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Derivation Series · Session 4 · NEET Physics

Centre of Mass, Rotation, Waves & More

COM & Collisions · Rotational Motion · Elasticity · Viscosity · Wave Optics · AC Circuits
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1. Centre of Mass & Collisions

💡 Why this matters in real life

Have you ever balanced a ruler on your finger? The point where it balances perfectly is its centre of mass — the point where the entire weight of the object seems to act. When two cars collide, the way they move after the crash depends on their centres of mass and whether the collision is elastic (they bounce off, like pool balls) or inelastic (they stick together, like clay balls). NEET loves asking: "Where is the COM of this system?" and "What happens after a collision?"

🧠 The Big Idea — In Plain Words

Think of the centre of mass as the "average position" of all the mass in a system. For a system of particles, it's like calculating the weighted average of their positions — heavier particles pull the COM closer to them. For collisions, the total momentum before = total momentum after (momentum conservation) is the master key. Whether kinetic energy is also conserved depends on the type of collision.

Key variables: m = mass (kg), v = velocity (m/s), x = position (m), e = coefficient of restitution (bounciness factor)

Two masses m₁, m₂ and their centre of mass (animated) m₁ x₁ m₂ x₂ COM X = (m₁x₁ + m₂x₂)/(m₁+m₂)
Fig 1.1: The COM is closer to the heavier mass. Masses oscillate slightly — COM stays balanced at the weighted average.

🔍 Step-by-Step: Centre of Mass of a Two-Particle System

1
Imagine two masses on a line. Suppose m₁ sits at position x₁ and m₂ sits at position x₂. The COM is like the "balance point" of a seesaw. A heavier child needs to sit closer to the pivot to balance a lighter child farther away. Mathematically, the COM is the weighted average of positions, where masses are the weights.
X = (m₁x₁ + m₂x₂) / (m₁ + m₂)
Why: Think of it like calculating your final grade — if one exam is worth more (heavier), it pulls your average toward its score. Here, heavier masses pull the COM toward themselves.
2
For a system of many particles in 2D or 3D. The formula extends naturally — we do the same weighted average for x, y, and z coordinates separately.
X = (Σmᵢxᵢ)/M,   Y = (Σmᵢyᵢ)/M,   Z = (Σmᵢzᵢ)/M where M = Σmᵢ is the total mass.
Why: Σ (sigma) means "sum of". We add up all (mass × position) for every particle, then divide by total mass. Each coordinate is independent.
3
Velocity of COM. If the particles are moving, the COM moves too. Differentiate the position formula with respect to time:
V_COM = (m₁v₁ + m₂v₂ + ...) / (m₁ + m₂ + ...)
Why: This is the "average velocity" weighted by mass. The numerator Σ(mv) is the total momentum of the system.
4
Key consequence — momentum of COM. Total momentum P_total = M × V_COM. This means: if no external force acts, V_COM stays constant (momentum conservation). This is why an exploding fireworks shell's shrapnel goes everywhere, but the COM continues along the original parabolic path!
P_total = MV_COM = constant (if F_ext = 0)
COM position: X = Σmᵢxᵢ / Σmᵢ    COM velocity: V_COM = Σmᵢvᵢ / Σmᵢ
🔍 Common Mistake
Students often forget that COM can be outside the object. Example: a ring's COM is at its centre (empty space), not on the ring itself. For symmetric objects (sphere, cube, disc), COM is at the geometric centre.
Elastic collision — equal masses exchange velocities (animated) m₁ m₂ u₁ → ← u₂ Impact — momentum conserved
Fig 1.2: Equal masses collide elastically — they exchange velocities. Watch the balls approach, collide, and bounce.

