question, the words diffraction and interference are sometimes used close together. The best approach is to identify what physical feature the question asks you to explain or calculate.

A few common mistakes are worth avoiding.

Mistake 1: Assuming one slit means no interference.

A single slit contains contributions from many positions across its width. These contributions can arrive with different phases and cancel. That is how its dark minima form.

Mistake 2: Assuming two slits produce interference only.

Each slit has a finite width. Each one diffracts light, and the resulting waves interfere with each other. Both effects contribute to the final pattern.

Mistake 3: Treating path difference as phase difference.

Path difference is a distance measured in metres. Phase difference describes the relative positions of waves within their cycles, measured in radians or degrees.

For sources that begin in phase, the relationship is:

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

If the sources have an initial phase difference, you must include it before deciding whether the result is constructive or destructive.

Mistake 4: Assuming every dark fringe is a diffraction minimum.

A double-slit pattern has destructive interference fringes even where the diffraction envelope is bright. A dark fringe does not automatically mean the single-slit diffraction condition has been met.

Mistake 5: Thinking that a wider slit produces more spreading.

A wider opening generally produces a narrower diffraction pattern for the same wavelength. The relevant comparison is between slit width and wavelength, not simply how much light gets through.

For a more detailed explanation of the underlying wave behaviour, see Wave Superposition and Interference for HSC Physics.

If you want to explore how wavelength changes the width of a single-slit pattern, see How to Compare Diffraction at Two Wavelengths Through One Slit.

04Questions and solutions

Question 1

Monochromatic light of wavelength \(500\text{ nm}\) passes through a single slit of width \(0.25\text{ mm}\). A screen is positioned \(1.5\text{ m}\) behind the slit.

Calculate:

a) The distance from the centre of the diffraction pattern to the first dark minimum.

b) The width of the central bright maximum.

Assume small diffraction angles.

Solution 1

The first dark minima are \(3.0\text{ mm}\) from the centre, giving a central maximum width of \(6.0\text{ mm}\).

Step 1: Convert the measurements into SI units.

\[ \lambda=500\times10^{-9}\text{ m} \]

\[ a=0.25\times10^{-3}\text{ m} \]

\[ L=1.5\text{ m} \]

Step 2: Calculate the position of the first minimum.

For a single slit:

\[ y_1=\frac{L\lambda}{a} \]

Substitute:

\[ y_1=\frac{1.5(500\times10^{-9})} {0.25\times10^{-3}} \]

\[ \boxed{y_1=0.003\text{ m}=3.0\text{ mm}} \]

Step 3: Calculate the central maximum width.

The pattern is symmetrical about the centre, so the central maximum extends \(3.0\text{ mm}\) in each direction.

\[ W=2y_1 \]

\[ \boxed{W=6.0\text{ mm}} \]

The central maximum is twice as wide as the distance from its centre to either first minimum.

Question 2

Two identical slits, each \(0.10\text{ mm}\) wide, are separated by \(0.40\text{ mm}\) between their centres. They are illuminated by coherent, monochromatic light of wavelength \(600\text{ nm}\). The sources are initially in phase.

Two points, P and Q, are observed on a distant screen. Their observation angles satisfy:

\[ \sin\theta_P=1.5\times10^{-3} \]

\[ \sin\theta_Q=2.25\times10^{-3} \]

For each point:

a) Calculate the geometric path difference.

b) Calculate the phase difference.

c) Identify whether the two-slit interference is constructive or destructive.

d) Explain why Q is not a single-slit diffraction minimum.

Solution 2

P experiences constructive interference, while Q experiences destructive interference. Neither point is a single-slit diffraction minimum.

Step 1: Identify the relevant quantities.

\[ d=0.40\times10^{-3}\text{ m} \]

\[ \lambda=600\times10^{-9}\text{ m} \]

For two slits, the geometric path difference is:

\[ \delta=d\sin\theta \]

Step 2: Calculate the path difference at P.

\[ \delta_P=(0.40\times10^{-3}) (1.5\times10^{-3}) \]

\[ \delta_P=600\times10^{-9}\text{ m} \]

Therefore:

\[ \delta_P=\lambda \]

Step 3: Calculate the phase difference at P.

\[ \Delta\phi_P=\frac{2\pi\delta_P}{\lambda} \]

\[ \Delta\phi_P=2\pi \]

This corresponds to one complete cycle. The waves arrive in phase, producing constructive interference.

