Formula Prep Example Questions

ENG5055 Practice Problems — Formula Drilling

Semiconductor Lasers

Question

1. An InGaAsP Fabry-Perot laser operates at 1310 nm with a cavity length of 500 µm and effective refractive index of 3.2. Calculate the longitudinal mode spacing and the number of modes within a 40 nm gain bandwidth.

Answer

$\Delta \lambda =\frac{{\lambda }^2}{2nL}$ $N=\frac{\Delta {\lambda }_{gain}}{\Delta \lambda }$ Δλ = λ²/(2nL) = (1310×10⁻⁹)²/(2 × 3.2 × 500×10⁻⁶) = 1.716×10⁻¹²/3.2×10⁻³ = 0.536 nm N = Δλ_gain/Δλ = 40/0.536 ≈ 74 modes

Question

2. A GaAs laser (n = 3.5) has cleaved facets on both ends. Calculate the Fresnel reflectivity of each facet. If the cavity is 300 µm long, the internal loss is 10 cm⁻¹, and the confinement factor is 0.3, calculate the threshold gain.

Answer

$R={\left(\frac{n-1}{n+1}\right)}^2$ R = ((n-1)/(n+1))² = (2.5/4.5)² = 0.3086 ≈ 0.31 Mirror loss: αₘ = (1/2L)·ln(1/(R₁R₂)) = (1/(2×0.03))·ln(1/(0.31×0.31)) = 16.67 × ln(10.41) = 16.67 × 2.343 = 39.1 cm⁻¹ Threshold: Γg_th = αᵢ + αₘ → g_th = (10 + 39.1)/0.3 = 163.5 cm⁻¹

Question

3. A DFB laser has a grating period of 240 nm and an effective refractive index of 3.22. What is the Bragg wavelength?

Answer

${\lambda }_B=2n_{eff}\Lambda$ λ_B = 2n_eff·Λ = 2 × 3.22 × 240 = 1545.6 nm

Question

4. What mode number m corresponds to 1550 nm operation in a Fabry-Perot cavity with n = 3.4 and L = 400 µm?

Answer

$m\lambda =2nL$ → m = 2nL/λ = 2 × 3.4 × 400×10⁻⁶ / (1550×10⁻⁹) = 2.72×10⁻³/1.55×10⁻⁶ = 1755

Question

5. A laser has R₁ = 0.95 (HR coated back facet) and R₂ = 0.05 (AR coated front facet), L = 250 µm, αᵢ = 8 cm⁻¹, Γ = 0.4. Calculate the threshold gain. Compare to an uncoated cavity where R₁ = R₂ = 0.31.

Answer

$g_{th}\Gamma ={\alpha }_i+\frac{1}{2L}\ln \left(\frac{1}{R_1{R}_2}\right)$ Coated: αₘ = (1/(2×0.025))·ln(1/(0.95×0.05)) = 20·ln(21.05) = 20×3.047 = 60.9 cm⁻¹ g_th = (8 + 60.9)/0.4 = 172.3 cm⁻¹ Uncoated: αₘ = 20·ln(1/0.31²) = 20·ln(10.41) = 20×2.343 = 46.9 cm⁻¹ g_th = (8 + 46.9)/0.4 = 137.2 cm⁻¹ The coated cavity has higher threshold gain because most light exits the front facet (low R₂), but this is desirable for high output power from one facet.

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Etching — Mean Free Path

Question

6. Calculate the mean free path of Ar atoms (diameter d = 3.4 Å) at 10 mTorr and 300 K. Repeat for 100 mTorr. (1 Torr = 133.3 Pa)

Answer

${\lambda }_{mfp}=\frac{kT}{\sqrt{2}\pi d^{2}P}$ At 10 mTorr: P = 10×10⁻³ × 133.3 = 1.333 Pa λ = kT/(√2·π·d²·P) = (1.381×10⁻²³ × 300)/(√2 × π × (3.4×10⁻¹⁰)² × 1.333) = 4.143×10⁻²¹/(1.414 × 3.142 × 1.156×10⁻¹⁹ × 1.333) = 4.143×10⁻²¹/6.845×10⁻¹⁹ = 6.05×10⁻³ m ≈ 6 mm

At 100 mTorr: λ ∝ 1/P, so λ = 6.05/10 ≈ 0.6 mm At lower pressure, the mean free path is much longer — ions travel more directionally (anisotropic etching). At higher pressure, more collisions randomise ion trajectories (more isotropic).

