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Diffraction Grating

Key idea: Use the diffraction grating equation d sinθ = nλ, convert line density to grating spacing, and find maximum order (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use diffraction gratings to analyse principal maxima and determine wavelength.

1. Definitions (Must Know)

A. Diffraction grating

A diffraction grating is a large number of equally spaced parallel slits (or lines) that produces sharp interference maxima.

B. Grating spacing, d (m)

d is the distance between adjacent slits/lines (sometimes called the grating spacing).

If a grating has N lines per metre, then: d = 1/N

C. Order, n

n = 0,1,2,… labels the principal maxima:

  • n = 0 is the central maximum
  • n = 1 is the first-order maximum, etc.

D. Grating equation

For principal maxima: d sin θ = nλ

2. Key Ideas (What Earns Marks)

  • Convert line density to spacing correctly:
    • if the grating is L lines per mm, then N = 1000L lines per m, so d = 1/N.
  • Use the grating equation: d sin θ = nλ.
  • Maximum possible order is limited by sin θ ≤ 1: nₘₐₓ = ⌊ d/λ ⌋
  • A grating gives sharper maxima than a double slit because it has many slits.
Don’t confuse symbols
  • Grating: d = grating spacing.
  • Double slit: a = slit separation; x ≈ λ D/a.
  • Single slit: b = slit width; minima at b sin θ = mλ.

3. Detailed Explanations

A. Why the grating equation works (path difference idea)

For waves from adjacent slits at angle θ, the path difference is d sin θ.

Constructive interference occurs when that path difference is an integer multiple of the wavelength: d sin θ = nλ

Grating maxima: sinθ vs order n (example)

For a fixed grating spacing and wavelength, sinθ increases linearly with order until no more orders are possible.

Scroll across the graph to read all labels.

For a fixed grating spacing and wavelength, sinθ increases linearly with order until no more orders are possible.For a fixed grating spacing and wavelength, sinθ increases linearly with order until no more orders are possible.
Example with λ = 600 nm and d = 2.00 × 10⁻⁶ m (500 lines/mm), so λ/d = 0.30. The next order would need sinθ = 1.20, so it is not possible.
Open full-size graph
View figure data
Values for Grating maxima: sinθ vs order n (example)
Order, n (unitless)sinθ = nλ/d
00
10.3
20.6
30.9

B. Why a grating gives sharp maxima

With many slits, the condition d sin θ = nλ makes waves from many slits line up in phase, giving a strong maximum.

At angles slightly away from that condition, waves from different slits cancel more strongly, so the maxima are narrow and well-defined.

4. Common Mistakes

  • Using degrees/radians inconsistently on the calculator (angles are usually in degrees in optics questions).
  • Using the wrong spacing (confusing “lines per mm” with “lines per m”).
  • Forgetting to check if an order is possible (must satisfy sin θ ≤ 1).
  • Mixing up d (grating spacing) with D (screen distance).

5. Exam Tips

  • Always do a feasibility check: compute sin θ = nλ/d and verify it is ≤ 1.
  • If the question gives θ for a specific order, solve for λ directly: λ = (d sin θ)/n
  • If the grating is given as “L lines/mm”, convert first, then proceed.

6. Worked Examples

Modelled example 1

Find wavelength from a first-order maximum

Core

Problem

A grating has 500 lines per mm. Its first-order maximum occurs at θ = 19.5°. Find λ.
Study the worked solution

  1. Convert line density

    Method

    The grating has 5.00 × 10⁵ lines per metre.

    Reason

    There are 1000 mm in one metre.

    Working

    N = 500 × 1000 = 5.00 × 10⁵ m⁻¹

  2. Find the grating spacing

    Method

    d = 2.00 × 10⁻⁶ m.

    Reason

    Line density is the reciprocal of the distance between adjacent lines.

    Working

    d = 1/N = 1/(5.00 × 10⁵) = 2.00 × 10⁻⁶ m

  3. Apply the first-order condition

    Method

    λ ≈ 668 nm.

    Reason

    For n = 1, the grating equation gives λ = d sin θ.

    Working

    λ = (2.00 × 10⁻⁶) sin 19.5° = 6.68 × 10⁻⁷ m = 668 nm

Guided practice 2

Find the angle for a given order

About 4 min

Problem

Light of wavelength 532 nm is incident on a grating with d = 1.67 × 10⁻⁶ m. Find the angle for the second-order maximum.

Try this before viewing the solution

Unit: °

Hints

Hint 1: find sine before angle
Calculate sin θ = 2λ/d and verify it does not exceed one.
View solution step by step

  1. Calculate the sine ratio

    Method

    sin θ = 0.637.

    Reason

    The second-order condition uses n = 2.

