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Rayleigh Criterion (Resolving Power of a Single Aperture)

Key idea: Use the Rayleigh criterion θ ≈ λ/b to solve resolving power questions for a single aperture (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain single-aperture diffraction and apply first-minimum and Rayleigh criteria.
  • Relate phase difference to separations in time and position.
  • Explain coherent two-source interference using phase and path difference.

1. Definitions (Must Know)

  • Diffraction limits how sharply an aperture can form an image.
  • Resolving power / resolution: the ability to distinguish two close point sources as separate.
  • Angular separation, θ (rad): the angle between the directions to the two sources (as seen by the observer/instrument).
  • Aperture width, b (m): characteristic width of a single slit/aperture.
  • Wavelength, λ (m): wavelength of the wave used (e.g. light).

2. Key Ideas (What Earns Marks)

  • Rayleigh criterion (single aperture of width b):
    • θₘᵢₙ ≈ λ/b
    • θₘᵢₙ is the smallest angular separation that is just resolvable.
  • Therefore:
    • smaller λ → better resolution
    • larger b → better resolution
  • Units check: λ/b is dimensionless (radians).

3. Detailed Explanations

A. What “just resolved” means

Two point sources form diffraction patterns. They are just resolved when the central maximum of one pattern coincides with the first minimum of the other. This gives the limiting angular separation.

For a single slit/aperture (syllabus form): θₘᵢₙ ≈ λ/b

4. Common Mistakes

  • Using θ in degrees instead of radians (use radians for λ/b unless you convert).
  • Confusing b (aperture width) with a (double-slit separation) or D (screen distance).
  • Forgetting that improving resolution means smaller θₘᵢₙ.

5. Exam Tips

  • If you are given a linear separation s at a distance L (far away), use the small-angle approximation:
    • θ ≈ s/L
    • then compare with θₘᵢₙ ≈ λ/b.
  • State the conclusion explicitly: “resolvable” if θ ≥ θₘᵢₙ.

6. Worked Examples

Modelled example 1

Smallest resolvable angle

Core

Problem

Green light of wavelength λ = 550 nm passes through an aperture of width b = 2.0 mm. Find θₘᵢₙ.
Study the worked solution

  1. Select the Rayleigh relation

    Method

    Use θₘᵢₙ ≈ λ/b.

    Reason

    The question asks for the diffraction-limited angular separation of one aperture.

    Working

    θₘᵢₙ ≈ λ/b

  2. Convert both lengths

    Method

    Use λ = 550 × 10⁻⁹ m and b = 2.0 × 10⁻³ m.

    Reason

    The ratio is dimensionless and is interpreted as an angle in radians.

    Working

    550 nm = 550 × 10⁻⁹ m; 2.0 mm = 2.0 × 10⁻³ m

  3. Calculate the limit

    Method

    θₘᵢₙ = 2.75 × 10⁻⁴ rad.

    Reason

    Sources separated by at least this angle meet the stated Rayleigh criterion.

    Working

    θₘᵢₙ = (550 × 10⁻⁹)/(2.0 × 10⁻³) = 2.75 × 10⁻⁴ rad

Guided practice 2

Required aperture width

About 4 min

Problem

Two stars have angular separation θ = 1.0 × 10⁻⁵ rad. What minimum aperture width is needed to resolve them using light of wavelength 500 nm?

Try this before viewing the solution

Unit: cm

Hints

Hint 1: use the just-resolved boundary
The smallest acceptable aperture occurs when the actual separation equals θₘᵢₙ.
View solution step by step

  1. Set the boundary condition

    Method

    At minimum aperture, θ = λ/b.

    Reason

    A still smaller aperture would make the Rayleigh limit exceed the stars’ separation.

    Working

    θ = θₘᵢₙ = λ/b

  2. Rearrange and calculate

    Method

    b = 5.0 × 10⁻² m.

    Reason

    Solve the limiting relation for aperture width.

    Working

    b = λ/θ = (500 × 10⁻⁹)/(1.0 × 10⁻⁵) = 5.0 × 10⁻² m

  3. State the inequality

    Method

    The aperture must be at least 5.0 cm wide.

    Reason

    Larger apertures reduce θₘᵢₙ and therefore also resolve the stars.

    Working

    b ≥ 5.0 cm

Common misconception 3

Resolving two objects with given separation and distance

Find and correct the mistake

Learner claim

Two headlights are 1.5 m apart and 3.0 km away. An observer uses a 3.0 mm aperture at 600 nm. A learner finds θ = 5.0 × 10⁻⁴ rad and θₘᵢₙ = 2.0 × 10⁻⁴ rad but says the lights are unresolved because the limit is smaller. Diagnose the conclusion.