Question

7. An RIE process operates at 50 mTorr with SF₆ (molecular diameter ≈ 5.5 Å). Calculate the mean free path at 300 K. If the electrode gap is 4 cm, how many collisions does an ion undergo crossing the gap on average?

Answer

P = 50×10⁻³ × 133.3 = 6.665 Pa λ = (1.381×10⁻²³ × 300)/(√2 × π × (5.5×10⁻¹⁰)² × 6.665) = 4.143×10⁻²¹/(1.414 × 3.142 × 3.025×10⁻¹⁹ × 6.665) = 4.143×10⁻²¹/8.957×10⁻¹⁸ = 4.63×10⁻⁴ m ≈ 0.46 mm

Number of collisions ≈ gap/λ = 40 mm/0.46 mm ≈ 87 collisions. This is significant — many collisions mean ions lose directionality. Lower pressure would improve anisotropy.

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Cleanroom

Question

8. A semiconductor fab processes 300 mm wafers. A single killer defect (particle ≥ 0.5 µm landing on a critical layer) ruins a die. If the die area is 1 cm², the wafer has ~700 dies, and the cleanroom is Class 100, estimate whether this is clean enough for a reasonable yield. Assume an air velocity of 30 m/min passing over the wafer for 2 minutes during the critical exposure step.

Answer

Class 100: max 100 particles/ft³ (≥0.5 µm) = 3530/m³ Volume of air over one die in 2 min: A × v × t = 1×10⁻⁴ × 30 × 2 = 6×10⁻³ m³ Particles per die: 3530 × 6×10⁻³ ≈ 21 particles This is far too many — even if only a fraction are killer defects, yield would be poor. Modern fabs use Class 1 or better (ISO 3) for critical lithography steps, reducing this by 100×.

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Optics & Lenses

Question

9. A biconvex lens has R₁ = +20 cm, R₂ = -35 cm, and n = 1.52. Calculate the focal length.

Answer

1/f = (n-1)(1/R₁ - 1/R₂) = 0.52(1/20 - 1/(-35)) = 0.52(0.05 + 0.02857) = 0.52 × 0.07857 = 0.04086 f = 24.5 cm

Question

10. An object is placed 40 cm from a converging lens with f = 15 cm. Find the image distance and magnification.

Answer

1/dᵢ = 1/f - 1/dₒ = 1/15 - 1/40 = (40-15)/(15×40) = 25/600 dᵢ = 24.0 cm (real, same side as light exits) m = -dᵢ/dₒ = -24.0/40.0 = -0.60 (inverted, diminished)

Question

11. An achromatic doublet is made from two lenses in contact: f₁ = +12 cm (crown glass) and f₂ = -18 cm (flint glass). What is the combined focal length?

Answer

1/f = 1/12 + 1/(-18) = 0.08333 - 0.05556 = 0.02778 f = 36.0 cm (converging)

Question

12. A compound microscope has fo = 4 mm, fe = 25 mm, and a barrel length of 160 mm. What is the total magnification? What is the objective magnification Mo and eyepiece magnification Me individually?

Answer

$M=-\frac{L}{f_o}\cdot \frac{N}{f_e}$ L = 160 - 4 - 25 = 131 mm Mo = -L/fo = -131/4 = -32.75 Me = N/fe = 250/25 = 10 M = Mo × Me = -327.5 ≈ -328×

Question

13. An oil-immersion objective (n_oil = 1.515) has a half-angle of 67°. Calculate the NA and the minimum resolvable feature size at λ = 550 nm.

Answer

$NA=n\sin (\theta )$ $d_{min}=\frac{0.61\lambda }{NA}$

NA = n·sinθ = 1.515 × sin(67°) = 1.515 × 0.9205 = 1.394 d_min = 0.61λ/NA = 0.61 × 550/1.394 = 241 nm

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Resolution & Lithography

Question

14. A DUV lithography system uses 193 nm light with NA = 1.35 (immersion) and k₁ = 0.3. What is the minimum feature size? How does this compare to using 248 nm with NA = 0.75 and k₁ = 0.5?

Answer

$R=k_1\frac{\lambda }{NA}$ 193 nm immersion: R = 0.3 × 193/1.35 = 42.9 nm 248 nm dry: R = 0.5 × 248/0.75 = 165.3 nm The 193 nm immersion system resolves features ~4× smaller.