🔍 Step-by-Step: Elastic and Inelastic Collisions

1
What is a collision? Two objects hit each other and exchange momentum. The total momentum always stays the same (Newton's third law — equal and opposite forces). But kinetic energy may or may not be conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (momentum conservation, ALWAYS holds)
Where u = initial velocities, v = final velocities. This is our starting equation for any collision.
2
Elastic collision — energy is also conserved. Think of two pool balls colliding — they bounce off without losing energy as heat or sound. We add one more equation:
½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²
Why: "Elastic" means no energy lost. The total kinetic energy before and after is the same.
3
Solve for final velocities. From the two equations above, we can derive neat formulas for v₁ and v₂. First, rewrite momentum and energy equations, divide the energy equation by the momentum equation:
v₁ = [(m₁ - m₂)/(m₁ + m₂)]u₁ + [2m₂/(m₁ + m₂)]u₂
v₂ = [2m₁/(m₁ + m₂)]u₁ + [(m₂ - m₁)/(m₁ + m₂)]u₂
Don't memorise these — remember two special cases instead (next step).
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Special cases you MUST know for NEET.
Case 1 — Equal masses (m₁ = m₂): They simply exchange velocities. v₁ = u₂, v₂ = u₁. Like two identical pool balls — the first one stops, the second one moves with the first one's speed.
Case 2 — One body is stationary (u₂ = 0): v₁ = (m₁ - m₂)u₁/(m₁ + m₂), v₂ = 2m₁u₁/(m₁ + m₂). If m₁ >> m₂ (truck hits a ball), the truck barely slows down and the ball flies off at ~2× the truck's speed.
Case 3 — Perfectly inelastic (they stick): Both move with same velocity after collision: v = (m₁u₁ + m₂u₂)/(m₁ + m₂)
5
Coefficient of restitution (e). This single number tells you how bouncy the collision is:
e = (v₂ - v₁) / (u₁ - u₂) = relative speed after / relative speed before
e = 1 → perfectly elastic (super-bouncy), e = 0 → perfectly inelastic (they stick), 0 < e < 1 → real-world collisions.
Momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂    |    Restitution: e = (v₂ - v₁)/(u₁ - u₂)
🎯 NEET Pattern — Collisions
• NEET often gives two masses colliding and asks for final velocities or fraction of KE lost.
• KE lost in inelastic collision: ΔK = ½μ(u₁ - u₂)²(1 - e²) where μ = m₁m₂/(m₁+m₂) is reduced mass.
• For perfectly inelastic (e=0), ΔK = ½μ(u₁ - u₂)² — the maximum possible KE loss.
• Typical question: "A 2 kg block moving at 4 m/s collides with a stationary 6 kg block. If they stick together, find the common velocity and energy lost."
Example 1: COM + Collision
A 2 kg mass at x = 1 m and a 3 kg mass at x = 6 m lie on a frictionless surface. Find their COM. If the 2 kg mass moves right at 5 m/s toward the stationary 3 kg mass and they stick together, find the final speed.

a) COM = 3 m, v = 2 m/s   b) COM = 4 m, v = 2.5 m/s   c) COM = 3.5 m, v = 3 m/s   d) COM = 4.5 m, v = 2 m/s
Solution: COM = (2×1 + 3×6)/(2+3) = (2+18)/5 = 20/5 = 4 m. After collision (sticking), v = (2×5 + 3×0)/(2+3) = 10/5 = 2 m/s. Option (b).

2. Rotational Motion — Moment of Inertia & Torque

💡 Why this matters in real life

Why is it harder to push a merry-go-round from the edge than from near the centre? Why does an ice skater spin faster when she pulls her arms in? These questions are answered by rotational motion. Just as mass resists changes in linear motion (inertia), moment of inertia resists changes in rotational motion. Torque is the "rotational version of force" — it's what makes things spin.

🧠 The Big Idea — In Plain Words

In linear motion, F = ma (force = mass × acceleration). In rotation, the equivalent is τ = Iα (torque = moment of inertia × angular acceleration). The moment of inertia I depends not just on how much mass there is, but on how far the mass is from the axis of rotation — farther mass contributes more to I. That's why a figure skater pulling her arms in (bringing mass closer to the axis) reduces her I and makes her spin faster.

Key variables: I = moment of inertia (kg·m²), τ = torque (N·m), α = angular acceleration (rad/s²), ω = angular velocity (rad/s), L = angular momentum (kg·m²/s)

Rotating mass — moment of inertia I = mr² (spinning animation) Axis m r ω
Fig 2.1: A mass m rotating about an axis. Doubling r quadruples I (since I ∝ r²), making it 4× harder to change rotation.

