AQA A-Level Physics Paper 1, June 2025: Question 17

1 mark · Easy difficulty · Multiple Choice

Determine the expression for the number of fringes observed on a screen of width Y in a double-slit experiment.

Practise this question

Question

Question 17 asks: Monochromatic light of wavelength lambda is used in a double-slit experiment. The slits are vertical and have a separation s. A narrow screen of width Y is placed a distance D from the slits. Which gives the number of fringes observed on the screen? Options: A is lambda D / (s Y), B is lambda s / (D Y), C is D Y / (s lambda), D is Y s / (lambda D).
Question text

17 Monochromatic light of wavelength λ is used in a double-slit experiment.

The slits are vertical and have a separation s.

A narrow screen of width Y is placed a distance D from the slits.

Which gives the number of fringes observed on the screen?

[1 mark]

λD

A

sY

λs

B

DY

DY

C

sλ

Ys

D

λD

Mark scheme

Show the mark scheme Mark scheme for Question 17: indicates correct response is D, with expression Y s / (lambda D), assessed under AO1.

Ys

17 D AO1

λD

How to answer it

Double-Slit Interference: Calculating Fringe Count on a Screen

📌 What this question tests

This question tests your ability to manipulate the standard Young's double-slit interference equation ( w = λD / s ) and combine it with geometric constraints to find the total number of interference fringes fitting within a finite screen width.

Question 17 • 1 Mark

Young's Double-Slit Algebraic Manipulation

AQA A-Level Physics • Section A: Multiple Choice

✅ Correct Answer

Option D: Ys / λD

Mark Award: 1 mark for selecting D (AO1 - recall and manipulation of physical formulae).

💡 Key Knowledge

  • Fringe Spacing Formula: w = λD / s , where:
    • w = fringe spacing (distance between adjacent maxima)
    • λ = wavelength of light
    • D = distance from slits to screen
    • s = slit separation
  • Total Fringe Count: If the total width of the screen is Y , the number of fringes N that can fit across the screen is given by N = Y / w .

📐 Step-by-Step Algebraic Derivation

  1. Identify the fringe spacing ( w ):
    From the formula sheet: w = λD / s
  2. Set up the relationship for number of fringes ( N ):
    The number of fringes is the total screen width divided by the width of one fringe interval:
    N = Y / w
  3. Substitute w into the equation:
    N = Y / (λD / s)
  4. Simplify the complex fraction:
    Dividing by a fraction is the same as multiplying by its reciprocal:
    N = Y × (s / λD) = Ys / λD

🧠 Exam Technique & Sanity Checks

  • Dimensional Analysis: N must be a dimensionless number (a pure count).
    Units of Ys / λD :
    [m] × [m] / ([m] × [m]) = 1 (dimensionless).
    Notice options A, B, and C also have balanced units, so dimensional analysis alone narrows options down to valid ratios, but physical intuition confirms:
  • Physical Proportionality:
    • Wider screen ( Y ↑ ) → more fringes ( Y must be on top).
    • Larger slit separation ( s ↑ ) → narrower fringes ( w ↓ ) → more fringes fit ( s must be on top).
    • Longer wavelength ( λ ↑ ) or larger distance ( D ↑ ) → wider fringes ( w ↑ ) → fewer fringes fit ( λ and D must be on the bottom).

❌ Common Errors & Traps

  • Option A ( λD / sY ): Inverting the fringe count! Students calculated w / Y instead of Y / w .
  • Option C ( DY / sλ ): Swapping slit separation s and slit-to-screen distance D during algebraic rearrangement.
  • Fraction Division Errors: Forgetting that dividing by (a / b) flips the term into × (b / a) .

Topics

Physics · Required Practicals · 3.3 Waves · AS practicals (1–6)

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.