AQA AS Level Physics Paper 1, June 2023: Question 2

12 marks · Medium difficulty · Short Answer

Explain destructive interference for a given layer thickness, calculate the resultant of two waves on a graph, determine a time interval from phase difference, and calculate the refractive index of a transparent layer.

Practise this question

Question

Figure 1 shows a light ray incident on a transparent layer of thickness 89 nm coated on a glass block, splitting into ray A reflecting from the upper surface and ray B reflecting from the lower surface. Figure 2 and Figure 3 show displacement-time graphs of waves A and B for different frequencies of light. Questions 0.2.1 through 0.2.4 ask to explain destructive interference, calculate the resultant wave at point P, show that time interval t is approximately 8 x 10^-16 s, and calculate the refractive index of the layer.
Question text

02 A glass block is coated with a layer of transparent material.

Figure 1 shows the incident ray and the reflected rays when monochromatic light is

shone onto the upper surface of the transparent layer.

Figure 1

A is light reflecting from the upper surface of the layer.

B is light that leaves the layer after reflection from the lower surface.

When light reflects at the upper and lower surfaces, there is a change of phase.

In this case, the change of phase is the same at each surface and so can be ignored.

When the monochromatic light is incident normally on the upper surface of the layer,

A and B meet and interfere.

Assume that the light is incident normally on the upper surface throughout this

question.

02.1 Figure 2 shows how A and B vary with time at the upper surface.

Figure 2

In the layer, the light has a wavelength of 356 nm.

The thickness of the layer is 89 nm.

Explain why destructive interference occurs at the upper surface for this thickness.

[3 marks]

The frequency of the monochromatic light incident on the layer is changed.

Figure 3 shows how A and B vary with time at the upper surface for this light.

Figure 3

02.2 Calculate the resultant of the waves at time P in Figure 3.

[2 marks]

9 resultant =

The frequency of the light in Figure 3 is 4.72 × 1014 Hz.

02.3 The phase difference between A and B shown in Figure 3 is 137°.

Show that the time interval labelled t in Figure 3 is approximately 8 × 10−16 s.

[3 marks]

02.4 89 nm is the minimum thickness that will produce a phase difference of 137° between

A and B.

Calculate the refractive index of the material of the layer.

[4 marks]

refractive index =

Mark scheme

Show the mark scheme The mark scheme details points for question 0.2.1 (antiphase condition and path difference equal to half a wavelength), 0.2.2 (reading off graph values and summing them), 0.2.3 (using T = 1/f and calculating time from phase angle), and 0.2.4 (calculating wave speed, wavelength in the layer, and refractive index).

Question Answers Additional Comments/Guidance Mark AO

02.1 3 3 ×

A and B are in antiphase or π rad out of phase or 180° out of AO2.1a

Condone: A and B are completely out of

phase

phase. Allow a description of one being a peak

when other is a trough.

(difference in distance travelled =) 178 (nm) or 2 x 89 (nm)

MP2 Alternative: thickness of layer is ¼ of

OR wavelength (of light in layer) or 89 nm is a ¼ of

the wavelength (of light in layer) or journey

A and B travel different distances through layer is ½ of wavelength (of light in

OR layer)

There is a path difference

OR

states the path difference = (n +1) λ MP2 and MP3 time-based alternative:

2 MP2 it takes (some) time for B to travel

through medium (before meets A)

MP3

The path difference is half of a wavelength (of the light in the time taken to travel this (half wavelength)

thin layer) equals half of the period.

Do not accept half of a wavelength out of

phase.

02.2 Correctly reads off 0.93 and (–)0.37 Allow a range of 0.90 to 0.95 for A’s read off 2 AO1.1b

Or and a range of (–)0.35 to (−)0.40 for B’s read AO2.1f

10 off.

Adds their two read-offs provided one is negative

Allows values of 0.9 and (–)0.4 for read-offs.

Look to graph for read-offs.

0.56 Answer in range 0.59 to 0.53

Answer to 2 sf answer here.

