Derivations of Physics Class 12 | Convex classes Jaipur

Convex Classes ki taraf se ek complete aur exam-oriented blog jisme Class 12 Physics (CBSE/NCERT) ke sabhi board exam mein baar-baar puche jaane wale derivations step-by-step, easy language mein cover kiye gaye hain. Yeh blog specially un students ke liye banaya gaya hai jo last-minute revision karna chahte hain ya derivations ko concept ke saath samajhna chahte hain — na ki sirf ratt-na.

1. Electrostatics

1.1 Electric Field Due to a Dipole – Axial Position (End-on Position)

Ek electric dipole mein do equal and opposite charges (+q aur −q) hote hain jo 2a distance par separated hote hain. Dipole moment: p = q × 2a

Point P, dipole ke axial line par center O se distance r par hai.

• Field due to +q (at distance r − a): E₁ = kq / (r − a)² (P ki taraf)
• Field due to −q (at distance r + a): E₂ = kq / (r + a)² (−q ki taraf)

Dono fields same line par hain lekin opposite direction mein, aur E₁ > E₂ (kyunki r−a < r+a), isliye resultant dipole moment ki direction mein hoga:

E_axial = E₁ − E₂ = kq [1/(r−a)² − 1/(r+a)²]

Simplify karne par:

E_axial = kq × [4ar / (r² − a²)²]

Since p = 2aq:

E_axial = 2kpr / (r² − a²)²

Jab r >> a (short dipole approximation), a² ko ignore kar sakte hain:

E_axial = 2kp / r³ = (1/4πε₀) × (2p/r³)

1.2 Electric Field Due to a Dipole – Equatorial Position (Broadside-on)

Point P dipole ke equatorial line par O se distance r par hai. Dono charges se distance equal hoga = √(r² + a²)

• E₁ (due to +q) aur E₂ (due to −q) dono magnitude mein equal: kq/(r²+a²)
• In dono ke components jo axis ke perpendicular hain wo cancel ho jaate hain, aur jo axis ke parallel hain (anti-parallel to p direction) wo add ho jaate hain.

Resultant:

E_equatorial = 2 × [kq/(r²+a²)] × cos θ, jahan cos θ = a/√(r²+a²)

E_equatorial = 2kqa / (r² + a²)^(3/2) = kp / (r² + a²)^(3/2)

Jab r >> a:

E_equatorial = kp/r³ = (1/4πε₀) × (p/r³)

📌 Important result: E_axial = 2 × E_equatorial (same distance r par)

1.3 Torque on an Electric Dipole in a Uniform Electric Field

Dipole (charges +q aur −q, separation 2a) ko uniform field E mein rakha gaya hai, jisme dipole moment p field se θ angle par hai.

• Force on +q = qE (field ki direction mein)
• Force on −q = qE (opposite direction mein)

Ye do equal and opposite forces ek couple banate hain jo alag-alag lines of action par act karte hain — isliye torque generate hota hai.

Perpendicular distance between the two forces = 2a sin θ

Torque, τ = Force × perpendicular distance = qE × 2a sin θ

Since p = q × 2a:

τ = pE sin θ

Vector form: τ = p × E

1.4 Capacitance of a Parallel Plate Capacitor (Without Dielectric)

Do parallel plates, area A, separation d, charge Q aur −Q.

Electric field between plates (using Gauss’s law): E = σ/ε₀ = Q/(Aε₀)

Potential difference between plates:

V = E × d = Qd/(Aε₀)

Capacitance definition: C = Q/V

C = ε₀A/d

1.5 Capacitance with a Dielectric Slab (thickness t < d)

Dielectric slab (dielectric constant K) plates ke beech rakha jaata hai, thickness t.

Field inside dielectric: E’ = E/K, isliye potential difference:

V = E(d − t) + (E/K)t = E[d − t + t/K]

Since E = Q/(Aε₀):

C = Aε₀ / [d − t(1 − 1/K)]

Agar poora gap dielectric se bhara ho (t = d):

C = Kε₀A/d

1.6 Energy Stored in a Charged Capacitor

Capacitor ko charge karte waqt chhote-chhote dq charge ko transfer karna padta hai jab plates ke beech potential difference V’ = q/C ho.

Small work done: dW = V’ dq = (q/C) dq

Total work (0 se Q tak integrate karke):

W = ∫₀^Q (q/C) dq = Q²/(2C)

Since Q = CV:

U = Q²/2C = ½CV² = ½QV

2. Current Electricity

2.1 Relation Between Current and Drift Velocity

Conductor mein free electrons random motion karte hain, lekin field lagne par unme ek net drift velocity vd generate hoti hai.

