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CIE AS/A Level Physics 8.4 The diffraction grating Study Notes- 2025-2027 Syllabus

CIE AS/A Level Physics 8.4 The diffraction grating Study Notes – New Syllabus

CIE AS/A Level Physics 8.4 The diffraction grating Study Notes at  IITian Academy  focus on  specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:

  1. recall and use d sin θ = nλ
  2. describe the use of a diffraction grating to determine the wavelength of light (the structure and use of the spectrometer are not included)

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Diffraction Grating Equation

A diffraction grating consists of a large number of equally spaced, parallel slits that diffract light. The diffracted waves from these slits interfere to produce sharp and distinct maxima (bright fringes) at specific angles.

Grating Equation: 

\( \mathrm{d \sin \theta = n\lambda} \)

Where:

  • \( \mathrm{d} \): grating spacing (distance between adjacent slits, in meters)
  • \( \mathrm{\theta} \): angle of diffraction for the bright fringe (measured from the central maximum)
  • \( \mathrm{n} \): order of the maximum (\( \mathrm{n = 0, 1, 2, 3, \dots} \))
  • \( \mathrm{\lambda} \): wavelength of the light

Derivation:

  • Each slit in the grating acts as a secondary source of diffracted waves.
  • Constructive interference occurs when the path difference between adjacent slits equals an integer multiple of the wavelength.
  • Path difference = \( \mathrm{d \sin \theta} \).
  • Hence, for bright fringes: \( \mathrm{d \sin \theta = n\lambda} \).

Key Features of the Diffraction Pattern:

  • The central maximum (\( \mathrm{n = 0} \)) is the brightest and occurs at \( \mathrm{\theta = 0°} \).
  • Higher-order maxima (\( \mathrm{n = 1, 2, 3, …} \)) appear symmetrically on both sides of the central maximum.
  • Each order corresponds to a specific value of \( \mathrm{\theta} \).
  • The pattern is sharper and more widely spaced than that of a double slit because the number of slits is very large, increasing constructive reinforcement at specific angles.

Example

A diffraction grating has 500 lines per millimetre. Calculate the angle of the first-order maximum for light of wavelength \( \mathrm{600 \ nm} \).

▶️ Answer / Explanation

Step 1: Convert grating spacing:

\( \mathrm{500 \ lines/mm = 500 \times 10^3 \ lines/m} \)

\( \mathrm{d = \dfrac{1}{500 \times 10^3} = 2.0 \times 10^{-6} \ m.} \)

Step 2: Apply \( \mathrm{d \sin \theta = n\lambda} \):

\( \mathrm{2.0 \times 10^{-6} \sin \theta = 1 \times 600 \times 10^{-9}} \)

\( \mathrm{\sin \theta = 0.3 \Rightarrow \theta = 17.5°} \)

Therefore, the first-order maximum occurs at \( \mathrm{17.5°.} \)

Determining the Wavelength of Light Using a Diffraction Grating

Principle: A diffraction grating can be used to determine the wavelength of light by measuring the angle at which the bright interference fringes (maxima) occur. The relationship between angle, order, and wavelength is given by \( \mathrm{d \sin \theta = n\lambda} \).

Apparatus:

  • Monochromatic light source (e.g., sodium lamp)
  • Diffraction grating (known number of lines per mm)
  • Screen or angular measurement setup (e.g., protractor or angular scale)

Procedure (Conceptual Outline):

  1. Illuminate the diffraction grating with the monochromatic light source.
  2. Observe the diffraction pattern of bright fringes (spectral lines) on a screen or through a detector.
  3. Measure the angle \( \mathrm{\theta} \) between the central maximum (\( \mathrm{n = 0} \)) and a higher-order maximum (\( \mathrm{n = 1, 2, …} \)).
  4. Determine the grating spacing \( \mathrm{d} \) using: \( \mathrm{d = \dfrac{1}{N}} \), where \( \mathrm{N} \) = number of lines per meter.
  5. Substitute \( \mathrm{d} \), \( \mathrm{\theta} \), and \( \mathrm{n} \) into the equation \( \mathrm{d \sin \theta = n\lambda} \) to find \( \mathrm{\lambda} \).

Example of Grating Measurement:

  • Suppose the grating has 600 lines/mm, giving \( \mathrm{d = 1.67 \times 10^{-6} \ m.} \)
  • For first-order maximum (\( \mathrm{n = 1} \)), the measured angle is \( \mathrm{\theta = 22°}. \)
  • Substitute in \( \mathrm{d \sin \theta = n\lambda} \):

\( \mathrm{1.67 \times 10^{-6} \sin 22° = 1 \times \lambda} \)

\( \mathrm{\lambda = 6.26 \times 10^{-7} \ m = 626 \ nm.} \)

Result: The wavelength of the light is \( \mathrm{626 \ nm} \) (orange-red region of visible light).

Advantages of Diffraction Grating Over Double-Slit:

  • Produces sharper and brighter maxima due to many slits reinforcing constructive interference.
  • Allows accurate measurement of wavelengths, including in multiple spectral orders.
  • Used in spectroscopic analysis to identify elements based on their characteristic wavelengths.
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