AQA AS Level Physics Paper 1, November 2021: Question 2

9 marks · Hard difficulty · Extended Answer

Analyze diffraction grating interference patterns using a DVD and laser pointers, including path difference, formation of maxima, and maximum number of spots on a screen.

Practise this question

Question

A multipart physics question about using a DVD as a transmission diffraction grating with laser pointers. Figure 3 shows a laser beam passing through a DVD onto a circular screen producing bright spots. Table 2 gives slit spacings for Blu-ray, DVD, and CD, and Table 3 gives wavelengths for laser pointers A and B. Figure 4 shows an experimental setup with a CD and laser pointer B producing a diffraction pattern on a 30 cm diameter screen positioned 15 cm away.
Question text

02 A student removes the reflective layer from a DVD. She uses the DVD as a

transmission diffraction grating.

Figure 3 shows light from a laser pointer incident normally on a small section of this

diffraction grating. The grooves on this section act as adjacent slits of the

transmission diffraction grating.

A vertical pattern of bright spots (maxima) is observed on a circular screen behind the

disc.

Figure 3

02.1 Light of wavelength λ travels from each illuminated slit, producing maxima on the

screen.

State the path difference between light from adjacent slits when this light produces a

first-order maximum on the screen.

[1 mark]

02.2 Explain how light from the diffraction grating forms a maximum on the screen.

[3 marks]

The student has three discs: a Blu-ray disc, a DVD and a CD. She removes the

reflective coating from the discs so that they act as transmission diffraction gratings.

These diffraction gratings have different slit spacings.

The student also has two laser pointers A and B that emit different colours of visible

light.

Table 2 and Table 3 show information about the discs and the laser pointers.

Table 2

Disc Slit spacing / μm

Blu-ray disc 0.32

DVD 0.74

CD 1.60

Table 3

Laser pointer Wavelength of light emitted / 10−7 m

A 4.45

B 6.36

02.3 Deduce the combination of disc and laser pointer that will produce the greatest

possible number of interference maxima.

[2 marks]

02.4 The student uses the CD and laser pointer B as shown in Figure 4. A diffraction

pattern is produced on the screen. Laser pointer B and the CD are in fixed positions.

The laser beam is horizontal and incident normally on the CD. The height of the

screen can be adjusted.

Figure 4

The screen has a diameter of 30 cm and is positioned behind the CD at a fixed

horizontal distance of 15 cm.

The student plans to adjust the height of the screen until she observes the greatest

number of spots.

The student predicts that, using this arrangement, the greatest number of spots on the

screen will be 3.

Determine whether the student’s prediction is correct.

[3 marks]

Mark scheme

Show the mark scheme A detailed mark scheme providing answers for questions 02.1 through 02.4. Question 02.1 requires stating the path difference as one wavelength (1 mark). Question 02.2 requires explaining light overlapping, path difference as a whole number of wavelengths, and superposition (3 marks). Question 02.3 requires deducing pointer A and CD produce the greatest number of maxima using n = d/lambda (2 marks). Question 02.4 requires using n lambda = d sin theta and evaluating screen dimensions to determine whether the prediction of 3 spots is correct (3 marks).

Question Answers Additional Comments/Guidance Mark AO

02.1 One wavelength Accept λ 1 AO1.1a

02.2 3 AO2.1c

Light from slits overlap / undergo diffraction AO2.1c

AO2.1c

Path difference is a whole number of wavelengths

Or

Arrive at screen in phase / zero phase difference

(Meet and) undergo superposition / waves superpose If no other mark awarded allow one mark for:

interfere constructively / Produces

reinforcement / produces constructive

interference

02.3 Pointer A and CD 2 AO3.1b

Condone:

AO3.1b

smallest wavelength and greatest slit spacing

(Smallest) angular spread for each order (θmin) is given by

n 𝜆𝜆

sin θ(min) =

d Max 1 mark for:

OR An argument that links spacing to slit width and

𝑑𝑑 its effect on diffraction.

Greatest number of maxima is given by nmax =

𝜆𝜆

02.4 use of n λ = d sin θ For example, where: 3 AO3.1b

n = 1, λ = 6.36 × 10–7 m and d = 1.6 × 10–6 m AO3.1b

AO3.1b

(θ = 23°)

or

n = 2, λ = 6.36 × 10–7 m and d = 1.6 × 10–6 m

(θ = 53°)

use of tan θ = r / 15

or Allow use of tan θ = r / 15

adds θ2 and θ1 and compares to 90° for any combination of θ, r and 15 where

or

unknown has been made subject.

adds θ2 and θ2 and compares to 90°

12 or

adds θ1 and θ1 and compares to 90°

θ = 45° and 3 bright spots therefore yes is a

maximum of max 2 marks

No, can see 4 (bright spots)

𝑑𝑑

Allow use of 𝑛𝑛 = where they have reached a

𝜆𝜆

conclusion for 1 mark maximum.