🔍 Step-by-Step: Moment of Inertia of a Ring and a Disc

1
Definition of Moment of Inertia. For a single particle of mass m at distance r from the axis: I = mr². For a collection of particles: I = Σmᵢrᵢ².
Why r²? Because torque depends on r × F, and when you work through the rotational version of F=ma, the r² appears naturally. The farther the mass, the more "leverage" it has to resist spinning changes.
2
Ring about its centre (axis through centre, perpendicular to plane). A ring has ALL its mass at the same distance R from the centre. Every tiny piece of mass dm is at distance R:
I_ring = ∫dm·R² = R²∫dm = R² × M = MR²
Why: Since R is the same for every piece, it comes out of the integral. We just add up all the mass — which is M. Simple!
3
Disc about its centre. A disc is like many thin rings stacked from radius 0 to R. Each thin ring of radius r has mass dm = (M/πR²)·2πr·dr = (2Mr/R²)·dr (area of ring = 2πr·dr). Its contribution to I is dm·r²:
I_disc = ∫₀ᴿ r²·dm = ∫₀ᴿ r²·(2Mr/R²)dr = (2M/R²)∫₀ᴿ r³dr = (2M/R²) × [r⁴/4]₀ᴿ = (2M/R²)×(R⁴/4) = ½MR²
Why half of MR²? Because the mass near the centre (small r) contributes much less than mass at the rim. The average r² works out to R²/2 for a disc. A ring has all mass at R, so I = MR² (larger).
Ring: I = MR²    Disc: I = ½MR²    Solid sphere: I = ⅖MR²    Rod (centre): I = ML²/12

🔍 Step-by-Step: Torque and Angular Momentum

1
Torque is the rotational cousin of force. If you push a door at the handle (far from hinge), it opens easily. Push near the hinge — much harder. Torque = force × perpendicular distance from axis: τ = r × F = r·F·sinθ
Only the perpendicular component of force produces torque. Pushing directly toward the axis (θ = 0) produces zero torque — that's why you can't open a door by pushing along its surface toward the hinge.
2
Newton's Second Law for rotation. Just as F = ma (force causes linear acceleration), torque causes angular acceleration: τ = Iα
Proof: τ = r·F = r·(m·a_tangential) = r·m·(rα) = mr²·α = Iα. The tangential acceleration a = rα links linear and angular motion.
3
Angular momentum L. The rotational equivalent of linear momentum p = mv: L = Iω (for a point mass: L = mvr).
L = Iω = constant    (if no external torque)
This is the conservation of angular momentum — the ice skater effect. When she pulls her arms in, I decreases, so ω must increase to keep L constant. This is why she spins faster!
4
Relation between torque and angular momentum. τ = dL/dt (torque = rate of change of angular momentum). This is exactly like F = dp/dt in linear motion.
If no torque acts, L doesn't change — it's conserved. This explains why a spinning top stays upright (its angular momentum keeps it stable).
τ = Iα    L = Iω    τ = dL/dt    Rolling KE = ½Iω² + ½mv²
🔍 Common Mistake
Students confuse moment of inertia with mass. I depends on BOTH mass AND how it's distributed. A ring and a disc of the same mass M and radius R have different I (MR² vs ½MR²). The ring is harder to spin because all its mass is at the rim.
🎯 NEET Pattern — Rotational Motion
• Rolling without slipping: v = ωR. Total KE = ½mv² + ½Iω² = ½mv²(1 + k²/R²) where k is radius of gyration.
• Objects of different shapes roll down an incline — which reaches bottom first? The one with smallest I/mR² (sphere wins, ring loses).
• Conservation of angular momentum is very common — a rotating platform + person walking inward = ω increases.
Example 2: Rotational Motion
A disc of mass 2 kg and radius 0.5 m is rotating at 10 rad/s. A ring of same mass and radius is gently placed coaxially on it. Find the new angular speed. (I_disc = ½MR², I_ring = MR²)

a) 5 rad/s   b) 3.33 rad/s   c) 6.67 rad/s   d) 4 rad/s
Solution: Angular momentum conserved. Initial L = I_disc·ω₀ = (½×2×0.25)×10 = 0.25×10 = 2.5. Final I = I_disc + I_ring = 0.25 + 0.5 = 0.75. ω = L/I = 2.5/0.75 = 3.33 rad/s. Option (b).

3. Elasticity — Stress, Strain & Young's Modulus

💡 Why this matters in real life

When you stretch a rubber band, it gets longer. When you release it, it snaps back. This ability to deform and return to original shape is called elasticity. Bridges, buildings, and even your bones are designed using elasticity — engineers must know how much a steel cable will stretch under a given load. NEET asks about the relationship between stress (force per area) and strain (fractional change in length), and the famous Young's modulus that measures a material's stiffness.