Step 4: Calculate the path difference at Q.

\[ \delta_Q=(0.40\times10^{-3}) (2.25\times10^{-3}) \]

\[ \delta_Q=900\times10^{-9}\text{ m} \]

Therefore:

\[ \delta_Q=1.5\lambda \]

Step 5: Calculate the phase difference at Q.

\[ \Delta\phi_Q= \frac{2\pi(1.5\lambda)}{\lambda} \]

\[ \Delta\phi_Q=3\pi \]

This is an odd multiple of \(\pi\), so the waves arrive in opposite phases.

The result is destructive interference.

Step 6: Check the single-slit diffraction condition.

The first single-slit minimum occurs when:

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

Rearranging:

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

\[ \sin\theta_1= \frac{600\times10^{-9}} {0.10\times10^{-3}} \]

\[ \sin\theta_1=0.006 \]

Both P and Q have smaller sine values than \(0.006\). They therefore lie inside the central diffraction envelope.

Q is dark because the contributions from the two slits destructively interfere, not because each individual slit has reached a diffraction minimum.

This is an important distinction: a dark interference fringe and a diffraction minimum can occur for entirely different reasons.

Question 3

A double-slit arrangement has the following properties:

• Individual slit width: \(0.20\text{ mm}\)
• Centre-to-centre slit separation: \(0.50\text{ mm}\)
• Wavelength: \(500\text{ nm}\)
• Screen distance: \(2.0\text{ m}\)

Calculate:

a) The spacing between adjacent bright interference fringes.

b) The distance from the central axis to the first diffraction minimum.

c) The number of bright interference maxima within the central diffraction envelope.

d) Determine whether the first diffraction minimum coincides with a bright or dark interference fringe.

Solution 3

There are five bright interference maxima within the central diffraction envelope. The first diffraction minimum coincides with a destructive interference fringe, not a bright maximum.

Step 1: Calculate the interference fringe spacing.

\[ \Delta y=\frac{L\lambda}{d} \]

Substitute:

\[ \Delta y= \frac{2.0(500\times10^{-9})} {0.50\times10^{-3}} \]

\[ \boxed{\Delta y=2.0\text{ mm}} \]

Adjacent bright fringes are separated by \(2.0\text{ mm}\).

Step 2: Calculate the first diffraction minimum.

\[ y_1=\frac{L\lambda}{a} \]

Substitute:

\[ y_1= \frac{2.0(500\times10^{-9})} {0.20\times10^{-3}} \]

\[ \boxed{y_1=5.0\text{ mm}} \]

The central envelope extends from \(-5.0\text{ mm}\) to \(+5.0\text{ mm}\).

Step 3: Determine the bright fringe positions.

Starting from the central maximum, the bright fringes occur at multiples of \(2.0\text{ mm}\):

\[ y_n=n\Delta y \]

The positions inside the central envelope are:

\[ -4,\ -2,\ 0,\ 2,\ 4\text{ mm} \]

There are therefore:

\[ \boxed{5\text{ bright maxima}} \]

The next bright fringes would occur at \(\pm6.0\text{ mm}\), outside the central envelope.

Step 4: Determine the interference condition at the first diffraction minimum.

At the first diffraction minimum:

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

Rearrange:

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

The path difference between the two slit centres is:

\[ \delta=d\sin\theta \]

Substituting the diffraction condition:

\[ \delta=\frac{d\lambda}{a} \]

\[ \delta=\frac{0.50}{0.20}\lambda \]

\[ \boxed{\delta=2.5\lambda} \]

A path difference of \(2.5\lambda\) corresponds to destructive interference.

Therefore, the first diffraction minimum coincides with an interference minimum.