Consider conductor with length L, area A, free electron density n.

Time t mein electrons jitni distance cover karte hain: vd × t

Volume jisme electrons current mein contribute karte hain: A × vd × t

Total electrons in this volume = n × A × vd × t

Total charge, Q = n × A × vd × t × e

Current: I = Q/t

I = nAevd

2.2 Ohm’s Law – Microscopic (Derivation from Drift Velocity)

Electric field E electron par force F = eE lagata hai.

Acceleration: a = eE/m

Average relaxation time τ ke baad drift velocity:

vd = aτ = eEτ/m

Substituting in I = nAevd:

I = nAe × (eEτ/m) = nAe²Eτ/m

Since E = V/L:

I = nAe²τV/(mL)

V = I × [mL/(nAe²τ)] = IR

Jahan Resistance, R = mL/(nAe²τ), aur Resistivity, ρ = m/(ne²τ)

Isse yeh prove hota hai ki V ∝ I, jo Ohm’s Law hai.

2.3 Wheatstone Bridge – Balance Condition

Wheatstone bridge mein 4 resistances P, Q, R, S is tarah arranged hote hain ki galvanometer (G) do junctions ke beech connected hota hai, aur bridge “balanced” condition mein galvanometer se koi current nahi guzarta (Ig = 0).

Kirchhoff’s junction aur loop rule use karke, balanced condition mein:

Loop ABDA: I₁P = I₂R Loop BCDB: I₁Q = I₂S

In dono equations ko divide karne par:

P/Q = R/S

Yahi Wheatstone bridge ki balance condition hai, jo unknown resistance nikalne mein use hoti hai.

3. Moving Charges and Magnetism

3.1 Magnetic Field Due to a Circular Current Loop (On its Axis) – Biot-Savart Law

Circular loop of radius R, current I, point P axis par center se distance x par.

Biot-Savart Law: dB = (μ₀/4π) × (I dl × r̂)/r²

Har current element dl se dB perpendicular to r hota hai, jiske do components hote hain: dB cos φ (axis ke along) aur dB sin φ (axis ke perpendicular, jo symmetry ki wajah se cancel ho jaata hai)

Sirf axial components add hote hain:

B = ∫ dB cos φ = ∫ (μ₀I dl)/(4π(R²+x²)) × R/√(R²+x²)

Integrating dl around full loop (2πR):

B = μ₀IR² / [2(R² + x²)^(3/2)]

Center par (x = 0): B = μ₀I/2R

3.2 Magnetic Field Due to a Long Straight Current-Carrying Wire (Ampere’s Circuital Law)

Ampere’s Law: ∮ B·dl = μ₀I_enclosed

Wire ke around radius r ka circular Amperian loop lo. Symmetry ki wajah se B har point par same magnitude aur tangential direction mein hota hai.

∮ B·dl = B × (2πr) = μ₀I

B = μ₀I / (2πr)

3.3 Magnetic Field Inside a Long Solenoid

Solenoid: length L, total turns N, current I, n = N/L (turns per unit length).

Rectangular Amperian loop lo jiska ek side solenoid ke andar (length l) aur doosra bahar (jahan B ≈ 0).

∮ B·dl = B × l (sirf andar wale side se contribution, kyunki bahar B = 0 aur sides B ke perpendicular)

Enclosed current = (n × l) × I [kyunki loop ke andar n×l turns hain]

B × l = μ₀ × n × l × I

B = μ₀nI

3.4 Force Between Two Parallel Current-Carrying Conductors

Do parallel wires, currents I₁ aur I₂, distance d apart.

Wire 1 ki wajah se wire 2 ki location par field: B₁ = μ₀I₁/(2πd)

Wire 2 (length L) par force: F = B₁I₂L = [μ₀I₁/(2πd)] × I₂L

Force per unit length, F/L = μ₀I₁I₂ / (2πd)

Same direction ki currents mein force attractive hoti hai, opposite direction mein repulsive.

3.5 Torque on a Current Loop in a Uniform Magnetic Field (Moving Coil Galvanometer Principle)

Rectangular loop (sides l aur b, area A = l×b), N turns, current I, magnetic field B, normal to loop plane ka field ke saath angle θ.

Do sides (length l) par forces F = BIl lagti hain, jo opposite direction mein hoti hain aur ek couple banati hain.