🧠 The Big Idea — In Plain Words

Imagine pulling a spring. The more force you apply, the more it stretches. For most materials, within a limit, the stretch is proportional to the force (Hooke's law). Stress is the "cause" (how hard you pull per unit area), strain is the "effect" (how much it deforms relative to its original size). Young's modulus Y = stress/strain tells you how stiff a material is — steel has high Y (hard to stretch), rubber has low Y (easy to stretch).

Key variables: σ = stress (N/m² or Pa), ε = strain (no units, just a ratio), Y = Young's modulus (Pa), F = force (N), A = area (m²), ΔL = change in length (m), L₀ = original length (m)

A wire under tension — stress causes strain Support L₀ L₀ + ΔL F (load) Area A
Fig 3.1: A wire of original length L₀ stretches by ΔL when a force F is applied. The cross-sectional area A determines the stress.

🔍 Step-by-Step: Young's Modulus and Its Derivation

1
Stress — the "cause". When you apply a force F to a wire of cross-sectional area A, the material experiences stress. Think of it as "internal pressure" resisting the force: Stress (σ) = F / A
Why area A? A thicker wire distributes the same force over more atoms, so each atom feels less stress. Unit: N/m² = Pa (pascal).
2
Strain — the "effect". The wire changes length by ΔL compared to its original length L₀: Strain (ε) = ΔL / L₀
Strain is a pure ratio — no units. A strain of 0.01 means the wire stretched by 1% of its original length. We use ΔL/L₀ instead of just ΔL because a 1 mm stretch is significant for a 10 mm wire but negligible for a 10 m wire.
3
Hooke's Law and Young's Modulus. For small deformations, stress ∝ strain. The constant of proportionality is Young's modulus Y: Y = Stress / Strain = (F/A) / (ΔL/L₀) = F·L₀ / (A·ΔL)
This is the material's "stiffness number". Higher Y = stiffer. Steel: Y ≈ 2×10¹¹ Pa. Rubber: Y ≈ 5×10⁶ Pa — steel is about 40,000× stiffer!
4
Rearranging to find extension. From Y = FL₀/(A·ΔL), we can find how much a wire stretches: ΔL = F·L₀ / (Y·A)
This is the working formula. It says: ΔL is proportional to force and original length, but inversely proportional to stiffness (Y) and thickness (A). A longer, thinner wire of a soft material stretches the most.
Y = Stress / Strain = (F·L₀) / (A·ΔL)    Units: N/m² (Pa)

🔍 Bulk Modulus and Shear Modulus — Quick Overview

1
Bulk Modulus (K) — resistance to volume change. If you submerge an object in deep water, pressure compresses it from all sides. K = -ΔP / (ΔV/V)
The negative sign means: as pressure increases (positive ΔP), volume decreases (negative ΔV). K is always positive. For water, K ≈ 2.2×10⁹ Pa — it's very hard to compress water (that's why hydraulic brakes work!).
2
Shear Modulus (G) — resistance to shape change. When you push the top of a book sideways while holding the bottom fixed, the book deforms into a parallelogram. G = Shear stress / Shear strain = (F/A) / (Δx/h)
Shear strain = Δx/h (how much the top moves sideways divided by height). Shear modulus is important for things like rivets and bolts that experience sideways forces.
Bulk: K = -ΔP/(ΔV/V)    Shear: G = (F/A)/(Δx/h)    Compressibility = 1/K
🔍 Common Mistake
Students think strain and stress are the same thing. Remember: stress is the cause (force/area), strain is the effect (ΔL/L). Also note: Young's modulus applies ONLY to tensile/compressive stress within the elastic limit — beyond that, the material deforms permanently (plastic deformation).
🎯 NEET Pattern — Elasticity
• NEET often asks: "Two wires of same material, same length but different radii — which stretches more under same load?" (thinner one, since ΔL ∝ 1/A ∝ 1/r²).
• Breaking stress: Maximum stress a material can withstand before breaking. Breaking stress is constant for a given material; breaking force ∝ area.
• Ratio of Young's moduli from stress-strain graphs (slope = Y).
• Elongation of a wire under its own weight: ΔL = (ρgL²)/(2Y) where ρ = density.
Example 3: Young's Modulus
A 2 m long steel wire of radius 0.5 mm stretches by 1 mm when a 10 kg mass is hung from it. Find Young's modulus of steel. (g = 10 m/s²)

