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Single-Slit Diffraction (First Minimum)

Key idea: Use b sin θ = λ (and small-angle approximations) to solve single-slit diffraction questions, including the width of the central maximum (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: spreading of waves when they pass through an aperture or around an obstacle.
  • Single slit: an aperture of width b.
  • Minima: angles where intensity is (approximately) zero.
  • First minimum: the first dark fringe on either side of the central maximum.
  • Geometry symbols (typical exam setup):
    • b = slit width (m)
    • λ = wavelength (m)
    • D = slit-to-screen distance (m)
    • y = distance from central maximum to a minimum on the screen (m)
    • θ = angle to the minimum (rad or °)

2. Key Ideas (What Earns Marks)

  • Single-slit minima occur at:
    • b sin θ = mλ (m = 1,2,3,…)
  • For the first minimum (m = 1):
    • b sin θ = λ
  • For small angles (θ small, and D≫ b):
    • sin θ ≈ tan θ ≈ y/D
    • so the first-minimum position is:
      • y ≈ (λ D)/b
  • Width of the central maximum:
    • from -y to + y:
      • w ≈ 2y ≈ (2λ D)/b

Central maximum width vs slit width (example)

For fixed wavelength and screen distance, the central maximum width is inversely proportional to slit width.

Scroll across the graph to read all labels.

For fixed wavelength and screen distance, the central maximum width is inversely proportional to slit width.For fixed wavelength and screen distance, the central maximum width is inversely proportional to slit width.
Example with λ = 600 nm and D = 2.0 m: w ≈ 2λ D/b, so halving b doubles w.
Open full-size graph
View figure data
Values for Central maximum width vs slit width (example)
Slit width, b (mm)w ∝ 1/b
0.124
0.1516
0.212
0.38
0.54.8
12.4

3. Detailed Explanations

A. Why minima happen (idea-level explanation)

Different parts of the slit act like many wave sources. At certain angles, contributions from different parts of the slit arrive out of phase and cancel.

For the first minimum, the phase difference across the slit corresponds to one full wavelength, giving: b sin θ = λ

B. Using the small-angle approximation on a screen

If the screen is far away: tan θ ≈ y/D

and for small θ: sin θ ≈ tan θ

So: b(y/D) ≈ λ ⇒ y ≈ (λ D)/b

4. Common Mistakes

  • Using the double-slit formula x = λ D/a (wrong symbol: a is slit separation, not slit width).
  • Forgetting that m = 1 for the first minimum.
  • Using y/D when the question gives θ directly (then use b sin θ = λ).
  • Mixing mm and m (convert everything to SI).

5. Exam Tips

  • Write the correct starting equation first:
    • single slit: b sin θ = mλ
    • diffraction grating: d sin θ = nλ
  • If a diagram shows a screen distance D and a position y, explicitly state “small angle, so sin θ ≈ tan θ ≈ y/D”.

6. Worked Examples

Modelled example 1

Find wavelength from first minimum position

Core

Problem

A slit has width b = 0.20 mm and the screen is D = 2.0 m away. The first minimum is at y = 6.0 mm from the centre. Find λ.
Study the worked solution

  1. Connect angle to screen position

    Method

    For a distant screen, use sin θ ≈ y/D.

    Reason

    The first minimum is observed at a small angle.

    Working

    b sin θ = λ ⇒ λ ≈ by/D

  2. Convert the measured lengths

    Method

    Use y = 6.0 × 10⁻³ m and b = 0.20 × 10⁻³ m.

    Reason

    Consistent SI units are required with D = 2.0 m.

    Working

    6.0 mm = 6.0 × 10⁻³ m; 0.20 mm = 0.20 × 10⁻³ m

  3. Calculate wavelength

    Method

    λ = 6.0 × 10⁻⁷ m.

    Reason

    The measured y is the centre-to-first-minimum distance, not the full central width.

    Working

    λ ≈ yb/D = ((6.0 × 10⁻³)(0.20 × 10⁻³))/2.0 = 6.0 × 10⁻⁷ m

Guided practice 2

Find the central maximum width

About 4 min

Problem

Light of wavelength 550 nm passes through a slit of width 0.10 mm onto a screen D = 1.5 m away. Find the width of the central maximum.

Try this before viewing the solution

Unit: mm

Hints

Hint 1: include both sides
The central maximum extends from the first minimum on one side to the first minimum on the other: w = 2y.
View solution step by step

  1. Use the full-width relation

    Method

    w ≈ 2λ D/b.

    Reason

    λ D/b gives the distance from the centre to one first minimum only.

