How to Compare Diffraction at Two Wavelengths Through One Slit

Put a 650 nm red laser and a 450 nm blue laser through the same \(0.10\ \mathrm{mm}\) slit, one at a time. Which colour produces the wider central bright band?

Red. With the opening and surrounding medium unchanged, the longer wavelength places the first dark minima at larger angles. This makes the central diffraction maximum wider.

The comparison is straightforward. Understanding why it happens requires you to separate three ideas that are easy to mix up: path difference, phase difference, and superposition. The boundaries of the opening also play an important role.

01The direct comparison: same opening, different wavelengths

Start with the two quantities:

• \(a\) is the width of the slit, measured in metres.
• \(\lambda\) is the wavelength of the light, measured in metres.

For a single slit, the first dark minima occur at angles satisfying

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

Here, \(\theta_1\) is the angle between the central axis and the first dark minimum on either side of the central maximum.

Rearranging gives

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

If the slit width stays constant, increasing the wavelength increases \(\sin\theta_1\), and therefore increases \(\theta_1\).

The central bright maximum extends from \(-\theta_1\) to \(+\theta_1\). Its angular width is consequently \(2\theta_1\).

This gives us the comparison:

For a screen a fixed distance \(L\) behind the slit, the central maximum has width

\[ W=2L\tan\theta_1 \]

where \(W\) is the distance between the two first dark minima.

For small diffraction angles, \(\sin\theta\approx\tan\theta\), so we can use

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

This approximation is particularly useful for visible light passing through slits much wider than the wavelength.

The key result: If the opening and screen distance are unchanged, the central maximum width is approximately proportional to wavelength.

But this relationship tells us what happens, not yet why. To understand that, we need to look inside the slit.

02Path difference: what does the geometry actually change?

Imagine light travelling through a narrow slit and reaching a point \(P\) on a distant screen.

The light passing through the top of the slit travels a slightly different distance to \(P\) than the light passing through the bottom.

That difference is called the path difference.

For a slit of width \(a\), the magnitude of the path difference between contributions from opposite edges is

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

where:

• \(\delta\) is the path difference, measured in metres;
• \(a\) is the slit width, measured in metres;
• \(\theta\) is the observation angle measured from the central axis.

Notice something important.

The path difference depends on the geometry of the opening and the direction in which we observe the light.

It does not directly depend on the wavelength.

For two different wavelengths travelling through the same slit, the path difference at a particular observation angle is therefore identical.

For example, if the slit width and observation angle are unchanged, red and blue light have the same geometric path difference.

So why do they produce different diffraction patterns?

Because path difference and phase difference are not the same thing.

03Phase difference: the missing part of the comparison

Phase describes the position of a wave within its repeating cycle.

Two waves can have:

• The same phase: their corresponding points in the cycle line up.
• Opposite phases: a crest of one aligns with a trough of the other.
• An intermediate phase difference: they are somewhere between these two situations.

The phase difference associated with a path difference \(\delta\) is

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

where:

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

Now return to our two colours.

At the same observation angle, both colours have the same path difference. However, the shorter wavelength produces a larger phase difference.

This follows directly from the equation: wavelength is in the denominator.

For example, suppose the geometric path difference is \(300\ \mathrm{nm}\).

For blue light with wavelength \(450\ \mathrm{nm}\),

\[ \Delta\phi_{\mathrm{blue}} = \frac{2\pi(300)}{450} = \frac{4\pi}{3} \]

For red light with wavelength \(650\ \mathrm{nm}\),

\[ \Delta\phi_{\mathrm{red}} = \frac{2\pi(300)}{650} = 0.923\pi \]

The blue light has accumulated a greater phase difference over the same path difference.

This is the important distinction:

• Path difference describes the difference in distances travelled.
• Phase difference describes how that distance difference translates into positions within the wave cycles.

A longer wavelength needs a greater path difference to accumulate the same phase change.

That is why its first diffraction minimum occurs at a larger angle.

04Superposition: why does the light actually become dark?

There is one more complication.

You might reasonably expect the first dark minimum to occur when the light from opposite edges of the slit is half a wavelength out of phase.

After all, a half-wavelength path difference produces a phase difference of \(\pi\) radians, which gives destructive interference.