There is no bright interference maximum to remove at this position. The two conditions happen to produce darkness at the same location.

Question 4

A student investigates a double-slit arrangement using monochromatic light of wavelength \(600\text{ nm}\).

The screen is \(2.4\text{ m}\) from the slits. The student measures:

Calculate:

a) The centre-to-centre separation of the slits.

b) The width of each slit.

c) Whether the fourth-order interference maximum is visible.

d) Predict what happens to the interference fringe spacing and diffraction envelope if the centre-to-centre separation is doubled while the individual slit widths remain unchanged.

Solution 4

The slit separation is \(0.40\text{ mm}\), and each slit is \(0.10\text{ mm}\) wide. The fourth-order bright fringe is missing. Doubling the slit separation halves the interference fringe spacing but leaves the diffraction envelope unchanged.

Step 1: Calculate the centre-to-centre slit separation.

The interference fringe spacing is:

\[ \Delta y=\frac{L\lambda}{d} \]

Rearrange:

\[ d=\frac{L\lambda}{\Delta y} \]

Substitute:

\[ d= \frac{2.4(600\times10^{-9})} {3.6\times10^{-3}} \]

\[ \boxed{d=4.0\times10^{-4}\text{ m}} \]

\[ \boxed{d=0.40\text{ mm}} \]

Step 2: Calculate the individual slit width.

The first diffraction minimum satisfies:

\[ y_1=\frac{L\lambda}{a} \]

Rearrange:

\[ a=\frac{L\lambda}{y_1} \]

Substitute:

\[ a= \frac{2.4(600\times10^{-9})} {14.4\times10^{-3}} \]

\[ \boxed{a=1.0\times10^{-4}\text{ m}} \]

\[ \boxed{a=0.10\text{ mm}} \]

Step 3: Determine whether the fourth-order maximum is visible.

The two-slit constructive interference condition is:

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

The first diffraction minimum occurs when:

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

Divide the two relationships:

\[ \frac{d}{a}=n \]

Substitute:

\[ n=\frac{0.40}{0.10}=4 \]

The fourth-order bright interference maximum coincides with the first diffraction minimum.

Therefore:

\[ \boxed{\text{The fourth-order maximum is missing.}} \]

The individual slit contributions cancel at that angle, so the two-slit pattern cannot produce a bright maximum there.

Step 4: Predict the effect of doubling the slit separation.

The new slit separation is:

\[ d’=0.80\text{ mm} \]

The new fringe spacing is:

\[ \Delta y’=\frac{L\lambda}{d’} \]

Because the separation has doubled:

\[ \Delta y’=\frac{3.6}{2} \]

\[ \boxed{\Delta y’=1.8\text{ mm}} \]

The interference fringes become twice as closely spaced.

However, the individual slit width has not changed.

The first diffraction minimum therefore remains at:

\[ \boxed{y_1=14.4\text{ mm}} \]

The broad diffraction envelope stays the same, but more closely spaced interference fringes appear within it.

This is exactly what we expect when the two slit dimensions are treated separately.

Question 5

Two identical slits each have width \(a\), and their centres are separated by \(d=4a\). They are illuminated by the same monochromatic, coherent light source. The waves reaching the two slits are initially in phase.

A distant screen displays the usual combination of fine interference fringes inside a broad single-slit diffraction envelope.

A student makes the following observations:

• The first diffraction minimum coincides with the fourth-order interference maximum.
• The second diffraction minimum coincides with the eighth-order interference maximum.
• Every fourth interference maximum is therefore missing.

The student then places a thin transparent plate over one slit. Assume the plate introduces an additional optical path of exactly \(\lambda/2\) for the light passing through that slit.

The plate does not change the slit width, amplitude, wavelength outside the plate, or diffraction envelope.

Answer the following:

a) Explain why every fourth interference maximum was originally missing.

b) Derive the new condition for constructive interference after the plate is introduced.

c) Predict what happens to the central bright maximum.

d) Does every fourth bright interference order remain missing? Explain why or why not.

e) Identify which parts of the overall pattern remain unchanged and which parts shift.

