Waves and Light Problem Solving

This lesson brings together all the key equations and concepts from the waves and particle nature of light topic. The focus is on multi-step exam-style problems — the kind that require you to select the right equations, convert units, and combine ideas from different parts of the topic. Work through each example carefully, paying attention to method and units.

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Key Equations Summary

Before tackling problems, here is a consolidated list of the equations you may need:

Equation	Context	Key variables
v = fλ	Wave speed, frequency, wavelength	v in m s⁻¹, f in Hz, λ in m
T = 1/f	Period and frequency	T in s, f in Hz
n = c/v	Refractive index	c = 3.00 × 10⁸ m s⁻¹
n₁ sin θ₁ = n₂ sin θ₂	Snell's law	Angles from normal
sin θc = n₂/n₁	Critical angle (n₁ > n₂)	Requires n₁ > n₂
w = λD/a	Double-slit fringe spacing	w = spacing, D = screen distance, a = slit separation
d sin θ = nλ	Diffraction grating	d = slit spacing, n = order
E = hf = hc/λ	Photon energy	h = 6.63 × 10⁻³⁴ J s
KE_max = hf − φ	Photoelectric equation	φ = work function
eVs = hf − φ	Stopping voltage	Vs in V, e = 1.60 × 10⁻¹⁹ C
λ = h/(mv) = h/p	de Broglie wavelength	p = momentum
ΔE = hf = hc/λ	Energy level transitions	ΔE in J or eV

Constants you must know:

• Speed of light: c = 3.00 × 10⁸ m s⁻¹
• Planck constant: h = 6.63 × 10⁻³⁴ J s
• Electron charge: e = 1.60 × 10⁻¹⁹ C
• Electron mass: mₑ = 9.11 × 10⁻³¹ kg
• 1 eV = 1.60 × 10⁻¹⁹ J

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Problem Type 1: Double-Slit Interference

Problem: In a Young's double-slit experiment, the slit separation is 0.25 mm and the screen is 1.8 m from the slits. The fringe spacing is measured as 4.3 mm. Calculate the wavelength of the light and identify the colour.

Solution:

Rearranging w = λD/a:

λ = wa/D = (4.3 × 10⁻³ × 0.25 × 10⁻³) / 1.8

= 1.075 × 10⁻⁶ / 1.8 = 5.97 × 10⁻⁷ m ≈ 600 nm

This corresponds to orange light in the visible spectrum.

Problem: A student repeats the experiment with a blue filter. The new fringe spacing is 3.4 mm. Calculate the wavelength of the blue light.

λ = wa/D = (3.4 × 10⁻³ × 0.25 × 10⁻³) / 1.8 = 4.72 × 10⁻⁷ m ≈ 472 nm ✓ (blue)

The fringes are closer together because blue light has a shorter wavelength (w ∝ λ).

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Problem Type 2: Diffraction Grating

Problem: A diffraction grating has 600 lines per mm. Light of wavelength 546 nm is passed through it. Calculate: (a) the angle of the second-order maximum, (b) the highest order visible.

Solution:

(a) First find d: d = 1/(600 × 10³) = 1.667 × 10⁻⁶ m

d sin θ = nλ → sin θ = nλ/d = (2 × 546 × 10⁻⁹) / (1.667 × 10⁻⁶) = 0.6551

θ = sin⁻¹(0.6551) = 40.9°

(b) Maximum order when sin θ = 1:

n_max = d/λ = 1.667 × 10⁻⁶ / (546 × 10⁻⁹) = 3.05

Since n must be a whole number, n_max = 3.

Problem: A grating with 300 lines/mm is used with light containing two wavelengths: 589 nm and 590 nm (the sodium doublet). Calculate the angular separation of these lines in the second order.

d = 1/(300 × 10³) = 3.333 × 10⁻⁶ m

For λ₁ = 589 nm: sin θ₁ = 2 × 589 × 10⁻⁹ / 3.333 × 10⁻⁶ = 0.3534 → θ₁ = 20.697°

For λ₂ = 590 nm: sin θ₂ = 2 × 590 × 10⁻⁹ / 3.333 × 10⁻⁶ = 0.3540 → θ₂ = 20.732°

Angular separation = 20.732° − 20.697° = 0.035° ≈ 0.04°

This tiny separation is resolvable because diffraction gratings produce very sharp maxima. The sodium doublet was historically important for testing the resolving power of spectrometers.