    Working

    w = 2y ≈ 2λ D/b

  2. Substitute in SI units

    Method

    w = 1.65 × 10⁻² m.

    Reason

    Convert nanometres and millimetres to metres.

    Working

    w = (2(550 × 10⁻⁹)(1.5))/(0.10 × 10⁻³) = 1.65 × 10⁻² m

  3. Convert to millimetres

    Method

    w = 16.5 mm.

    Reason

    The screen measurement is conveniently expressed in millimetres.

    Working

    1.65 × 10⁻² m = 16.5 mm

Common misconception 3

Using the angle form

Find and correct the mistake

Learner claim

Monochromatic light of wavelength 600 nm produces a first minimum at θ = 0.20°. A learner divides λ by 0.20 as though the degree value were sin θ. Diagnose the method and find the slit width b.

Try this before viewing the solution

Unit: mm

View solution step by step

  1. Keep the trigonometric function

    Method

    Use b = λ/sin θ.

    Reason

    The exact first-minimum relation is b sin θ = λ; a numerical degree measure is not itself a sine value.

    Working

    b sin θ = λ ⇒ b = λ/sin θ

  2. Evaluate the angle correctly

    Method

    sin 0.20° = 3.49 × 10⁻³.

    Reason

    The calculator must interpret the supplied angle in degrees.

    Working

    0.20° = 3.49 × 10⁻³ rad for this small angle.

  3. Calculate and convert

    Method

    b = 1.72 × 10⁻⁴ m = 0.172 mm.

    Reason

    Divide the SI wavelength by the dimensionless sine ratio.

    Working

    b = (600 × 10⁻⁹)/(3.49 × 10⁻³) = 1.72 × 10⁻⁴ m

Examiner practice 4

Find the first-minimum angle

3 marks

Examination question

Light of wavelength 500 nm passes through a single slit of width b = 0.15 mm. Find the angle to the first minimum in degrees. [3 marks]

Try this before viewing the solution

Unit: °

View solution step by step

  1. Select the first-minimum condition

    1 mark

    Method

    b sin θ = λ.

    Reason

    The first minimum has m = 1.

    Working

    b sin θ = mλ with m = 1

  2. Calculate the sine ratio

    1 mark

    Method

    sin θ = 3.33 × 10⁻³.

    Reason

    Both wavelength and slit width are converted to metres.

    Working

    sin θ = (500 × 10⁻⁹)/(0.15 × 10⁻³) = 3.33 × 10⁻³

  3. State the requested angle

    1 mark

    Method

    θ = 0.191°.

    Reason

    The inverse sine gives 3.33 × 10⁻³ rad, equivalent to 0.191°.

    Working

    θ = sin⁻¹ (3.33 × 10⁻³) = 0.191°

Challenge 5

Find wavelength from central maximum width

Minimal support

Independent transfer

A single slit produces a central maximum of width w = 12 mm on a screen D = 1.5 m away. The slit width is b = 0.20 mm. Estimate λ using the small-angle model.

Try this before viewing the solution

Unit: nm

Hints

Hint 1: distinguish width from half-width
Use w ≈ 2λ D/b, not y ≈ λ D/b.
View solution step by step

  1. Rearrange the central-width relation

    Method

    λ ≈ wb/(2D).

    Reason

    The measured w is twice the centre-to-first-minimum distance.

    Working

    w ≈ 2λ D/b ⇒ λ ≈ wb/(2D)

  2. Substitute consistent units

    Method

    λ = 8.0 × 10⁻⁷ m.

    Reason

    Convert both millimetre measurements to metres.

    Working

    λ = ((12 × 10⁻³)(0.20 × 10⁻³))/2(1.5) = 8.0 × 10⁻⁷ m

  3. Express the wavelength

    Method

    λ = 800 nm.

    Reason

    This is the same physical result in a convenient optical unit.

    Working

    8.0 × 10⁻⁷ m = 800 × 10⁻⁹ m

7. Mind Stretchers

Mind stretcher 1: How does changing slit width affect the pattern?Extension

Explain what happens to the central maximum width if the slit width b is halved.

Show Answer

Since w ≈ (2λ D)/b, halving b makes w double. The diffraction pattern spreads out.

Mind stretcher 2: Comparing first and second minima (small-angle)Extension

For a single slit, minima occur at b sin θ = mλ (m = 1,2,…).

For small angles, compare the angle of the second minimum (m = 2) with the first minimum (m = 1).

Show Answer

For small angles, sin θ ≈ θ, so: bθₘ ≈ mλ ⇒ θₘ ≈ mλ/b Therefore θ₂ ≈ 2θ₁ (second minimum is at about twice the angle of the first).

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