But the single-slit diffraction condition says

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

That is a full wavelength of path difference across the opening, not half a wavelength.

So why does the first minimum occur there?

The answer is that we must consider contributions from the entire opening, not just its two edges.

Pairing contributions across the slit

At the first minimum, the path difference between the extreme edges is \(\lambda\).

Now divide the slit into two equal halves.

Consider two points at matching positions in opposite halves. They are separated by a distance \(a/2\).

Their path difference is

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

Using the first-minimum condition,

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

The two contributions therefore have a phase difference of

\[ \Delta\phi_{\mathrm{pair}} = \frac{2\pi(\lambda/2)}{\lambda} = \pi \]

They are exactly out of phase.

If the slit is uniformly illuminated, the contributions from these corresponding points have equal amplitudes and cancel.

Every point in one half can be paired with a point in the other half. The entire transmitted wave therefore cancels at this observation angle.

This is destructive interference.

This explains an important misconception.

The opposite edges do not need to cancel each other. In fact, at the first minimum, their contributions are in phase because their path difference is one complete wavelength.

It is the contributions from corresponding points across the two halves that cancel.

This is why considering only two rays from the edges gives an incomplete explanation of single-slit diffraction.

Where the boundaries of the opening matter

The slit boundaries determine which parts of the incoming wavefront can pass through.

Inside the opening, the wave contributes to the outgoing pattern. Outside it, the opaque screen blocks the light.

The finite width of the opening therefore determines the range of path differences available across the transmitted wavefront.

The boundaries are important, but they do not act like tiny mirrors that simply redirect light sideways. The diffraction pattern comes from the superposition of contributions across the whole opening.

If the opening becomes narrower, the maximum path difference across it becomes smaller at any fixed angle. A larger angle is then needed to reach the first-minimum condition for the same wavelength.

This is the physical reason that changing the slit boundaries changes the diffraction pattern.

For the higher-order dark minima, the corresponding condition is

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

where \(m=1,2,3,\ldots\) identifies the minimum.

The first minimum has \(m=1\). The same principle of cancellation across the aperture applies to the higher minima, with the opening divided into a greater number of smaller regions.

06Testing the comparison experimentally

How could you check that the longer wavelength really produces greater spreading?

A simple two-colour investigation can do this.

The apparatus and prediction

Arrange the equipment in a straight line:

1. Place a monochromatic laser source on the left.
2. Position a narrow, vertical slit directly in front of the beam.
3. Place a screen behind the slit, a fixed distance \(L\) away.
4. Observe the diffraction pattern on the screen.

Use a red laser and a blue laser separately. Both must pass through the same physical slit.

A simple ray model would predict that the two colours travel along the same geometrical paths through the same opening.

The wave model predicts something different: the red light should produce a greater angular spread because its wavelength is longer.

We can test these predictions by measuring the width of the central bright maximum.

Controlling the experiment

An initial comparison using two different slit cards might seem convenient. However, if the slit widths are not identical, we have changed two variables at once.

The result would not tell us whether the different pattern came from the wavelength or the opening width.

The improved design uses:

• The same physical slit for both colours.
• The same screen distance.
• The same alignment.
• One colour at a time.
• The same method for identifying the first dark minima.

The independent variable is wavelength. The dependent variable is the angular spread, or the central maximum width when the screen distance is fixed.

There is another important experimental detail.

If the screen is moved further away, the measured central maximum becomes wider even though the diffraction angle remains unchanged.

For that reason, comparing angles is more reliable than comparing screen widths when the screen distances differ.

Illustrative experimental data

Suppose the slit width is \(0.100\ \mathrm{mm}\), and the screen is \(2.00\ \mathrm{m}\) away.

An illustrative set of measurements gives:

These are illustrative values, not measurements from an actual experiment.

Source B produces a wider central maximum. We can use the measurements to estimate its wavelength relative to source A.

07A reliable method for comparing two wavelengths

When a question asks you to compare diffraction through the same opening, work through these decisions.

1. Check whether the opening is actually the same.

If the slit width \(a\) is unchanged, compare the wavelengths directly.

The longer wavelength produces the greater angular spread.

2. If the slit widths differ, compare the ratios.

Use

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

A larger ratio means a greater first-minimum angle, provided the single-slit model applies.