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Problem Type 3: Photoelectric Effect

Problem: Light of wavelength 320 nm shines on a metal surface with a work function of 3.0 eV. Calculate: (a) the photon energy in eV, (b) the maximum kinetic energy of the photoelectrons, (c) the stopping voltage.

Solution:

(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (320 × 10⁻⁹) = 6.22 × 10⁻¹⁹ J

Convert to eV: 6.22 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.89 eV

(b) KE_max = hf − φ = 3.89 − 3.0 = 0.89 eV

(c) eVs = KE_max → Vs = KE_max / e = 0.89 eV / e = 0.89 V

The stopping voltage numerically equals the maximum kinetic energy when KE is expressed in eV. This is a useful shortcut.

Problem: When the wavelength is changed to 450 nm, no photoelectrons are emitted. Explain why and calculate the threshold wavelength.

E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (450 × 10⁻⁹) = 4.42 × 10⁻¹⁹ J = 2.76 eV

Since 2.76 eV < φ = 3.0 eV, the photon energy is less than the work function. The photon does not have enough energy to free an electron from the surface.

Threshold wavelength: λ₀ = hc/φ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (3.0 × 1.60 × 10⁻¹⁹) = 1.989 × 10⁻²⁵ / 4.80 × 10⁻¹⁹ = 4.14 × 10⁻⁷ m = 414 nm

Any wavelength longer than 414 nm (lower frequency) will not cause emission.

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Problem Type 4: de Broglie Wavelength and Electron Diffraction

Problem: Electrons are accelerated from rest through a potential difference of 5.0 kV. Calculate: (a) the kinetic energy, (b) the speed, (c) the de Broglie wavelength.

Solution:

(a) KE = eV = 1.60 × 10⁻¹⁹ × 5000 = 8.00 × 10⁻¹⁶ J

(b) v = √(2KE/m) = √(2 × 8.00 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = √(1.757 × 10¹⁵) = 4.19 × 10⁷ m s⁻¹

(Note: this is ~14% of c. At this speed a relativistic correction would improve accuracy, but for A-Level the non-relativistic formula is acceptable.)

(c) λ = h/(mv) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 4.19 × 10⁷) = 6.63 × 10⁻³⁴ / 3.82 × 10⁻²³ = 1.74 × 10⁻¹¹ m = 0.017 nm

Or using the direct formula: λ = h/√(2meV) = 6.63 × 10⁻³⁴ / √(2 × 9.11 × 10⁻³¹ × 8.00 × 10⁻¹⁶) = 6.63 × 10⁻³⁴ / 3.82 × 10⁻²³ = 0.017 nm ✓

This wavelength is much shorter than visible light, which is why electron microscopes can achieve much higher resolution.

Problem: In an electron diffraction experiment, the accelerating voltage is doubled. By what factor does the de Broglie wavelength change?

Since λ = h/√(2meV), and λ ∝ 1/√V:

λ₂/λ₁ = √(V₁/V₂) = √(1/2) = 1/√2 ≈ 0.71

The wavelength decreases by a factor of √2. The diffraction rings become smaller.

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Problem Type 5: Energy Levels and Photon Transitions

Problem: A hydrogen atom is excited to the n = 4 level. (a) How many distinct photon wavelengths could be emitted as the atom de-excites? (b) Calculate the wavelength of the photon emitted in the n = 4 → n = 2 transition.

Solution:

(a) Number of possible transitions = n(n−1)/2 = 4 × 3 / 2 = 6

The transitions are: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1

(b) ΔE = E₄ − E₂ = (−0.85) − (−3.40) = 2.55 eV = 2.55 × 1.60 × 10⁻¹⁹ = 4.08 × 10⁻¹⁹ J

λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.08 × 10⁻¹⁹) = 1.989 × 10⁻²⁵ / 4.08 × 10⁻¹⁹ = 4.87 × 10⁻⁷ m = 487 nm

This is the Hβ line — blue-green light in the Balmer series.

All Transitions from n = 4

Transition	ΔE / eV	λ / nm	Spectral region
4 → 3	0.66	1879	Infrared (Paschen)
4 → 2	2.55	487	Visible — blue-green (Balmer)
4 → 1	12.75	97.5	UV (Lyman)
3 → 2	1.89	658	Visible — red (Balmer)
3 → 1	12.09	103	UV (Lyman)
2 → 1	10.20	122	UV (Lyman)

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Problem Type 6: Multi-Step Combined Problems

Problem: A sodium street lamp emits light predominantly at 589 nm. The lamp has a power output of 70 W (of light). Calculate: (a) the energy of each photon, (b) the number of photons emitted per second.