    Working

    sin θ = (2(532 × 10⁻⁹))/(1.67 × 10⁻⁶) = 0.637

  2. Recover the angle

    Method

    θ ≈ 39.6°.

    Reason

    The maximum angle is the inverse sine of the feasible ratio.

    Working

    θ = sin⁻¹ (0.637) ≈ 39.6°

Common misconception 3

Maximum possible order

Find and correct the mistake

Learner claim

Light of wavelength 600 nm is incident on a grating with 600 lines per mm. A learner calculates d/λ = 2.78 and rounds to claim that order 3 is visible. Diagnose the claim and find the maximum order.

Try this before viewing the solution

View solution step by step

  1. Find the spacing

    Method

    d = 1.67 × 10⁻⁶ m.

    Reason

    Convert 600 lines per millimetre to 6.00 × 10⁵ m⁻¹ before taking the reciprocal.

    Working

    d = 1/(600 × 1000) = 1.67 × 10⁻⁶ m

  2. Apply the feasibility limit

    Method

    n ≤ 2.78.

    Reason

    Since sin θ = nλ/d ≤ 1, the order cannot exceed d/λ.

    Working

    d/λ = (1.67 × 10⁻⁶)/(600 × 10⁻⁹) = 2.78

  3. Choose the largest allowed integer

    Method

    The maximum observable order is nₘₐₓ = 2.

    Reason

    Order 3 would require sin θ > 1; ordinary rounding is not valid.

    Working

    nₘₐₓ = ⌊2.78⌋ = 2

Examiner practice 4

Find line density from a maximum angle

4 marks

Examination question

A grating produces a first-order maximum at θ = 30° for light of wavelength λ = 500 nm. Find the line density in lines per mm. [4 marks]

Try this before viewing the solution

Unit: lines/mm

View solution step by step

  1. Use the first-order equation

    1 mark

    Method

    d = λ/sin θ.

    Reason

    The stated maximum has n = 1.

    Working

    d = nλ/sin θ

  2. Calculate spacing

    1 mark

    Method

    d = 1.00 × 10⁻⁶ m.

    Reason

    sin 30° = 0.50 and wavelength is converted to metres.

    Working

    d = (500 × 10⁻⁹)/0.50 = 1.00 × 10⁻⁶ m

  3. Take the reciprocal

    1 mark

    Method

    N = 1.00 × 10⁶ m⁻¹.

    Reason

    Line density is reciprocal spacing.

    Working

    N = 1/d

  4. Convert the density

    1 mark

    Method

    The grating has 1000 lines per mm.

    Reason

    One metre contains 1000 millimetres.

    Working

    N = (1.00 × 10⁶)/1000 = 1000 lines/mm

Challenge 5

Check if a higher order is possible

Minimal support

Independent transfer

Light of wavelength 600 nm is incident on a grating with 500 lines per mm. Determine whether the third order is possible and, if it is, find θ.

Try this before viewing the solution

Is third order possible?
Unit: °

Hints

Hint 1: test the sine ratio
Find d = 1/N, then calculate 3λ/d. A value no greater than one represents a possible angle.
View solution step by step

  1. Convert density to spacing

    Method

    d = 2.00 × 10⁻⁶ m.

    Reason

    500 lines per millimetre is 5.00 × 10⁵ lines per metre.

    Working

    d = 1/(5.00 × 10⁵) = 2.00 × 10⁻⁶ m

  2. Test feasibility

    Method

    The third order is possible because sin θ = 0.900.

    Reason

    A sine ratio between zero and one corresponds to a physical angle.

    Working

    sin θ = (3(600 × 10⁻⁹))/(2.00 × 10⁻⁶) = 0.900

  3. Calculate the angle

    Method

    θ ≈ 64.2°.

    Reason

    Take the inverse sine only after establishing that the order exists.

    Working

    θ = sin⁻¹ (0.900) = 64.2°

7. Mind Stretchers

Mind stretcher 1: Two wavelengths on the same gratingExtension

A grating is illuminated by two colours: blue (450 nm) and red (650 nm).

For the same order n, which colour appears at a larger angle? Explain.

Show Answer

From d sin θ = nλ, larger λ gives larger sin θ and therefore larger θ.

So red appears at a larger angle than blue for the same order.

Mind stretcher 2: Why does a grating give sharper maxima than a double slit?Extension

Explain (qualitatively) why increasing the number of slits makes the principal maxima narrower and sharper.

Show Answer

At the exact maxima condition, waves from many slits add in phase, giving a strong peak.

Slightly away from that angle, phase differences accumulate across the many slits, so cancellation becomes much stronger. This makes the maxima narrow and well-defined compared with a two-slit pattern.

Mind stretcher 3: Optional (Enrichment)Extension

Diffraction gratings are used to measure wavelengths by measuring the angles of the principal maxima and applying d sin θ = nλ.

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