3. Check whether the screen distances are equal.

A wider pattern on a more distant screen does not necessarily mean greater angular diffraction.

Use

\[ \theta_1=\tan^{-1}\left(\frac{W}{2L}\right) \]

to compare the angles.

4. If the question asks about phase, calculate path difference first.

Use

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

Then calculate

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

Do not confuse the geometric path difference with the phase difference it produces.

5. If the question asks about a dark minimum, consider superposition across the whole opening.

The first minimum occurs when

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

The contributions from corresponding points across the two halves of the slit then cancel.

For a broader explanation of the relationship between wavelength and opening size, see Diffraction of Light: Slit Width, Wavelength and HSC Physics.

The same wavelength-to-opening comparison also applies to water waves and other types of waves. See Wave Diffraction: How Wavelength and Gap Size Control Spreading.

08Questions and solutions

Question 1

Monochromatic light of wavelengths \(480\ \mathrm{nm}\) and \(600\ \mathrm{nm}\) is sent separately through the same single slit, which has a width of \(0.120\ \mathrm{mm}\).

A screen is placed \(1.50\ \mathrm{m}\) behind the slit.

Calculate the first-minimum angle and central maximum width for each wavelength. State which wavelength produces greater diffraction.

Assume normal incidence and a far-field diffraction pattern.

Solution 1

The \(600\ \mathrm{nm}\) wavelength produces greater diffraction. Its central maximum is approximately \(15.0\ \mathrm{mm}\) wide, compared with \(12.0\ \mathrm{mm}\) for the \(480\ \mathrm{nm}\) wavelength.

Step 1: Convert the slit width.

\[ a=0.120\times10^{-3} =1.20\times10^{-4}\ \mathrm{m} \]

Step 2: Calculate the first-minimum angle for \(480\ \mathrm{nm}\).

\[ \sin\theta_1 = \frac{480\times10^{-9}}{1.20\times10^{-4}} \]

\[ \sin\theta_1=0.00400 \]

\[ \theta_1=\sin^{-1}(0.00400) =0.229^\circ \]

Step 3: Calculate its central maximum width.

\[ W=2L\tan\theta_1 \]

\[ W=2(1.50)\tan(0.229^\circ) \]

\[ \boxed{W=0.0120\ \mathrm{m}=12.0\ \mathrm{mm}} \]

Step 4: Repeat for \(600\ \mathrm{nm}\).

\[ \sin\theta_1 = \frac{600\times10^{-9}}{1.20\times10^{-4}} \]

\[ \sin\theta_1=0.00500 \]

\[ \theta_1=\sin^{-1}(0.00500) =0.286^\circ \]

Step 5: Calculate the central maximum width.

\[ W=2(1.50)\tan(0.286^\circ) \]

\[ \boxed{W=0.0150\ \mathrm{m}=15.0\ \mathrm{mm}} \]

The longer wavelength produces a larger first-minimum angle and a wider central maximum.

This is exactly what we expect when the slit width and screen distance remain constant.

Question 2

A student investigates diffraction using the same single slit, which has a width of \(0.0750\ \mathrm{mm}\).

Two separate trials are performed:

• Trial A: wavelength \(520\ \mathrm{nm}\), screen distance \(2.00\ \mathrm{m}\).
• Trial B: wavelength \(650\ \mathrm{nm}\), screen distance \(1.50\ \mathrm{m}\).

Calculate the first-minimum angle and central maximum width for each trial.

Then answer both questions:

(a) Which wavelength produces greater angular diffraction?

(b) Which trial produces the wider central maximum on its screen?

Solution 2

The \(650\ \mathrm{nm}\) wavelength produces greater angular diffraction, but the \(520\ \mathrm{nm}\) trial produces the wider central maximum on its screen.

The difference occurs because the screens are at different distances.