Try this before viewing the solution

Correct conclusion

View solution step by step

  1. Find actual angular separation

    Method

    θ = 5.0 × 10⁻⁴ rad.

    Reason

    For a distant pair, θ ≈ s/L.

    Working

    θ = 1.5/3000 = 5.0 × 10⁻⁴ rad

  2. Find the diffraction limit

    Method

    θₘᵢₙ = 2.0 × 10⁻⁴ rad.

    Reason

    The Rayleigh limit is λ/b.

    Working

    θₘᵢₙ = (600 × 10⁻⁹)/(3.0 × 10⁻³) = 2.0 × 10⁻⁴ rad

  3. Apply the inequality in the correct direction

    Method

    The headlights are resolvable.

    Reason

    Their actual separation exceeds the smallest separation the aperture can resolve: θ > θₘᵢₙ.

    Working

    5.0 × 10⁻⁴ > 2.0 × 10⁻⁴

Examiner practice 4

Minimum separation at a given distance

4 marks

Examination question

An instrument has aperture width b = 5.0 mm and uses light of wavelength λ = 550 nm. Two objects are L = 2.0 km away. Estimate the minimum resolvable separation sₘᵢₙ. [4 marks]

Try this before viewing the solution

Unit: m

View solution step by step

  1. Use the Rayleigh limit

    1 mark

    Method

    θₘᵢₙ = λ/b.

    Reason

    This is the smallest just-resolved angular separation.

    Working

    θₘᵢₙ = λ/b

  2. Calculate the angular limit

    1 mark

    Method

    θₘᵢₙ = 1.10 × 10⁻⁴ rad.

    Reason

    Wavelength and aperture are converted to metres.

    Working

    θₘᵢₙ = (550 × 10⁻⁹)/(5.0 × 10⁻³) = 1.10 × 10⁻⁴ rad

  3. Connect angle and separation

    1 mark

    Method

    sₘᵢₙ ≈ θₘᵢₙL.

    Reason

    At small angles, θ ≈ s/L.

    Working

    θₘᵢₙ ≈ sₘᵢₙ/L ⇒ sₘᵢₙ ≈ θₘᵢₙL

  4. Calculate the linear limit

    1 mark

    Method

    sₘᵢₙ = 0.22 m.

    Reason

    2.0 km = 2000 m.

    Working

    sₘᵢₙ = (1.10 × 10⁻⁴)(2000) = 0.22 m

Challenge 5

Comparing two apertures

Minimal support

Independent transfer

Two telescopes use the same wavelength. Telescope A has aperture b_A = 40 mm and telescope B has aperture b_B = 20 mm. Find θ_(min,B)/θ_(min,A) and identify which telescope has better resolution.

Try this before viewing the solution

Better resolution

Hints

Hint 1: take the inverse aperture ratio
Because θₘᵢₙ = λ/b, the ratio of angular limits reverses the ratio of aperture widths.
View solution step by step

  1. Form the limit ratio

    Method

    θ_(min,B)/θ_(min,A) = b_A/b_B.

    Reason

    The common wavelength cancels and angular limit varies inversely with aperture.

    Working

    (λ/b_B)/(λ/b_A) = b_A/b_B

  2. Calculate

    Method

    Telescope B’s limiting angle is twice telescope A’s.

    Reason

    40/20 = 2.

    Working

    (θ_(min,B))/(θ_(min,A)) = 40/20 = 2

  3. Interpret resolution

    Method

    Telescope A has better resolution.

    Reason

    Its larger aperture produces the smaller minimum resolvable angle and therefore finer detail.

    Working

    θ_(min,A) = (1/2)θ_(min,B)

7. Mind Stretchers

Mind stretcher 1: Effect of changing wavelengthExtension

An instrument just resolves two sources using red light. What happens to the resolution if the same aperture is used with blue light?

Show Answer

Blue light has smaller λ, so θₘᵢₙ ≈ λ/b becomes smaller. Resolution improves (it can resolve smaller angular separations).

Mind stretcher 2: Infrared vs visible for resolutionExtension

Suppose you image the same scene with the same aperture size, but you switch from visible light to infrared (longer wavelength).

How does the resolution change?

Show Answer

Infrared has a larger wavelength, so θₘᵢₙ ≈ λ/b increases. Resolution becomes worse (it can only resolve larger angular separations).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027