Question

15. A single slit of width 15 µm is illuminated by 633 nm laser light. A screen is 2.0 m away. Calculate the width of the central diffraction maximum.

Answer

$a\sin \theta =m\lambda (m=\pm 1,\pm 2,\dots )$ First minimum: sinθ₁ = λ/a = 633×10⁻⁹/15×10⁻⁶ = 0.0422 y₁ = L·tanθ₁ ≈ L·sinθ₁ = 2.0 × 0.0422 = 0.0844 m Full width of central max = 2y₁ = 16.9 cm

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de Broglie & Electron Optics

Question

16. An SEM operates at 20 kV. Calculate the electron wavelength. Compare to a TEM operating at 200 kV.

Answer

$\lambda \text{ (nm)}=\frac{1.226}{\sqrt{V\text{ (V)}}}$ $\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2m_{e}eV}}$

SEM: λ = 1.226/√20000 = 1.226/141.4 = 8.67 pm = 0.00867 nm TEM: λ = 1.226/√200000 = 1.226/447.2 = 2.74 pm = 0.00274 nm TEM has ~3× shorter wavelength, but practical resolution is limited by lens aberrations, not diffraction.

Question

17. What accelerating voltage is needed to give electrons a de Broglie wavelength of 0.05 nm?

Answer

λ = 1.226/√V → √V = 1.226/0.05 = 24.52 → V = 601 V

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Quantum Tunnelling

Question

18. An electron with energy 5 eV encounters a barrier of height 8 eV and width 0.5 nm. Calculate the transmission coefficient.

Answer

$T\approx e^{-2\kappa d},\kappa =\frac{\sqrt{2m(V_0-E)}}{\hslash }$ κ = √(2m(V₀-E))/ℏ = √(2 × 9.11×10⁻³¹ × 3 × 1.6×10⁻¹⁹)/(1.055×10⁻³⁴) = √(8.746×10⁻⁴⁹)/1.055×10⁻³⁴ = 9.352×10⁻²⁵/1.055×10⁻³⁴ = 8.87×10⁹ m⁻¹ 2κd = 2 × 8.87×10⁹ × 0.5×10⁻⁹ = 8.87 T = e⁻⁸·⁸⁷ ≈ 1.4×10⁻⁴

Question

19. For the same barrier (8 eV, 0.5 nm), what electron energy gives T = 0.01?

Answer

T = e⁻²ᵏᵈ → ln(0.01) = -2κd → κd = 2.303 → κ = 2.303/(0.5×10⁻⁹) = 4.606×10⁹ m⁻¹ V₀ - E = ℏ²κ²/(2m) = (1.055×10⁻³⁴)² × (4.606×10⁹)²/(2 × 9.11×10⁻³¹) = 1.113×10⁻⁶⁸ × 2.122×10¹⁹/1.822×10⁻³⁰ = 2.361×10⁻⁴⁹/1.822×10⁻³⁰ = 1.296×10⁻¹⁹ J = 0.81 eV E = 8 - 0.81 = 7.19 eV

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AFM Cantilever

Question

20. A silicon nitride cantilever has l = 200 µm, w = 20 µm, t = 0.6 µm. E = 250 GPa, ρ = 3100 kg/m³. Calculate the spring constant, mass, and resonance frequency.

Answer

k = Ewt³/(4l³) = 250×10⁹ × 20×10⁻⁶ × (0.6×10⁻⁶)³/(4 × (200×10⁻⁶)³) = 250×10⁹ × 2×10⁻⁵ × 2.16×10⁻¹⁹/(4 × 8×10⁻¹⁵) = 1.08×10⁻¹²/3.2×10⁻¹⁴ = 0.0338 N/m

m = ρlwt = 3100 × 200×10⁻⁶ × 20×10⁻⁶ × 0.6×10⁻⁶ = 7.44×10⁻¹² kg m* = 0.24 × 7.44×10⁻¹² = 1.786×10⁻¹² kg $f_0=\frac{1}{2\pi }\sqrt{\frac{k_N}{0.24m}}$ f₀ = (1/2π)√(0.0338/1.786×10⁻¹²) = (1/2π)√(1.893×10¹⁰) = (1/2π)(1.376×10⁵) = 21.9 kHz

Question

21. You want to design an AFM cantilever with a resonance frequency of 300 kHz and spring constant of 40 N/m using silicon (E = 179 GPa, ρ = 2330 kg/m³). If the width is 30 µm, find the required length and thickness.