Step 1: Convert the slit width.

\[ a=0.0750\times10^{-3} =7.50\times10^{-5}\ \mathrm{m} \]

Step 2: Calculate the first-minimum angle for trial A.

\[ \sin\theta_A = \frac{520\times10^{-9}}{7.50\times10^{-5}} \]

\[ \sin\theta_A=0.006933 \]

\[ \theta_A=\sin^{-1}(0.006933) \]

\[ \boxed{\theta_A=0.397^\circ} \]

Step 3: Calculate the width on the screen for trial A.

\[ W_A=2L_A\tan\theta_A \]

\[ W_A=2(2.00)\tan(0.397^\circ) \]

\[ \boxed{W_A=27.7\ \mathrm{mm}} \]

Step 4: Calculate the first-minimum angle for trial B.

\[ \sin\theta_B = \frac{650\times10^{-9}}{7.50\times10^{-5}} \]

\[ \sin\theta_B=0.008667 \]

\[ \theta_B=\sin^{-1}(0.008667) \]

\[ \boxed{\theta_B=0.497^\circ} \]

Step 5: Calculate the width on the screen for trial B.

\[ W_B=2L_B\tan\theta_B \]

\[ W_B=2(1.50)\tan(0.497^\circ) \]

\[ \boxed{W_B=26.0\ \mathrm{mm}} \]

Step 6: Explain the apparent contradiction.

The red light has a larger diffraction angle, so it has greater angular spreading.

However, its screen is closer to the slit. This reduces the distance over which the pattern can spread out.

Consequently, the blue trial produces a slightly wider measured central maximum despite having a smaller diffraction angle.

This is why screen width and diffraction angle must not be treated as interchangeable measurements.

Question 3

A single slit has a width of \(0.150\ \mathrm{mm}\). Consider a point on a distant screen at an angle of \(0.100^\circ\) from the central axis.

Blue light has a wavelength of \(450\ \mathrm{nm}\), while red light has a wavelength of \(600\ \mathrm{nm}\).

(a) Calculate the path difference between contributions from opposite edges of the slit at this observation angle.

(b) Calculate the phase difference across the slit for each wavelength.

(c) Calculate the first-minimum angle for each wavelength.

(d) Explain why the blue light has a larger phase difference at the chosen observation angle, yet the red light produces a wider central maximum.

Solution 3

The blue light has the larger phase difference at the chosen observation angle, but it also reaches its first dark minimum at a smaller angle. Red light therefore produces the wider central maximum.

Step 1: Calculate the path difference.

Use

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

Convert the slit width:

\[ a=0.150\times10^{-3} =1.50\times10^{-4}\ \mathrm{m} \]

Substitute:

\[ \delta = 1.50\times10^{-4}\sin(0.100^\circ) \]

\[ \boxed{\delta=2.62\times10^{-7}\ \mathrm{m}} \]

This is approximately \(262\ \mathrm{nm}\).

Notice that this path difference is identical for both colours because the slit width and observation angle are unchanged.

Step 2: Calculate the phase difference for blue light.

\[ \Delta\phi_B = \frac{2\pi\delta}{\lambda_B} \]

\[ = \frac{2\pi(262\times10^{-9})} {450\times10^{-9}} \]

\[ \boxed{\Delta\phi_B\approx3.66\ \mathrm{rad}} \]

Step 3: Calculate the phase difference for red light.

\[ \Delta\phi_R = \frac{2\pi(262\times10^{-9})} {600\times10^{-9}} \]

\[ \boxed{\Delta\phi_R\approx2.74\ \mathrm{rad}} \]

The blue light has the greater phase difference at this particular observation angle.

Step 4: Calculate the first-minimum angle for each colour.

For blue light:

\[ \sin\theta_{1,B} = \frac{450\times10^{-9}}{1.50\times10^{-4}} \]

\[ \theta_{1,B} = \sin^{-1}(0.00300) \]

\[ \boxed{\theta_{1,B}=0.172^\circ} \]

For red light:

\[ \sin\theta_{1,R} = \frac{600\times10^{-9}}{1.50\times10^{-4}} \]

\[ \theta_{1,R} = \sin^{-1}(0.00400) \]

\[ \boxed{\theta_{1,R}=0.229^\circ} \]

Step 5: Explain the difference.

The phase difference at a chosen observation point tells us how contributions from different parts of the slit are related at that particular angle.

The diffraction spread is determined by the angles at which destructive interference occurs.

Blue light accumulates phase difference more quickly as the observation angle increases. It therefore reaches its first minimum at a smaller angle.

Red light needs a greater path difference to reach the same phase condition. Its first minimum occurs further from the central axis.

There is no contradiction: the blue light has the larger phase difference at the same angle, but the red light has the wider central diffraction maximum.