An answer of 0.6 or 0.5 scores 1 mark

maximum.

11 –15

02.3 Use of T = (T =) 14 OR (T =) 2.12 × 10 (s) seen 3 AO1.1a

f 4.72 × 10

Condone use of their T in MP2 2 ×

OR

Equation has been rearranged and t would be the subject. AO2.1b

Fraction of a cycle determined: Condone use of their decimal fraction of a cycle in MP2.

137 8 (squares) Expect to see decimal fraction of 0.38 or 0.37

or or 0.38 seen

360 21(.3) (squares)

Answer to at least 2 significant figures

(t =) × 𝑇𝑇 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

360 Alternative:

OR MP1 finds wavelength above surface (636 nm) and determines

fraction of wavelength corresponding to fraction of cycle (242

8 (squares) nm)

(t =) × T seen

21(.3) (squares)

MP2 divides the fraction of the wavelength by the speed of the

OR light above the surface.

137 1 MP3 answer to at least 2 significant figures.

(t =) × 14 seen

360 4.72 × 10 Accept any answer to 2 significant figures that would round to

8.1 × 10–16 (s).

(t =) 8.1 × 10–16 (s)

must include valid supporting work

Use of 𝜆𝜆 = 356 nm is incorrect. Only MP1 is available for fraction

of a cycle.

• Do not allow answers where c = f λ is used to

determine c = 1.68 × 108 m s-1.

135.5

• Do not allow use × T OR

135.5 1

× 14 in MP2

356 4.72 × 10

02.4 distance 2 ×89 ×10−9 Their incorrect t must round to 8 x 10-16 (s). 4 2 ×

Use of speed = = −16

time 8.06 ×10 for ecf.

AO2.1b

(Speed =) 2.2 × 108 (m s–1)

Condone POT in MP1 OR condone use of 89 nm for 2 ×

c distance. (Will give an answer for n of 2.72 or 2.7)

Use of ns = AO3.1a

cs Condone use of their cs in MP3

ns = 1.36 or 1.4 Where 3 sf answer seen, range is 1.35 to 1.37

OR

Allow 2 marks maximum for an answer for n of 1.79

t –9 or 1.8 with working. (Error in use of of 𝜆𝜆 = 356 nm) .

Use of × λ = 178 × 10 Condone as an ECF on MP4.

T

λ = 468 nm Condone POT in MP1 OR condone use of 89 nm for

c fraction of wavelength. (Will give an answer for n of

Use of cs = f λ and ns = 2.72 or 2.7)

cs

Condone use of their λ in MP3

ns = 1.36 or 1.4

Where 3sf answer seen, range is 1.35 to 1.37

OR Condone POT in MP1 OR condone use of 89 nm

137 for fraction of wavelength. (Will give an answer for n

Use of × λ = 178 × 10–9 of 2.72 or 2.7)

Condone use of their λ in MP3

λ = 468 nm

c Where 3sf answer seen, range is 1.35 to 1.37

Use of cs = f λ and ns =

cs Allow an ECF in MP4 for one arithmetical error /

transcription error.

ns = 1.36 or 1.4

Where t = 8 × 10–16 is used:

Speed = 2.225 × 108 m s–1

Wavelength in layer = 471 nm

Refractive index = 1.35

Total 12

How to answer it

Interference and Phase Difference in Thin Films

What this question tests

This question assesses wave superposition, path difference, phase difference, interpreting displacement-time graphs, and calculating wave speed and refractive index in thin transparent layers.

Question 0.2.1 [3 marks]

Explaining Destructive Interference

✅ Correct Answer

  • Waves A and B are in antiphase / pi rad out of phase / 180 degrees out of phase.
  • Path difference = 2 × 89 nm = 178 nm.
  • This path difference corresponds to half a wavelength (lambda / 2) of the light in the layer.

💡 Key Knowledge

  • Light reflects from both the upper and lower surfaces, but ray B travels an extra distance equal to twice the thickness of the layer (2t).
  • Destructive interference occurs when superimposing waves have a path difference of (n + 0.5)λ, resulting in phase opposition (antiphase).