Perpendicular distance between forces = b sin θ

Torque (single turn): τ = BIl × b sin θ = BIA sin θ

N turns ke liye:

τ = NBIA sin θ

Moving coil galvanometer mein radial field use karne se θ = 90° hamesha rahta hai, isliye:

τ = NBIA (deflecting torque, jo spring ke restoring torque ke barabar hoti hai: kφ = NBIA)

4. Electromagnetic Induction

4.1 Motional EMF (Rod Moving in Magnetic Field)

Ek conducting rod PQ, length l, uniform magnetic field B (page ke andar) mein velocity v se move karti hai, B aur v dono perpendicular hain.

Rod ke free electrons par magnetic force lagti hai: F = qvB, jisse electrons ek end par accumulate hote hain aur ek potential difference banti hai.

Equilibrium par, electric force electron ko magnetic force ke against balance karti hai:

qE = qvB → E = vB

EMF = E × l (kyunki EMF = potential difference across length l)

ε = Bvl

4.2 Self Inductance of a Long Solenoid

Solenoid: N turns, length L, area A, current I.

Magnetic field inside: B = μ₀nI = μ₀(N/L)I

Flux through one turn: Φ₁ = B×A = μ₀(N/L)IA

Total flux linkage: NΦ₁ = μ₀N²IA/L

Self-inductance defined by: NΦ₁ = LI

L = μ₀N²A/L (yahan doosra L length hai, formula: L = μ₀N²A/l)

Agar core mein relative permeability μr ho:

L = μ₀μrN²A/l

4.3 Energy Stored in an Inductor

Jab inductor mein current 0 se I tak badhta hai, back EMF ke against work karna padta hai.

Instantaneous EMF: ε = L(di/dt)

Power = εi = Li(di/dt)

Total work done (energy stored):

W = ∫₀^I Li di = L[i²/2]₀^I

U = ½LI²

5. Alternating Current

5.1 Impedance of a Series LCR Circuit

R, L, C series mein connected, AC source V = V₀ sin ωt.

• Voltage across R: VR = IR (current ke saath in phase)
• Voltage across L: VL = IXL (current se 90° aage), XL = ωL
• Voltage across C: VC = IXC (current se 90° peeche), XC = 1/ωC

VL aur VC opposite phase mein hote hain, isliye net reactive voltage = I(XL − XC)

Phasor diagram se, resultant voltage:

V = √[VR² + (VL−VC)²] = I√[R² + (XL−XC)²]

Since V = IZ:

Z = √[R² + (XL − XC)²]

Phase angle: tan φ = (XL − XC)/R

5.2 Resonance Frequency of Series LCR Circuit

Resonance us condition mein hota hai jab XL = XC (impedance minimum, sirf R reh jaata hai):

ωL = 1/ωC

ω² = 1/LC

ω₀ = 1/√(LC), isliye f₀ = 1/(2π√(LC))

6. Ray Optics

6.1 Refraction at a Single Spherical Surface

Rarer medium (n₁) se denser medium (n₂) mein refraction ho raha hai, spherical surface radius R, object distance u, image distance v (sign convention ke saath).

Small angle approximation use karke, Snell’s law: n₁ sin i = n₂ sin r ≈ n₁i = n₂r

Geometry se angles ko object distance, image distance aur radius of curvature se relate karte hue:

i = angle NOM + angle NCM (approx), aur similar relations use karke, final result milta hai:

n₂/v − n₁/u = (n₂ − n₁)/R

Yeh formula lens maker’s formula derive karne ka base hai.

6.2 Lens Maker’s Formula

Thin lens do refracting surfaces se bana hota hai (radii R₁ aur R₂), refractive index n₂, medium n₁.

Surface 1 (refraction n₁ → n₂) ke liye: n₂/v₁ − n₁/u = (n₂−n₁)/R₁

Surface 2 (refraction n₂ → n₁, image v₁ ab object ban jaata hai) ke liye: n₁/v − n₂/v₁ = (n₁−n₂)/R₂

In dono equations ko add karne par v₁ cancel ho jaata hai:

n₁/v − n₁/u = (n₂−n₁)[1/R₁ − 1/R₂]

Divide by n₁, aur relative refractive index n₂₁ = n₂/n₁ use karke:

1/v − 1/u = (n₂₁ − 1)[1/R₁ − 1/R₂]

Jab u = ∞, v = f:

1/f = (n₂₁ − 1)[1/R₁ − 1/R₂]

Yehi Lens Maker’s Formula hai.

6.3 Refraction Through a Prism (Prism Formula)

Prism ka refracting angle A, incident ray se deviation δ.