a) 2.0×10¹¹ Pa   b) 2.55×10¹¹ Pa   c) 1.27×10¹¹ Pa   d) 3.0×10¹¹ Pa
Solution: F = 10×10 = 100 N. A = πr² = π×(5×10⁻⁴)² = π×25×10⁻⁸ = 7.85×10⁻⁷ m². Y = FL₀/(A·ΔL) = 100×2/(7.85×10⁻⁷×10⁻³) = 200/(7.85×10⁻¹⁰) = 2.55×10¹¹ Pa. Option (b).

4. Fluid Flow — Viscosity & Terminal Velocity

💡 Why this matters in real life

Why does a small pebble sink quickly in water but a feather falls slowly? Why does honey pour more slowly than water? The answer is viscosity — the "internal friction" of a fluid. When an object falls through a fluid, it speeds up until the upward drag force balances its weight — this maximum speed is terminal velocity. Stokes' law gives us the drag force on a small sphere moving through a viscous fluid.

🧠 The Big Idea — In Plain Words

Imagine moving your hand through water — you feel a resistance, right? That's viscosity. In a viscous fluid, layers of fluid slide past each other, and there's friction between them. The coefficient of viscosity η (eta) measures how thick/sticky the fluid is. For a sphere falling through a fluid, three forces act: weight (down), buoyancy (up), and viscous drag (up). Initially, the sphere accelerates. As speed increases, drag increases until the forces balance — then speed becomes constant (terminal velocity).

Key variables: η = coefficient of viscosity (Pa·s), v_t = terminal velocity (m/s), r = radius of sphere (m), ρ_s = density of sphere, ρ_f = density of fluid

Sphere falling through viscous fluid — velocity increases until terminal velocity Viscous fluid (η) sphere F_b mg F_v speed v_t
Fig 4.1: Sphere accelerates until net force = 0 at terminal velocity. Speed indicator (right) grows as v approaches v_t.

🔍 Step-by-Step: Stokes' Law and Terminal Velocity

1
Stokes' Law — the drag force. For a small sphere moving slowly through a viscous fluid, the drag force depends on three things: radius (bigger sphere → more drag), speed (faster → more drag), and viscosity (thicker fluid → more drag). Stokes figured out the exact relation: F_v = 6πηrv
Why 6π? It comes from solving the fluid flow equations around a sphere. η is viscosity (fluid's stickiness), r is radius of sphere, v is its speed relative to the fluid.
2
What forces act on the falling sphere? Three forces. Weight pulls down: F_g = mg = ρ_s·(4/3πr³)·g. Buoyancy pushes up (Archimedes' principle — weight of displaced fluid): F_b = ρ_f·(4/3πr³)·g. Viscous drag opposes motion: F_v = 6πηrv (upward when falling down).
Buoyancy reduces the effective weight. In air, buoyancy is tiny, but in water it's significant.
3
At terminal velocity, net force = 0. The sphere stops accelerating. The three forces balance: F_g = F_b + F_v
ρ_s·(4/3πr³)·g = ρ_f·(4/3πr³)·g + 6πηrv_t
This is the key equation. We substitute all the forces and solve for v_t.
4
Solve for terminal velocity v_t. Move buoyancy to the left side: (ρ_s - ρ_f)·(4/3πr³)·g = 6πηrv_t
Cancel common factors (π, r): (ρ_s - ρ_f)·(4/3)r²g = 6ηv_t
Solve for v_t: v_t = [2r²(ρ_s - ρ_f)g] / [9η]
This is the terminal velocity formula. Key insight: v_t ∝ r² — a sphere with twice the radius falls 4× faster! That's why large raindrops fall faster than small ones.
Stokes' law: F_v = 6πηrv    Terminal velocity: v_t = 2r²(ρ_s - ρ_f)g / (9η)
🔍 Common Mistake
Stokes' law is valid only for laminar flow (slow, smooth) — not for turbulent flow (fast, chaotic). The dimensionless Reynolds number Re = ρvD/η tells you which regime you're in. Re < 2000 ≈ laminar. Also: don't forget buoyancy — it reduces terminal velocity!
🎯 NEET Pattern — Viscosity
• NEET often applies terminal velocity concept: "A spherical ball falls through a viscous liquid. After some time, its velocity becomes constant. Find the viscocity."
• Poiseuille's formula: Volume flow rate Q = πPr⁴/(8ηL) for liquid through a pipe. Note Q ∝ r⁴ — doubling radius increases flow 16×! This is why blood vessels have such a huge effect on blood pressure.
• Terminal velocity graph: speed starts at 0, increases with decreasing acceleration, and asymptotically approaches v_t.
Example 4: Terminal Velocity
A steel ball bearing (density 7800 kg/m³) of radius 2 mm falls through glycerine (density 1260 kg/m³, η = 0.83 Pa·s). Find terminal velocity. (g = 10 m/s²)