🧠 Exam Technique

To secure all 3 marks, you must link the physical distance travelled inside the film to the wavelength of light. State clearly that ray B travels an extra 178 nm, which is half of the 356 nm wavelength.

❌ Common Errors

  • Forgetting to multiply the film thickness by 2 to account for the down-and-back journey through the layer.
  • Stating path difference as a full wavelength instead of half a wavelength.
Mark breakdown: 1 mark for antiphase/phase relationship, 1 mark for path difference calculation (178 nm), 1 mark for linking path difference to half a wavelength.
Question 0.2.2 [2 marks]

Calculating Wave Superposition Resultant

✅ Correct Answer

  • Read-off for wave A at time P ≈ 0.93 (accept 0.90 to 0.95).
  • Read-off for wave B at time P ≈ -0.37 (accept -0.35 to -0.40).
  • Resultant displacement = 0.93 + (-0.37) = 0.56 (accept range 0.53 to 0.59).

📐 Calculation Steps

  1. Step 1: Locate time P on Figure 3.
  2. Step 2: Determine the individual displacements of curve A and curve B from the vertical axis.
  3. Step 3: Apply the principle of superposition by adding the two displacement values together (accounting for the negative sign of curve B).

❌ Common Errors

  • Treating wave B's negative displacement as a positive value and erroneously adding their magnitudes.
  • Misreading grid line increments on the graph axes.
Mark breakdown: 1 mark for correct read-offs with one negative, 1 mark for correct addition leading to a value between 0.53 and 0.59.
Question 0.2.3 [3 marks]

Time Interval and Frequency Calculation

✅ Correct Answer

8.1 × 10⁻¹⁶ s (or 8 × 10⁻¹⁶ s as requested by the "show that" prompt).

📐 Step-by-Step Calculation

  1. Step 1: Find the period (T).
    T = 1 / f = 1 / (4.72 × 10¹⁴) = 2.118 × 10⁻¹⁵ s.
  2. Step 2: Determine the fraction of the cycle.
    Fraction = 137° / 360° = 0.3806.
  3. Step 3: Calculate time interval (t).
    t = (137 / 360) × (2.118 × 10⁻¹⁵) = 8.06 × 10⁻¹⁶ s ≈ 8.1 × 10⁻¹⁶ s.

🧠 Exam Technique

For "show that" questions, always calculate to at least 3 significant figures before rounding to match the stated value (8 × 10⁻¹⁶ s). Show all intermediate working clearly.

Mark breakdown: 1 mark for finding period T or fraction of cycle, 1 mark for scaling formula substitution, 1 mark for final evaluated answer with valid working.
Question 0.2.4 [4 marks]

Refractive Index Calculation

✅ Correct Answer

Refractive index n = 1.36 or 1.4 (Accepted range: 1.35 to 1.37 depending on rounding paths).

📐 Step-by-Step Calculation

  1. Step 1: Calculate speed of light in the layer (c_s).
    Distance = 2 × 89 nm = 178 × 10⁻⁹ m.
    c_s = distance / time = (178 × 10⁻⁹) / (8.06 × 10⁻¹⁶) = 2.21 × 10⁸ m s⁻¹.
  2. Step 2: Use refractive index formula.
    n = c / c_s = (3.00 × 10⁸) / (2.21 × 10⁸) = 1.36.

💡 Alternative Method

Alternatively, find wavelength in the layer: λ_s = (137 / 360) × λ = distance, then find frequency relationships. Both paths converge on n = 1.36.

❌ Common Errors

  • Using the speed of light in a vacuum (c) instead of calculating the reduced speed of light inside the transparent layer (c_s).
  • Failing to double the 89 nm thickness when working out the total distance traversed by the light wave.
Mark breakdown: 1 mark for speed/distance formulation, 1 mark for correct speed evaluation, 1 mark for applying refractive index equation (n = c / c_s), 1 mark for final correct value (1.35 – 1.37).

Topics

Physics · 3.3 Waves

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.