Solution:

(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (589 × 10⁻⁹) = 3.38 × 10⁻¹⁹ J = 2.11 eV

(b) Power = energy per second = nE/t, where n/t is the number of photons per second

n/t = P/E = 70 / (3.38 × 10⁻¹⁹) = 2.07 × 10²⁰ photons per second

That is over 200 billion billion photons per second — yet each one carries only a tiny energy.

Problem: A clean caesium surface (φ = 2.1 eV) is illuminated with the 589 nm sodium light. (a) Will photoelectrons be emitted? (b) If so, calculate the maximum kinetic energy and the stopping voltage.

(a) We already calculated E = 2.11 eV and φ = 2.1 eV. Since E > φ (just barely), yes, photoelectrons will be emitted.

(b) KE_max = hf − φ = 2.11 − 2.1 = 0.01 eV = 1.6 × 10⁻²¹ J

Vs = KE_max / e = 0.01 V = 10 mV

The electrons barely escape — they have very little kinetic energy. This makes sense because the photon energy is only marginally above the work function.

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Problem Type 7: Refraction and Critical Angle

Problem: Light enters a glass prism (n = 1.52) at an angle of incidence of 45°. The prism has a 60° apex angle. (a) Calculate the angle of refraction as the light enters. (b) Determine the angle of incidence at the second face.

Solution:

(a) At first face: n₁ sin θ₁ = n₂ sin θ₂

1.00 × sin 45° = 1.52 × sin θ₂

sin θ₂ = 0.7071/1.52 = 0.4652

θ₂ = 27.7°

(b) Inside the prism, using geometry with the 60° apex:

The angle of incidence at the second face = 60° − 27.7° = 32.3°

The critical angle for this glass: sin θc = 1/1.52 = 0.6579 → θc = 41.1°

Since 32.3° < 41.1°, the light will refract out (not TIR). At the second face:

1.52 × sin 32.3° = 1.00 × sin θ_exit

sin θ_exit = 1.52 × 0.5341 = 0.8119

θ_exit = 54.3°

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Common Exam Pitfalls — A Checklist

Pitfall	How to avoid it
Forgetting unit conversions	nm → m: × 10⁻⁹. MHz → Hz: × 10⁶. mm → m: × 10⁻³. eV → J: × 1.60 × 10⁻¹⁹
Confusing slit spacing d with lines per mm N	d = 1/N (in metres). If N is per mm, d = 1/(1000N)
Reading wavelength from a time graph	Displacement–time → period. Displacement–distance → wavelength
Counting fringes vs fringe spacings	n fringes span (n−1) spacings
Mixing up n (refractive index) and n (order)	Context dependent — read the question carefully
Using eV in SI equations	Either convert to J first, or use h = 4.14 × 10⁻¹⁵ eV s
Forgetting sin θ ≤ 1 for max order	n_max = floor(d/λ) for gratings
KE_max depends on intensity (wrong!)	KE_max = hf − φ depends on frequency only

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Practice Strategy

For exam preparation on this topic:

1. List the equations and practise rearranging each one for every variable
2. Practise unit conversions until they are automatic — especially nm ↔ m and eV ↔ J
3. Draw diagrams for refraction and diffraction problems — they help you identify the correct angles
4. Check your answer makes physical sense: wavelengths of visible light are 400–700 nm; the speed of light is 3 × 10⁸ m s⁻¹; energies of visible photons are 1.8–3.1 eV
5. Work multi-step problems that combine different parts of the topic — examiners love these because they test deep understanding, not just formula recall

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A-Level Deep Dive: Waves and Light Problem Solving

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and Particle Nature of Light is the synoptic capstone for the topic, drawing on wave behaviour (superposition, interference, diffraction, polarisation, refraction), the electromagnetic spectrum, the photoelectric effect, atomic line spectra, wave–particle duality and de Broglie's relation (refer to the official specification document for exact wording). Although Topic 5 sits inside Paper 1's content allocation, its synoptic style means it is examined throughout Paper 1 (Advanced Physics I), Paper 2 (Advanced Physics II — when waves underpin oscillations or thermal radiation) and especially Paper 3 (General and Practical Principles in Physics), where extended-response items deliberately fuse waves with quantum behaviour. The Edexcel data and formulae booklet supplies $v = f\lambda$v=fλ, $n = c/v$n=c/v, $E = hf$E=hf, $hf = \phi + E_{k(\max)}$hf=ϕ+Ek(max)​, $\lambda = h/p$λ=h/p and $d\sin\theta = n\lambda$dsinθ=nλ — but the combinations of these (e.g. predicting a photocurrent from a graded interference pattern) are not formula-booklet items and must be assembled by the candidate.