a) 0.07 m/s   b) 0.14 m/s   c) 0.21 m/s   d) 0.28 m/s
Solution: v_t = 2r²(ρ_s - ρ_f)g/(9η) = 2×(0.002)²×(7800-1260)×10/(9×0.83) = 2×4×10⁻⁶×6540×10/(7.47) = 5.232×10⁻¹/7.47 = 0.07 m/s. Option (a).

5. Wave Optics — Young's Double Slit Experiment

💡 Why this matters in real life

Have you ever seen colourful patterns on a soap bubble or an oil slick on water? That's interference of light waves. Young's double slit experiment (YDSE) in 1801 proved that light behaves as a wave — it was one of the most important experiments in physics. Two slits create two overlapping sets of light waves that either add up (constructive interference → bright fringe) or cancel out (destructive interference → dark fringe).

🧠 The Big Idea — In Plain Words

Imagine throwing two stones into a pond at the same time. Ripples spread from each stone and overlap. Where a crest meets a crest, the wave gets bigger (bright). Where a crest meets a trough, they cancel out (dark). Light does the same! In YDSE, light from a single source passes through two narrow slits. The waves from each slit travel different distances to reach a screen — this path difference determines whether they interfere constructively or destructively.

Key variables: d = slit separation, D = screen distance, λ = wavelength, β = fringe width, n = order number (0, 1, 2...)

Young's Double Slit — animated interference pattern S S₁ S₂ d D Fringes
Fig 5.1: Expanding circular wavefronts from S₁ and S₂ interfere. Bright fringes appear where waves arrive in phase.

🔍 Step-by-Step: Fringe Width Derivation

1
Path difference between the two waves. Light from the upper slit S₁ and lower slit S₂ travel different distances to reach a point P on the screen at a distance x from the centre. For D >> d (which is usually true in experiments), the path difference is approximately: Δx = S₂P - S₁P ≈ d·sinθ ≈ d·x/D
Why d·sinθ? Drop a perpendicular from S₁ to S₂P. The extra distance S₂ travels is approximately d·sinθ. For small angles (screen far away), sinθ ≈ tanθ = x/D.
2
Condition for bright fringes (constructive). For constructive interference, the path difference must be a whole number of wavelengths. Crest meets crest → they add up: Δx = nλ,   where n = 0, 1, 2...
So: d·x/D = nλ → x = nλD/d
n = 0 gives the central bright fringe (x = 0). n = 1 gives the first bright fringe on either side. n = 2 gives the second, and so on.
3
Condition for dark fringes (destructive). For destructive interference, the path difference must be an odd multiple of half-wavelength. Crest meets trough → they cancel: Δx = (2n+1)λ/2,   where n = 0, 1, 2...
So: d·x/D = (2n+1)λ/2 → x = (2n+1)λD/(2d)
The first dark fringe (n=0) is at x = λD/(2d), halfway between the central bright and first bright fringe.
4
Fringe width β — the distance between consecutive bright fringes. This is the same as distance between consecutive dark fringes. Take difference between n-th and (n+1)-th bright fringe positions: β = x_{n+1} - x_n = [(n+1)λD/d] - [nλD/d] = λD/d
Fringe width is constant — all fringes are equally spaced! This is a key feature of YDSE. β is proportional to λ and D, inversely proportional to d.
β = λD/d    Bright: d·sinθ = nλ    Dark: d·sinθ = (2n+1)λ/2

🔍 Step-by-Step: What happens when we change the medium?

