Single-aperture diffraction and Rayleigh resolution
Key idea: H2 Physics lessons on progressive-wave models, standing waves, interference, diffraction and resolution.
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The core idea
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Big question: When does an aperture spread a wave enough to limit resolution?
Diffraction becomes prominent when aperture size is comparable with wavelength. A narrower aperture produces a wider central pattern. For the single aperture specified in this course, the Rayleigh criterion gives the minimum resolvable angle θ ≈ λ/b, where b is the aperture width: shorter wavelength or larger aperture improves angular resolution.
Compare aperture size with wavelength
Diffraction is the spreading of a wave after an aperture or around an obstacle. It is most noticeable when the aperture dimension is comparable with the wavelength. In a ripple tank a narrow gap spreads waves more than a wide gap; sound diffracts around doorways and corners because its wavelength is large enough for the effect to be noticeable.
For a single slit of width b, first minima satisfy b sinθ = λ. Here b is an aperture width, not the slit separation used in interference or the grating spacing d. A narrower aperture gives a wider angular spread for fixed wavelength.
Check your understanding: A slit width halves at fixed wavelength. What happens to the first-minimum angle?
sinθ doubles, so the diffraction pattern spreads more widely, provided a minimum is still physically possible.
Use the Rayleigh criterion as a resolution rule
Two point sources form overlapping diffraction patterns through a finite aperture. By the Rayleigh criterion they are just resolved when the central maximum of either image falls at the first minimum of the other.
For the single aperture in this course, the limiting angular separation is θmin ≈ λ/b for small angles, where b is the aperture width. A wider aperture or shorter wavelength gives a smaller limit and therefore better resolution. Resolution is about distinguishing sources, not making an image brighter.
Check your understanding: Which aperture resolves closer sources: width b or 2b at the same wavelength?
The 2b aperture; its Rayleigh limiting angle is half as large.
Key ideas to keep
- Diffraction occurs at every aperture; its visibility changes with size-to-wavelength ratio.
- Resolution improves when the minimum resolvable angle becomes smaller.
- Rayleigh's criterion concerns angular separation, not image brightness.
See the reasoning
Worked example
Find the smallest resolvable separation
Question: A telescope is modelled as a single aperture of width 0.12 m and observes light of wavelength 550 nm. Find its Rayleigh angular limit and the smallest separation distinguishable at 2.0 km.
Step 1: Use the specified aperture limit
Why: The course models the aperture with θmin ≈ λ/b.
Working: θmin = λ/b = (550×10⁻⁹)/0.12 = 4.58×10⁻⁶ rad.
Step 2: Apply small-angle geometry
Why: At a distant object, separation s ≈ Lθ.
Working: L = 2.0×10³ m.
Step 3: Calculate the linear limit
Why: This converts angular resolving power to object separation.
Working: s = (2.0×10³)(4.58×10⁻⁶) = 9.17×10⁻³ m.
Answer: The angular limit is 4.6 × 10⁻⁶ rad and the smallest separation is about 9.2 mm.
Check: Increasing b would reduce this limit, matching the expectation that a wider aperture resolves finer detail.
Another worked model
Question
Light of wavelength 600 nm passes through a 0.250 mm slit onto a screen 2.00 m away. Estimate central-maximum width. Also find the Rayleigh limit for a 50.0 mm aperture at 550 nm.
Check the worked solution
First minima have θ ≈ λ/b = 2.40 × 10⁻³ rad. Central width ≈ 2Dθ = 2(2.00)(2.40 × 10⁻³) = 9.60 mm. Rayleigh limit = 550 × 10⁻⁹/0.0500 = 1.10 × 10⁻⁵ rad.
Use a hint if needed
Practise with support
Try this
A 0.100 mm slit produces first minima at ±0.0050 rad. Estimate wavelength, then state how doubling slit width changes the first-minimum angle.
Hint: For small angles, bθ ≈ λ.
Check your answer
λ ≈ bθ = (0.100 × 10⁻³)(0.0050) = 5.0 × 10⁻⁷ m. Since θ ≈ λ/b, doubling b halves the first-minimum angle and narrows the central maximum.
Now work without the hint
Practise independently
Your turn
Explain why sound is heard around a doorway more readily than light is seen around it, then compare first-minimum and Rayleigh equations.
Check your answer
Sound wavelengths are often comparable to doorway dimensions, giving pronounced diffraction; visible wavelengths are far smaller. For a single slit, b sinθ = λ locates the first intensity minimum. The Rayleigh estimate θ ≈ λ/b gives the angular separation at which one diffraction maximum falls at the other's first minimum.
Avoid these traps
Common mistakes
Common mistake
Diffraction begins only when an aperture is narrower than the wavelength.
What is wrong with this reasoning?
Show better thinking
Diffraction always occurs; its spreading becomes pronounced when aperture size is comparable to wavelength.
Common mistake
The grating spacing and single-slit width are interchangeable.
What is wrong with this reasoning?
Show better thinking
A grating uses adjacent-slit separation in a sinθ = nλ. A single slit uses its width in b sinθ = λ for the first minimum.
Write for the examiner
Exam guidance
Use radians for small-angle resolution calculations and explain the physical trend after computing.
Exam-style practice [8 marks]
A single aperture of width b is made narrower. Explain the effects on diffraction and the ability to resolve two distant point sources. State the Rayleigh condition for just resolving them and distinguish resolution from image brightness.
Plan before you answer
- Apply the wavelength-to-aperture comparison.
- Use the specified Rayleigh relation.
- Treat resolution and collected power separately.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Reducing b increases the diffraction spread and the Rayleigh limit θmin ≈ λ/b, so two sources need a larger angular separation to be distinguished: resolution worsens. Two sources are just resolved when the central maximum of either diffraction pattern lies at the first minimum of the other. A narrower aperture may also reduce collected power and image brightness, but brightness and resolution are separate properties.
Come back in three days
Check what stayed with you
Recall question 1
When is diffraction especially noticeable?
Check the answer
When the aperture or obstacle size is comparable with the wavelength.
Recall question 2
State the single-aperture Rayleigh limit used in this course.
Check the answer
θmin ≈ λ/b.
Recall question 3
What does a smaller limiting angle mean?
Check the answer
Better resolution.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 10 states no explicit exclusions. Topic 11(j) does not require knowledge of spectrometer structure or use. Graph interpretations distinguish time traces at one position from spatial profiles at one instant, and all small-angle equations are used only with their stated geometry.
- GCE A-Level H2 PhysicsTopic 11(k) / Topic 11(l) / Topic 11(m) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027