Question 4

A student compares two diffraction patterns using different slit widths.

The first experiment uses blue light with wavelength \(450\ \mathrm{nm}\) and a slit width of \(0.0800\ \mathrm{mm}\).

The second experiment uses red light with wavelength \(600\ \mathrm{nm}\) and a slit width of \(0.120\ \mathrm{mm}\).

Both screens are placed \(2.00\ \mathrm{m}\) behind their respective slits.

(a) Calculate the first-minimum angle for each experiment.

(b) Calculate the central maximum width on each screen.

(c) The student concludes that red light must always diffract more because it has a longer wavelength. Evaluate this conclusion.

Solution 4

Blue light produces the greater angular diffraction in this particular comparison, despite having the shorter wavelength.

The reason is that the blue light passes through the narrower opening.

Step 1: Calculate the first-minimum angle for blue light.

Convert the slit width:

\[ a_B=0.0800\times10^{-3} =8.00\times10^{-5}\ \mathrm{m} \]

Use

\[ \sin\theta_{1,B} = \frac{450\times10^{-9}}{8.00\times10^{-5}} \]

\[ \sin\theta_{1,B}=0.005625 \]

Therefore,

\[ \boxed{\theta_{1,B}=0.322^\circ} \]

Step 2: Calculate the blue central maximum width.

\[ W_B=2L\tan\theta_{1,B} \]

\[ W_B=2(2.00)\tan(0.322^\circ) \]

\[ \boxed{W_B=22.5\ \mathrm{mm}} \]

Step 3: Calculate the first-minimum angle for red light.

Convert the slit width:

\[ a_R=0.120\times10^{-3} =1.20\times10^{-4}\ \mathrm{m} \]

\[ \sin\theta_{1,R} = \frac{600\times10^{-9}}{1.20\times10^{-4}} \]

\[ \sin\theta_{1,R}=0.00500 \]

Therefore,

\[ \boxed{\theta_{1,R}=0.286^\circ} \]

Step 4: Calculate the red central maximum width.

\[ W_R=2(2.00)\tan(0.286^\circ) \]

\[ \boxed{W_R=20.0\ \mathrm{mm}} \]

Step 5: Evaluate the student’s conclusion.

The student has ignored the effect of changing the slit width.

The relevant comparison is the ratio \(\lambda/a\), not wavelength alone.

For blue light,

\[ \frac{\lambda_B}{a_B} = 0.005625 \]

For red light,

\[ \frac{\lambda_R}{a_R} = 0.00500 \]

The blue light has the larger wavelength-to-opening ratio, so it produces the greater angular spread.

The experiment cannot establish that blue light intrinsically diffracts more than red light. To isolate the effect of wavelength, both colours must pass through the same opening.

Question 5

Two monochromatic light sources illuminate the same single slit simultaneously.

The red light has a wavelength of \(600\ \mathrm{nm}\), and the blue light has a wavelength of \(450\ \mathrm{nm}\).

The slit width is \(0.120\ \mathrm{mm}\), and a screen is placed \(2.00\ \mathrm{m}\) behind it.

Assume:

• Both wavelengths illuminate the slit uniformly.
• The light is incident normally.
• The diffraction pattern is in the far field.
• The detector measures time-averaged intensity.
• The two colours do not maintain a fixed relative phase.

A student argues:

“Because the wavelengths are different, their destructive interference conditions must also be different. Therefore, there cannot be any off-axis position where the combined light intensity is exactly zero.”

Evaluate this claim.

(a) Calculate the first-minimum angle for each wavelength.

(b) Determine whether a non-central position can exist where both wavelengths have a dark minimum simultaneously.

(c) If such a position exists, find the smallest positive angle at which it occurs, identify the order of the minimum for each colour, and calculate its distance from the centre of the screen.

(d) Explain why a minimum for just one wavelength does not necessarily produce a dark position in the combined pattern.

Solution 5

The student’s claim is incorrect.

Although the first minima occur at different angles, a higher-order minimum of the red pattern can coincide with a higher-order minimum of the blue pattern.

In this case, the first shared off-axis dark position occurs at approximately \(0.859^\circ\), corresponding to the third red minimum and fourth blue minimum. It is approximately \(30.0\ \mathrm{mm}\) from the centre of the screen.