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Worked example with full mark scheme

Question (12 marks):

A laser of wavelength 532 nm illuminates a diffraction grating of 600 lines mm⁻¹ at normal incidence. The first-order maximum on the far side falls onto a small caesium photocell of work function 2.14 eV.

(a) Calculate the angle at which the first-order maximum emerges from the grating. (3)

(b) Show that photons in the incident beam have enough energy to liberate photoelectrons from caesium, and calculate the maximum kinetic energy of an emitted photoelectron in joules. (4)

(c) The laser delivers 2.5 mW of optical power and the grating diverts 18% of this power into the first-order maximum that strikes the photocell. Assuming a quantum efficiency of 6.0% (i.e. 6 photoelectrons released per 100 incident photons), calculate the expected photocurrent. (5)

Solution with mark scheme:

(a) Step 1 — find the slit spacing.

$d = 1/(600 \times 10^3 \text{ m}^{-1}) = 1.667 \times 10^{-6}$d=1/(600×103 m−1)=1.667×10−6 m.

M1 — converting "lines per mm" to a slit spacing in metres. Common error: leaving $d$d in mm or computing $d = 600/1$d=600/1 instead of its reciprocal — both lose this M1 immediately.

Step 2 — apply the grating equation.

$\sin\theta = n\lambda/d = (1)(532 \times 10^{-9})/(1.667 \times 10^{-6}) = 0.3192$sinθ=nλ/d=(1)(532×10−9)/(1.667×10−6)=0.3192.

M1 — correct substitution into $d\sin\theta = n\lambda$dsinθ=nλ with $n = 1$n=1 and consistent units (metres on both sides).

$\theta = \arcsin(0.3192) = 18.6°$θ=arcsin(0.3192)=18.6°.

A1 — answer to 3 sf with degree symbol.

(b) Step 1 — photon energy.

$E = hc/\lambda = (6.63 \times 10^{-34})(3.00 \times 10^8)/(532 \times 10^{-9}) = 3.74 \times 10^{-19}$E=hc/λ=(6.63×10−34)(3.00×108)/(532×10−9)=3.74×10−19 J.

M1 — using $E = hc/\lambda$E=hc/λ with SI values.

Step 2 — convert work function to joules.

$\phi = 2.14 \times 1.60 \times 10^{-19} = 3.42 \times 10^{-19}$ϕ=2.14×1.60×10−19=3.42×10−19 J.

M1 — eV → J conversion shown explicitly.

Step 3 — comparison and $E_{k(\max)}$Ek(max)​.

Because $E > \phi$E>ϕ, photoemission occurs. $E_{k(\max)} = hf - \phi = (3.74 - 3.42) \times 10^{-19} = 3.2 \times 10^{-20}$Ek(max)​=hf−ϕ=(3.74−3.42)×10−19=3.2×10−20 J.

A1 — explicit statement that $E > \phi$E>ϕ (the "show that" demand).

A1 — final $E_{k(\max)} = 3.2 \times 10^{-20}$Ek(max)​=3.2×10−20 J.

(c) Step 1 — power reaching the photocell.

$P_{\text{cell}} = 0.18 \times 2.5 \times 10^{-3} = 4.5 \times 10^{-4}$Pcell​=0.18×2.5×10−3=4.5×10−4 W.

M1 — applying the 18% diffraction efficiency.

Step 2 — incident photon rate.

$R_{\gamma} = P_{\text{cell}}/E = (4.5 \times 10^{-4})/(3.74 \times 10^{-19}) = 1.20 \times 10^{15}$Rγ​=Pcell​/E=(4.5×10−4)/(3.74×10−19)=1.20×1015 s⁻¹.

M1 — using photon energy from (b) — credit available even with carried-forward error.

Step 3 — photoelectron rate.