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TGBIE · 2nd Year · MPC & BiPC

Physics — Important Derivations

Step-by-step derivation outlines for the most frequently asked derivations in the TS Inter 2nd Year Physics board exam. Practise writing each derivation in full — including diagrams, numbered equations, and the final boxed result.

Optics (Ch 2 & 3)

1. Mirror formula: 1/f = 1/v + 1/u

  1. Draw concave mirror with pole P, centre C, focus F
  2. Take object AB beyond C; draw two rays — one parallel to principal axis (reflects through F), one through C (reflects back)
  3. Use similar triangles △APB and △APB′ to get A′B′/AB = PA′/PA
  4. Apply sign convention; obtain v and u in terms of f → derive 1/f = 1/v + 1/u
  5. Result: f = R/2

2. Lens maker's equation: 1/f = (n−1)(1/R₁ − 1/R₂)

  1. Apply refraction at first surface (radius R₁): n₁/u + n₂/v₁ = (n₂−n₁)/R₁
  2. Apply refraction at second surface (radius R₂) using v₁ as virtual object: n₂/−v₁ + n₁/v = (n₁−n₂)/R₂
  3. Add both equations; for n₁=1 (air), n₂=n: 1/f = (n−1)(1/R₁ − 1/R₂)

3. Young's double slit — fringe width β = λD/d

  1. S₁ and S₂ separated by d; screen at distance D
  2. Path difference Δ = S₂P − S₁P = d·y/D (for small angles)
  3. Bright fringe at y_n = nλD/d; Dark fringe at y_n = (2n−1)λD/2d
  4. Fringe width: β = y_(n+1) − y_n = λD/d

4. Prism — minimum deviation relation n = sin[(A+δₘ)/2]/sin(A/2)

  1. At minimum deviation: i₁ = i₂ and r₁ = r₂ = A/2
  2. i₁ = (A + δₘ)/2 from A = r₁ + r₂ and δₘ = 2i − A
  3. Apply Snell's law at first surface: sin(i₁) = n·sin(r₁)
  4. Substitute: n = sin[(A+δₘ)/2]/sin(A/2)

Electrostatics & Current (Ch 4–6)

1. Electric field due to an infinite sheet: E = σ/2ε₀

  1. Choose a Gaussian cylinder with flat ends perpendicular to the sheet
  2. By symmetry, E is perpendicular to the sheet on both sides
  3. Gauss's law: 2·E·A = σA/ε₀
  4. Therefore E = σ/(2ε₀)

2. Energy stored in a capacitor: U = ½CV²

  1. Work done dW to add charge dq against potential V = q/C is dW = (q/C)dq
  2. Total work: W = ∫₀^Q (q/C)dq = Q²/(2C)
  3. Since Q = CV: U = ½CV² = Q²/2C

3. Wheatstone bridge balance condition: P/Q = R/S

  1. Four resistors P, Q, R, S in a diamond; galvanometer across BD, battery across AC
  2. At balance: I_G = 0, so V_B = V_D
  3. Applying Kirchhoff: V_A − V_B = I₁P and V_B − V_C = I₁Q
  4. Also V_A − V_D = I₂R and V_D − V_C = I₂S
  5. Dividing: P/Q = R/S

Magnetic Effects (Ch 7)

1. Magnetic field at centre of circular current loop: B = μ₀I/(2R)

  1. Each element dl of the circular loop at distance R from centre
  2. Biot-Savart: dB = μ₀I dl sin90°/(4πR²) = μ₀I dl/(4πR²)
  3. Integrate over full circle (dl = 2πR): B = μ₀I(2πR)/(4πR²) = μ₀I/(2R)
  4. Direction: right-hand rule perpendicular to loop plane

2. Force on current in uniform magnetic field: F = BIL sinθ

  1. Current I in wire of length L; carriers each have drift velocity v_d
  2. Force on one carrier: f = qv_dB sinθ
  3. Number of carriers in length L: n·A·L (n = carrier density)
  4. Total force: F = nALq·v_d·B sinθ = IL·B sinθ (since I = nAqv_d)

3. Radius of circular motion in magnetic field: r = mv/(qB)

  1. Magnetic force provides centripetal force: qvB = mv²/r
  2. Solving for r: r = mv/(qB)
  3. Time period: T = 2πr/v = 2πm/(qB) — cyclotron frequency

EMI, AC & EM Waves (Ch 9–11)

1. Faraday's law: ε = −dΦ/dt

  1. Magnetic flux Φ = B·A·cosθ
  2. An induced EMF opposes the change in flux (Lenz's law)
  3. ε = −dΦ/dt; for N turns: ε = −N·dΦ/dt

2. Motional EMF: ε = Bvl

  1. Conducting rod of length l moving with velocity v perpendicular to field B
  2. Free charges experience force F = qvB along the rod
  3. Work done in moving charge q from one end to other: W = qvBl
  4. EMF = W/q = Bvl

3. LCR series circuit — impedance Z = √(R² + (X_L−X_C)²)

  1. V_R = IR (in phase with I)
  2. V_L = IX_L (leads current by 90°), V_C = IX_C (lags current by 90°)
  3. Phasor addition: V = √(V_R² + (V_L−V_C)²)
  4. Dividing by I: Z = √(R² + (X_L−X_C)²)

Modern Physics (Ch 12–14)

1. Bohr's model — radii: rₙ = n²a₀; energy: Eₙ = −13.6/n² eV

  1. Centripetal force = Coulomb attraction: mv²/r = e²/(4πε₀r²)
  2. Angular momentum quantisation: mvr = nh/(2π)
  3. From these two: r_n = n²ε₀h²/(πme²) = n² × 0.529 Å
  4. Total energy: E = KE + PE = −e²/(8πε₀r_n) = −13.6/n² eV

2. Radioactive decay law: N = N₀e^(−λt)

  1. Rate of disintegration: dN/dt = −λN
  2. Separate variables: dN/N = −λ dt
  3. Integrate: ln(N/N₀) = −λt
  4. Therefore: N = N₀e^(−λt)
  5. Half-life: T₁/₂ = ln2/λ = 0.693/λ

Semiconductors (Ch 15)

1. Transistor current relations: I_E = I_B + I_C; β = α/(1−α)

  1. Emitter current splits between base and collector: I_E = I_B + I_C
  2. DC current gain α = I_C/I_E (common-base, < 1)
  3. DC current gain β = I_C/I_B (common-emitter, large)
  4. Relation: α = β/(β+1); β = α/(1−α)
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