## ERQ · 12 marks · Topics: C.3 Wave phenomena + D.2 Electric and magnetic fields · Archetype: theory_application
**Integration:** primary=C.3 Wave phenomena, secondary=D.2 Electric and magnetic fields (strength: supporting)
**Stem.** A student investigates single-slit diffraction using a laser of unknown wavelength. The laser beam is directed at a narrow vertical slit of width b = 0.12 mm, and the resulting diffraction pattern is observed on a screen placed a distance D = 2.40 m from the slit. The student uses a light sensor mounted on a horizontal track to measure the relative intensity of the light across the central region of the pattern. The student records that the first minimum on either side of the central maximum is located at a horizontal distance of 12.6 mm from the centre of the pattern. After completing the optical measurement, the student considers a second experiment in which a narrow beam of electrons, of de Broglie wavelength equal to that of the laser light, is directed through a region of crossed (perpendicular) uniform electric and magnetic fields before reaching the slit.
### Part (a) State [2 marks] · AO1 · Topic: C.3 Wave phenomena
State two conditions that must be satisfied for a clear single-slit diffraction pattern to be observed on the screen.
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: C.3 Wave phenomena
Using the student's measurement of the first minimum, calculate the wavelength of the laser light.
### Part (b)(ii) Determine [2 marks] · AO2 · Topic: C.3 Wave phenomena
Determine the horizontal distance, in mm, between the central maximum and the second minimum of the diffraction pattern on the screen.
### Part (c) Explain [3 marks] · AO3 · Topic: C.3 Wave phenomena
The student replaces the slit with a new one of width b' = 0.06 mm, using the same laser and the same screen distance. Explain how the appearance of the central maximum on the screen changes compared with the original pattern.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: C.3+D.2
In the electron-beam version of the experiment, a uniform magnetic field is applied perpendicular to the velocity of the electrons before they reach the slit. The student claims that this magnetic field will not change the angular width of the central diffraction maximum. Suggest whether this claim is justified.
---
## Mark Scheme
### Part (a) [2 marks] — State
- M1: light must be monochromatic (single wavelength) [no ECF]
- M2: light must be coherent / slit width must be comparable to the wavelength of light / screen distance D ≫ b [no ECF; accept any one additional valid condition]
### Part (b)(i) [3 marks] — Calculate
- M1: uses sin θ ≈ θ ≈ y/D with y = 12.6 × 10⁻³ m, D = 2.40 m, giving θ ≈ 5.25 × 10⁻³ rad (or equivalent substitution into b sin θ = λ) [no ECF]
- M2: applies single-slit minimum condition λ = b sin θ ≈ b·y/D with b = 0.12 × 10⁻³ m [no ECF]
- M3: λ ≈ 6.3 × 10⁻⁷ m (accept 630 nm, range 625–635 nm) [ECF from M1/M2]
### Part (b)(ii) [2 marks] — Determine
- M1: identifies second minimum at b sin θ = 2λ, so y₂ = 2λD/b (or y₂ = 2 × y₁) [no ECF]
- M2: y₂ ≈ 25.2 mm [ECF from (b)(i)]
### Part (c) [3 marks] — Explain (causal chain per §4.4.1)
- M1: the angular position of the first minimum is given by sin θ = λ/b, so halving b doubles sin θ (the central maximum becomes wider / angular width doubles)
- M2: therefore the linear width of the central maximum on the screen approximately doubles (from ≈ 25.2 mm to ≈ 50.4 mm between first minima) [linking marking point — ECF from (b)(i)]
- M3: because the same incident power is now spread over a wider region (and less light passes through the narrower slit), the peak intensity of the central maximum decreases
### Part (d) [2 marks] — Suggest (proposal + reasoning per §4.4.1)
- M1: claim is justified — the angular width of the central maximum is unchanged [proposal]
- M2: reasoning must combine BOTH (i) the magnetic force qv×B is perpendicular to the velocity so it does no work / the electron's speed (and kinetic energy) is unchanged, AND (ii) therefore the de Broglie wavelength λ = h/(mv) is unchanged, so sin θ = λ/b is unchanged [both elements required for the mark]
### Marker notes
- Alternative method accepted for (b)(ii): direct doubling of y₁ from (b)(i) without re-substitution.
- (b)(i): accept small-angle approximation OR explicit sin θ calculation; answers in the range 625–635 nm acceptable.
- (d): if student claims the field DOES change the angular width, award 0; the physics in M2 must explicitly link "perpendicular force ⇒ no work ⇒ unchanged speed ⇒ unchanged λ". Award M2 only if both perpendicularity AND wavelength invariance are stated.
- ECF chain: (b)(i) λ feeds (b)(ii) and (c) M2 linear-width calculation.