Solution 5

The diffraction envelope remains unchanged, but the interference fringes shift by half a fringe spacing. The central bright maximum becomes dark, and the original rule that every fourth bright order is missing no longer applies.

The reason is that the plate changes the relative phase of the two slit contributions without changing the individual slit diffraction patterns.

Step 1: Establish why the original missing orders occur.

Before the plate is introduced, the two slits are in phase.

The diffraction minima satisfy:

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

where \(p=1,2,3,\ldots\).

The interference maxima satisfy:

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

We are told:

\[ d=4a \]

At the first diffraction minimum:

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

Multiply both sides by four:

\[ d\sin\theta=4\lambda \]

This is exactly the condition for the fourth-order interference maximum.

At the second diffraction minimum:

\[ a\sin\theta=2\lambda \]

Therefore:

\[ d\sin\theta=8\lambda \]

This coincides with the eighth-order interference maximum.

The same relationship continues for higher orders.

Every fourth interference maximum coincides with a diffraction minimum, so those bright orders cannot be observed.

Step 2: Account for the extra optical path.

The transparent plate adds an extra path difference of:

\[ \frac{\lambda}{2} \]

This corresponds to an additional phase difference of:

\[ \Delta\phi= \frac{2\pi}{\lambda} \left(\frac{\lambda}{2}\right) \]

\[ \Delta\phi=\pi \]

The two contributions now have an additional phase difference of \(180^\circ\).

Taking the total path difference to be the geometric path difference plus the additional path through the plate:

\[ \delta_{\text{total}}= d\sin\theta+\frac{\lambda}{2} \]

Step 3: Derive the new constructive interference condition.

Constructive interference requires:

\[ \delta_{\text{total}}=n\lambda \]

Substitute:

\[ d\sin\theta+\frac{\lambda}{2}=n\lambda \]

Rearrange:

\[ \boxed{ d\sin\theta= \left(n-\frac12\right)\lambda } \]

The bright fringes now occur at half-integer multiples of the wavelength in geometric path difference.

They have shifted by half a fringe spacing compared with the original pattern.

Step 4: Determine what happens at the centre.

At the central axis:

\[ \theta=0 \]

Therefore:

\[ d\sin\theta=0 \]

The total path difference becomes:

\[ \delta_{\text{total}}=\frac{\lambda}{2} \]

This produces a phase difference of \(\pi\).

The two contributions cancel, producing destructive interference.

Therefore, the central bright maximum disappears and is replaced by a dark fringe.

This is a useful result because the central axis is still at the centre of the diffraction envelope, where each individual slit would produce its greatest amplitude.

The darkness is caused by interference between the two slits, not by a diffraction minimum.

Step 5: Reconsider the missing orders.

The diffraction minima still satisfy:

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

Since \(d=4a\):

\[ d\sin\theta=4p\lambda \]

These positions are unchanged because the plate has not altered the slit widths or the diffraction envelope.

However, the new constructive interference condition requires:

\[ d\sin\theta= \left(n-\frac12\right)\lambda \]

The diffraction minima occur at integer multiples of \(4\lambda\), while the new interference maxima occur at half-integer multiples of \(\lambda\).

They can no longer coincide.

Therefore, the original statement that every fourth bright interference order is missing is no longer correct.

The diffraction minima remain dark, but no bright interference maximum is positioned exactly at those minima.

Step 6: Separate what changes from what stays fixed.

The tempting prediction was that the missing-order pattern would remain identical because the diffraction envelope had not moved.

The mistake is assuming that the envelope alone determines which bright fringes are missing. A missing order requires a diffraction minimum to coincide with a constructive interference maximum.

The plate changes the relative phase, shifting the interference maxima while leaving the diffraction envelope fixed.

That is the deeper relationship between diffraction and interference: the boundary determines how the wave contributions spread, while their relative phases determine how those contributions combine.

The next useful step is applying the same reasoning to diffraction gratings. With many closely spaced slits, interference produces sharp principal maxima, while the finite width of each slit still influences the overall diffraction envelope.