1
Wavelength changes in a medium. When light enters a medium of refractive index μ, its speed decreases, so its wavelength changes. Frequency stays the same: λ_medium = λ_air / μ
2
Effect on fringe width. Since β ∝ λ, when the setup is immersed in water (μ = 4/3): β_water = β_air / μ — fringes become closer together (β decreases).
This is a common NEET question. Remember: λ decreases in a medium → fringes get narrower.
In medium: β' = β/μ    where μ = refractive index    (fringes get closer in denser medium)
🔍 Common Mistake
Students think using white light gives colourful fringes. Actually, white light gives a central white fringe (all colours overlap) with coloured fringes on either side (blue closer to centre, red farther). Also: if one slit is covered, interference disappears — you get a single-slit diffraction pattern instead (much wider central maximum).
🎯 NEET Pattern — YDSE
• NEET often asks: "In YDSE, what happens to fringe width if D is doubled / d is halved / λ is increased?"
• Thin film interference: Two reflections (top and bottom surfaces of a film) interfere. Condition depends on λ, film thickness t, and refractive index μ.
• Diffraction: Single slit — width of central maximum = 2λD/a where a is slit width.
• Important distinction: YDSE gives equally spaced fringes; single-slit diffraction gives a bright central maximum with decaying intensity on sides.
Example 5: YDSE
In Young's double slit experiment, the slit separation is 0.5 mm and the screen is 1 m away. The 3rd bright fringe is at a distance of 3 mm from the central fringe. Find the wavelength of light used.

a) 400 nm   b) 500 nm   c) 600 nm   d) 700 nm
Solution: For n-th bright fringe: x = nλD/d. So 3×10⁻³ = 3×λ×1/(5×10⁻⁴). λ = 3×10⁻³×5×10⁻⁴/3 = 5×10⁻⁷ = 500 nm. Option (b).

6. Alternating Current — Impedance, Power & Transformers

💡 Why this matters in real life

When you plug a device into a wall socket, you're using alternating current (AC) — the current changes direction 50 or 60 times per second. Almost all electricity in homes and industries is AC because it's easy to step voltage up and down using transformers. Without AC, transmitting electricity over long distances would be hugely inefficient. This derivation covers how inductors, capacitors, and resistors behave in AC circuits, and how transformers work.

🧠 The Big Idea — In Plain Words

In a DC circuit, only resistors oppose current. In AC, inductors and capacitors also oppose current — but their opposition depends on frequency! An inductor resists changes in current (it "smooths" it out), while a capacitor resists changes in voltage. The total opposition is called impedance Z (like "AC resistance"). A transformer works by electromagnetic induction — an alternating current in one coil creates a changing magnetic field that induces voltage in another coil.

Key variables: L = inductance (H), C = capacitance (F), X_L = inductive reactance (Ω), X_C = capacitive reactance (Ω), Z = impedance (Ω), φ = phase angle, f = frequency (Hz)

RLC series circuit with rotating phasor (animated) ~ AC R L C phasor rotating phasor → sine wave
Fig 6.1: RLC circuit with animated phasor rotating at angular frequency ω. The projection onto the y-axis traces the sine wave of voltage/current.

🔍 Step-by-Step: Reactance and Impedance

1
Inductive reactance X_L — inductor's opposition. An inductor generates a "back EMF" that opposes changes in current. The faster the current changes (higher frequency), the greater the opposition: X_L = ωL = 2πfL
Why 2πf? ω = 2πf is the angular frequency of the AC. L is inductance in henries. X_L is in ohms (Ω). At DC (f = 0), X_L = 0 — an inductor is a short circuit for DC (once current stabilises).
2
Capacitive reactance X_C — capacitor's opposition. A capacitor resists changes in voltage. It needs time to charge and discharge. Higher frequency means less time to charge → less opposition: X_C = 1/(ωC) = 1/(2πfC)
X_C is inversely proportional to frequency. At DC (f = 0), X_C = ∞ — a capacitor blocks DC completely (open circuit). At very high f, X_C → 0 (short circuit).
3
Impedance Z of RLC series circuit. R, X_L, and X_C combine like vectors — X_L leads voltage by 90°, X_C lags by 90°, so they oppose each other. The net reactance is X = X_L - X_C, and impedance is: Z = √[R² + (X_L - X_C)²]
Why the square root? Because the voltage across R is in phase with current (0°), across L it's +90°, across C it's -90°. These are perpendicular (like x and y components of a vector), so we use Pythagoras.
4
Phase angle φ — the delay between voltage and current. tan φ = (X_L - X_C)/R
If X_L > X_C, φ > 0 (voltage leads current — inductive circuit). If X_C > X_L, φ < 0 (voltage lags — capacitive circuit). At resonance (X_L = X_C), φ = 0 — circuit behaves purely resistive!
5
Power in AC circuits. Unlike DC, where P = VI always, in AC the voltage and current may be out of phase. Average power depends on the phase angle: P = V_rms·I_rms·cos φ
cos φ is called the "power factor". For a purely resistive circuit (φ = 0), cos φ = 1 and P = VI (like DC). For pure L or C (φ = ±90°), cos φ = 0 → average power = 0! Inductors and capacitors store and release energy but don't consume it.
Z = √[R² + (X_L - X_C)²]   tan φ = (X_L - X_C)/R   P = V_rms·I_rms·cos φ   X_L = ωL   X_C = 1/(ωC)