Geometry se: A = r₁ + r₂ (dono refraction angles ka sum, prism ke andar)

Total deviation: δ = (i − r₁) + (e − r₂) = i + e − A

Minimum deviation (δm) ki condition mein i = e aur r₁ = r₂ = A/2, aur δ = δm:

A + δm = 2i → i = (A+δm)/2

r = A/2

Snell’s law apply karke:

n = sin[(A+δm)/2] / sin(A/2)

Yeh Prism Formula hai, jo minimum deviation experiment mein refractive index nikalne ke liye use hota hai.

7. Wave Optics

7.1 Young’s Double Slit Experiment – Fringe Width Derivation

Do slits S₁ aur S₂, separation d, screen distance D (D >> d), wavelength λ.

Point P screen par center se distance y par hai. Path difference:

Δ = S₂P − S₁P

Geometry (D >> d approximation) se:

Δ = yd/D

Constructive interference (bright fringe): Δ = nλ → y_n = nλD/d

Destructive interference (dark fringe): Δ = (n+½)λ → y_n = (n+½)λD/d

Fringe width (distance between two consecutive bright/dark fringes):

β = y_(n+1) − y_n = λD/d

8. Dual Nature of Radiation and Matter

8.1 de Broglie Wavelength

Einstein ke mass-energy relation aur Planck’s quantum theory ko combine karke, de Broglie ne propose kiya ki matter bhi wave properties dikhata hai.

Photon ke liye: E = hν = hc/λ, aur E = pc (photon momentum se energy)

Isse: p = h/λ

De Broglie ne extend kiya ki yeh relation kisi bhi particle (mass m, velocity v) ke liye valid hai:

λ = h/p = h/(mv)

Agar particle ko potential V se accelerate kiya jaaye (KE = qV):

p = √(2mqV)

λ = h/√(2mqV)

Electron ke liye (q = e):

λ = h/√(2meV) = 12.27/√V Å

9. Atoms and Nuclei

9.1 Bohr Model – Radius of nth Orbit

Electron nucleus ke around circular orbit mein ghoomta hai. Coulomb force provides centripetal force:

kZe²/r² = mv²/r → mv² = kZe²/r … (1)

Bohr’s quantization condition: mvr = nh/2π → v = nh/(2πmr) … (2)

(2) ko (1) mein substitute karke:

m × [n²h²/(4π²m²r²)] = kZe²/r

r = n²h²/(4π²mkZe²)

r_n = n²h²ε₀ / (πmZe²) (using k = 1/4πε₀)

Hydrogen (Z=1) ke liye n=1 par, r₁ = 0.529 Å (Bohr radius)

9.2 Bohr Model – Energy of Electron in nth Orbit

Total Energy = Kinetic Energy + Potential Energy

KE = ½mv² = kZe²/2r (from equation 1 above)

PE = −kZe²/r (electrostatic potential energy, attractive hone ki wajah se negative)

Total Energy, E = KE + PE = −kZe²/2r

r_n ki value substitute karne par:

E_n = −(mZ²e⁴)/(8ε₀²n²h²) = −13.6 Z²/n² eV (hydrogen ke liye)

Negative sign yeh dikhata hai ki electron nucleus se bound hai.

9.3 Radioactive Decay Law

Radioactive substance mein disintegration rate, us waqt maujood nuclei ki sankhya N ke proportional hoti hai:

−dN/dt ∝ N → −dN/dt = λN

Jahan λ = decay constant

Variables separate karke integrate karne par (N₀ se N tak, 0 se t tak):

∫(dN/N) = −λ∫dt

ln(N/N₀) = −λt

N = N₀e^(−λt)

Half-life (T½) us waqt milta hai jab N = N₀/2:

T½ = 0.693/λ

Conclusion

Physics ke derivations sirf yaad karne ki cheez nahi hain — inhe concept ke saath samajhna zaroori hai, kyunki board exam mein aksar values change karke ya diagram ke saath application-based question puche jaate hain. Har derivation ko:

1. Diagram bana kar samjho
2. Steps ko logically follow karo (kis law ya principle se start ho raha hai)
3. Khud se 2-3 baar likh kar practice karo
4. Final formula ko underline/highlight karke yaad rakho

Convex Classes mein hum Class 12 Physics ke sabhi chapters ko concept-clarity ke saath padhate hain, taaki derivations rattene ki jagah samajh mein aayein. Agar aapko in derivations par doubt clear karne hain ya live classes join karni hain, toh Convex Classes se judiye.