Total 9

How to answer it

Diffraction Gratings Using Optical Discs

What this question tests

This question assesses your understanding of wave optics, specifically transmission diffraction gratings, path difference, constructive interference, and the application of the diffraction grating equation ( nλ = d sin θ ). It tests your ability to reason about slit spacing, wavelength, maximum possible orders, and geometric trigonometry on a screen.

Question Part 02.1

Path Difference for First-Order Maximum

✅ Correct Answer

One wavelength (or λ )

💡 Key Knowledge

  • For constructive interference to occur producing the n -th order maximum, the path difference from adjacent slits must be an integer multiple of the wavelength ( nλ ).
  • For the first-order maximum, n = 1 , making the path difference exactly 1λ .
Mark allocation: 1 mark for stating "one wavelength" or λ .
Question Part 02.2

Explaining Diffraction Grating Maxima Formation

✅ Correct Answer

  • Light from the slits overlaps and undergoes diffraction.
  • Path difference between light from adjacent slits is a whole number of wavelengths (or waves arrive in phase / zero phase difference).
  • Waves superpose constructively (reinforcement).

🧠 Exam Technique

To secure all 3 marks, you must use standard physics terminology in a logical sequence: Diffraction/Spreading → Path/Phase Difference → Superposition/Reinforcement.

Mark allocation: 3 marks total. (1 mark per correct linking bullet point, with alternative allowance for constructive interference if other points hit).
Question Part 02.3

Deducing Maximum Number of Interference Maxima

✅ Correct Answer

Pointer A and CD

💡 Key Knowledge

  • The maximum number of orders is given by the formula n_max = d / λ (rounded down to the nearest integer).
  • To get the greatest number of maxima, you need the greatest slit spacing (d) and the smallest wavelength (λ).
  • From the tables: CD has the largest d ( 1.60 µm ) and Pointer A has the shortest λ ( 4.45 × 10⁻⁷ m ).
Mark allocation: 2 marks (1 for correct choice of Pointer A and CD, 1 for justifying using sin θ = nλ / d or n_max = d / λ ).
Question Part 02.4

Determining Screen Spot Capacity (Calculation)

📐 Step-by-Step Calculation

  1. Identify parameters for CD and Pointer B:
    d = 1.60 µm = 1.60 × 10⁻⁶ m
    λ = 6.36 × 10⁻⁷ m
  2. Calculate angles for each maximum ( n ):
    • For n = 1 : sin θ₁ = (1 × 6.36 × 10⁻⁷) / (1.60 × 10⁻⁶) = 0.3975 → θ₁ = 23.4°
    • For n = 2 : sin θ₂ = (2 × 6.36 × 10⁻⁷) / (1.60 × 10⁻⁶) = 0.7950 → θ₂ = 52.7°
    • For n = 3 : sin θ₃ = 1.1925 (> 1, so no 3rd order exists). Max orders are n = 0, ±1, ±2 .
  3. Relate angles to screen geometry using tan:
    Distance to screen x = 15 cm . Radius to spot r = 15 × tan(θ) .
    • For n = 1 : r₁ = 15 × tan(23.4°) = 6.5 cm
    • For n = 2 : r₂ = 15 × tan(52.7°) = 19.6 cm
  4. Check against screen dimensions:
    The screen has a diameter of 30 cm , meaning its radius is 15 cm .
    The n = 2 spots occur at 19.6 cm from the center, which exceeds the 15 cm screen radius! Therefore, second-order spots miss the screen entirely.
  5. Final Conclusion:
    Visible spots on the screen are the central zero-order ( n = 0 ) and the first-order pair ( n = ±1 ), totalling 3 bright spots. The student's prediction is correct.

❌ Common Calculation Traps

  • Unit Mismatch: Forgetting to convert micrometres ( µm ) and nanometres ( 10⁻⁷ m ) into standard metres.
  • Screen Radius vs Diameter: Confusing the 30 cm screen diameter with its 15 cm radius limit when checking if spots fit on the physical screen.
  • Forgetting Symmetry: Not accounting for both positive and negative orders ( ±1 ) when counting total visible spots.
Mark allocation: 3 marks total. (1 mark for grating equation calculation of angles, 1 mark for geometric trigonometric relation to screen dimensions, 1 mark for correct final conclusion of 3 spots).

Topics

Physics · 3.3 Waves

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.