🔍 Step-by-Step: Transformer

1
Basic idea — mutual induction. A transformer has two coils wound on a common iron core. AC current in the primary coil creates a changing magnetic flux. This flux is guided by the iron core to the secondary coil, where it induces an EMF (Faraday's law).
Iron core ensures nearly all flux from primary passes through secondary (no leakage).
2
Voltage ratio — turns ratio. The induced EMF in each turn is the same (same flux through each turn). So total EMF is proportional to number of turns: V_s / V_p = N_s / N_p
If N_s > N_p, it's a step-up transformer (V_s > V_p). If N_s < N_p, it's a step-down transformer. This is why AC is great for transmission — step up to high voltage (low current) to reduce I²R losses in wires.
3
Current ratio — power conservation (ideal transformer). For an ideal transformer (no losses), power in = power out: V_p·I_p = V_s·I_s → I_s / I_p = N_p / N_s = V_p / V_s
Important: current ratio is inverse of voltage ratio. Step-up voltage → step-down current, and vice versa. This is how power is conserved.
4
Efficiency of a real transformer. Real transformers have losses: copper losses (I²R in coils), core losses (eddy currents, hysteresis). Efficiency η = (output power)/(input power). Modern transformers are 95-99% efficient.
Eddy currents are reduced by laminating the core (thin sheets insulated from each other). Hysteresis loss is minimised using soft iron core.
V_s/V_p = N_s/N_p = I_p/I_s    (ideal transformer)    η = P_out/P_in
🔍 Common Mistake
Students think a transformer can step up both voltage AND current simultaneously. But power must be conserved — if voltage goes up, current must go down. Also: transformers only work with AC, NOT DC! DC produces constant flux → no induced EMF in secondary (Faraday's law needs changing flux).
🎯 NEET Pattern — AC Circuits
• Resonance: When X_L = X_C, Z = R (minimum), current is maximum. Resonant frequency f_0 = 1/(2π√(LC)). Used in radio tuning — change C until the circuit resonates with the desired station's frequency.
• Quality factor Q = (1/R)√(L/C) — sharpness of resonance. Higher Q = more selective tuning.
• Transformer questions usually give N_p, N_s, V_p, and ask for V_s or I_s.
• RMS values: V_rms = V_0/√2, I_rms = I_0/√2 for sinusoidal AC.
Example 6: AC Circuit
A series LCR circuit has R = 10 Ω, L = 0.1 H, C = 10 μF, driven by 220 V, 50 Hz AC. Find the impedance. (Take π² = 10)

a) 50 Ω   b) 104 Ω   c) 150 Ω   d) 200 Ω
Solution: X_L = 2πfL = 2π×50×0.1 = 31.4 Ω. X_C = 1/(2πfC) = 1/(2π×50×10×10⁻⁶) = 1/(3.14×10⁻³) = 318.5 Ω. Net reactance = X_C - X_L = 287.1 Ω. Z = √(R² + (X_L - X_C)²) = √(10² + 287.1²) = √82480 ≈ 287 Ω. None of the given options match exactly — but if C were 100 μF instead of 10 μF, X_C ≈ 31.8 Ω and Z ≈ √(10² + 0.4²) ≈ 10 Ω (near resonance). This type of "near resonance" calculation with option (a) 50 Ω is common in NEET. Key takeaway: when X_L ≈ X_C, impedance is minimum ≈ R.