Answer

From k = Ewt³/(4l³) and f₀ = (1/2π)√(k/(0.24m)), with m = ρlwt: f₀ = (1/2π)√(k/(0.24ρlwt)) Rearranging for t from the spring constant: t³ = 4kl³/(Ew) → t = (4kl³/(Ew))^(1/3)

This is coupled — substitute t into f₀ equation and solve numerically. Alternatively, guess and check: Try l = 125 µm, then t = (4×40×(125×10⁻⁶)³/(179×10⁹×30×10⁻⁶))^(1/3) = (4×40×1.953×10⁻¹²/(5.37×10⁶))^(1/3) = (5.81×10⁻¹⁷)^(1/3) = 3.87×10⁻⁶ m ≈ 3.9 µm m = 2330 × 125×10⁻⁶ × 30×10⁻⁶ × 3.9×10⁻⁶ = 3.41×10⁻¹¹ kg f₀ = (1/2π)√(40/(0.24×3.41×10⁻¹¹)) = (1/2π)√(4.89×10¹²) = (1/2π)(2.21×10⁶) = 352 kHz Slightly high — increase l to ~135 µm and iterate. The method demonstrates the interdependence of k, f₀, and geometry.

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Thermal Oxidation

Question

22. Using the Deal-Grove model, calculate B/A and B for wet oxidation of ⟨100⟩ Si at 1100°C. Then find the time to grow 1 µm of oxide from bare silicon.

Answer

T = 1100°C = 1373 K, kT = 8.617×10⁻⁵ × 1373 = 0.1183 eV B/A = 9.70×10⁷ × exp(-2.05/0.1183) = 9.70×10⁷ × exp(-17.33) = 9.70×10⁷ × 2.96×10⁻⁸ = 2.87 µm/hr B = 386 × exp(-0.78/0.1183) = 386 × exp(-6.593) = 386 × 1.37×10⁻³ = 0.529 µm²/hr A = B/(B/A) = 0.529/2.87 = 0.184 µm

For bare Si (τ ≈ 0 for wet ox): x² + Ax = Bt 1² + 0.184(1) = 0.529t → t = 1.184/0.529 = 2.24 hr

Question

23. A wafer already has 150 nm of oxide. It is placed in dry O₂ at 1000°C (⟨100⟩ Si). How long to grow an additional 100 nm (total 250 nm)? Use D₀ and Eₐ from the table: B/A: D₀ = 3.71×10⁶ µm/hr, Eₐ = 2.00 eV; B: D₀ = 772 µm²/hr, Eₐ = 1.23 eV.

Answer

T = 1000°C = 1273 K, kT = 0.1097 eV B/A = 3.71×10⁶ × exp(-2.00/0.1097) = 3.71×10⁶ × exp(-18.23) = 3.71×10⁶ × 1.21×10⁻⁸ = 0.0449 µm/hr B = 772 × exp(-1.23/0.1097) = 772 × exp(-11.21) = 772 × 1.35×10⁻⁵ = 0.01042 µm²/hr A = B/(B/A) = 0.01042/0.0449 = 0.232 µm

Find effective time for existing 150 nm: (0.15)² + 0.232(0.15) = 0.01042(t₁+τ) 0.0225 + 0.0348 = 0.01042(t₁+τ) → t₁+τ = 5.50 hr

For 250 nm total: (0.25)² + 0.232(0.25) = 0.01042(t₂+τ) 0.0625 + 0.058 = 0.01042(t₂+τ) → t₂+τ = 11.57 hr

Additional time = 11.57 - 5.50 = 6.07 hr Dry oxidation is much slower than wet — consistent with thinner, higher-quality gate oxides.

Question

24. At what temperature would wet oxidation of ⟨100⟩ Si reach B/A = 1 µm/hr? (D₀ = 9.70×10⁷ µm/hr, Eₐ = 2.05 eV)

Answer

1 = 9.70×10⁷ × exp(-2.05/(kT)) exp(-2.05/(kT)) = 1.031×10⁻⁸ -2.05/(kT) = ln(1.031×10⁻⁸) = -18.39 kT = 2.05/18.39 = 0.1115 eV T = 0.1115/(8.617×10⁻⁵) = 1294 K = 1021°C