Step 1: Calculate the first minimum for red light.

The slit width is

\[ a=0.120\times10^{-3} =1.20\times10^{-4}\ \mathrm{m} \]

For red light,

\[ \sin\theta_{1,R} = \frac{600\times10^{-9}}{1.20\times10^{-4}} \]

\[ \sin\theta_{1,R}=0.00500 \]

Therefore,

\[ \boxed{\theta_{1,R}=0.286^\circ} \]

Step 2: Calculate the first minimum for blue light.

\[ \sin\theta_{1,B} = \frac{450\times10^{-9}}{1.20\times10^{-4}} \]

\[ \sin\theta_{1,B}=0.00375 \]

Therefore,

\[ \boxed{\theta_{1,B}=0.215^\circ} \]

The first minima are clearly at different angles.

However, this does not mean that higher-order minima cannot coincide.

Step 3: Find the condition for a shared minimum.

For a single slit, the dark minima occur when

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

For red light, write

\[ a\sin\theta=m_R\lambda_R \]

For blue light,

\[ a\sin\theta=m_B\lambda_B \]

At a shared minimum, both expressions must be equal:

\[ m_R\lambda_R=m_B\lambda_B \]

Substitute the wavelengths:

\[ 600m_R=450m_B \]

Divide by \(150\):

\[ 4m_R=3m_B \]

The smallest positive whole-number solution is

\[ m_R=3,\qquad m_B=4 \]

Therefore, the third red minimum coincides with the fourth blue minimum.

This is the first possible shared minimum away from the centre.

Step 4: Calculate the shared minimum angle.

The common path-difference condition is

\[ a\sin\theta=3\lambda_R \]

Substitute:

\[ 1.20\times10^{-4}\sin\theta = 3(600\times10^{-9}) \]

Rearranging,

\[ \sin\theta = \frac{1.80\times10^{-6}} {1.20\times10^{-4}} \]

\[ \sin\theta=0.0150 \]

Therefore,

\[ \boxed{\theta=0.859^\circ} \]

The same result is obtained using the fourth blue minimum.

Step 5: Find the position on the screen.

The distance from the central axis is

\[ y=L\tan\theta \]

Substitute:

\[ y=2.00\tan(0.859^\circ) \]

\[ y=0.0300\ \mathrm{m} \]

Therefore,

\[ \boxed{y=30.0\ \mathrm{mm}} \]

There is a shared dark position \(30.0\ \mathrm{mm}\) above the centre, and another \(30.0\ \mathrm{mm}\) below it.

Step 6: Explain why the combined pattern can be dark.

This is the most important conceptual part.

The red and blue light have different frequencies. Their relative phase therefore changes continuously over time.

For an ordinary detector measuring time-averaged intensity, the two colours contribute separate intensities rather than maintaining a fixed phase relationship that would allow one colour to cancel the other.

Consequently:

• If red light is at a minimum but blue light is not, blue light still reaches the screen.
• If blue light is at a minimum but red light is not, red light still reaches the screen.
• If both wavelengths are at minima simultaneously, both contributions vanish in the ideal single-slit model.

At the shared minimum, each wavelength undergoes its own destructive interference across the opening.

There is another interesting detail.

At this position, the edge-to-edge path difference is \(1800\ \mathrm{nm}\).

For red light, this equals three complete wavelengths. For blue light, it equals four complete wavelengths.

The edge contributions are therefore in phase for both colours. However, the complete aperture still produces cancellation because the contributions can be paired across smaller regions of the opening.

For the third red minimum, the slit can be divided into six equal regions and adjacent regions paired for cancellation. For the fourth blue minimum, it can be divided into eight equal regions and paired in the same way.

The pairings are different, but both wavelengths produce zero resultant amplitude at the same observation angle.

This is why the student’s assumption fails: different wavelengths have different first-minimum positions, but their higher-order minima can still coincide.

The deeper lesson is that diffraction is not determined by the behaviour of just two rays or by the wavelength alone. The pattern depends on the phase relationships and superposition of contributions across the entire opening.

That same reasoning becomes useful when studying multiple-slit interference and diffraction gratings, where the positions of bright and dark fringes depend on how contributions from different parts of an aperture or different openings combine.