Comprehensive physics lessons with theory, formulas, and solved examples for JEE, NEET, and board exams.
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Kinematics is the branch of mechanics that describes the motion of points, bodies, and systems of bodies without considering the forces that cause the motion.
Kinematics deals with concepts like displacement, velocity, acceleration, and their relationships through mathematical equations.
Distance: Total path length (scalar)
Displacement: Shortest path between points (vector)
Speed: Rate of change of distance (scalar)
Velocity: Rate of change of displacement (vector)
\( v = u + at \)
Final velocity
\( s = ut + \frac{1}{2}at^2 \)
Displacement
\( v^2 = u^2 + 2as \)
Velocity-position relation
\( s = \frac{u + v}{2}t \)
Average velocity
Slope = Velocity
Slope = Acceleration
Area = Displacement
Area = Change in velocity
Motion along a straight line
Motion under constant acceleration (gravity)
Horizontal Motion
\( x = u_xt \)
\( v_x = u_x \) (constant)
Vertical Motion
\( y = u_yt - \frac{1}{2}gt^2 \)
\( v_y = u_y - gt \)
Key Results:
Angular Quantities
\( \omega = \frac{d\theta}{dt} \)
\( \alpha = \frac{d\omega}{dt} \)
Linear Relations
\( v = r\omega \)
\( a_c = \frac{v^2}{r} = r\omega^2 \)
The motion of an object as observed from different reference frames
Position
\( \vec{r}_{AB} = \vec{r}_{AC} + \vec{r}_{CB} \)
Velocity
\( \vec{v}_{AB} = \vec{v}_{AC} + \vec{v}_{CB} \)
Examples:
Problem: A car accelerates from rest at 2 m/s² for 5 seconds. What is its final velocity and displacement?
Solution:
Using \( v = u + at \):
\( v = 0 + (2)(5) = 10 \) m/s
Using \( s = ut + \frac{1}{2}at^2 \):
\( s = 0 + \frac{1}{2}(2)(5)^2 = 25 \) m
Problem: A ball is thrown with initial velocity 20 m/s at 30° to horizontal. Find maximum height and range.
Solution:
Maximum height: \( H = \frac{u^2\sin^2\theta}{2g} = \frac{(20)^2\sin^2 30°}{2(9.8)} \)
\( H = \frac{400 \times 0.25}{19.6} = 5.1 \) m
Range: \( R = \frac{u^2\sin 2\theta}{g} = \frac{(20)^2\sin 60°}{9.8} \)
\( R = \frac{400 \times 0.866}{9.8} = 35.3 \) m
Problem: A boat crosses a 100m wide river flowing at 2 m/s. The boat can move at 5 m/s in still water. Find time to cross and drift.
Solution:
Time to cross: \( t = \frac{\text{width}}{\text{velocity perpendicular}} = \frac{100}{5} = 20 \) s
Drift: \( \text{drift} = v_{\text{river}} \times t = 2 \times 20 = 40 \) m
1. A train accelerates from 10 m/s to 30 m/s in 20 seconds. Find acceleration and distance covered.
[Answer: a = 1 m/s², s = 400 m]
2. A stone is dropped from a height of 80m. Find time to reach ground and final velocity.
[Answer: t = 4.04 s, v = 39.6 m/s]
3. A projectile is fired at 45° with velocity 50 m/s. Find maximum range and time of flight.
[Answer: R = 255.1 m, T = 7.21 s]
Newton's laws describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three fundamental laws form the foundation of classical mechanics.
Newton's Laws explain how objects move when forces act upon them, from everyday motion to planetary movements.
"An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."
Key Concept:
Inertia - resistance to change in motion
"The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass."
Formula:
\( \vec{F} = m\vec{a} \)
"For every action, there is an equal and opposite reaction."
Key Concept:
Forces always occur in pairs
Inertia is the tendency of an object to resist changes in its state of motion. The mass of an object is a quantitative measure of its inertia.
Examples:
Mathematical Expression:
If \( \sum \vec{F} = 0 \), then \( \vec{v} = \text{constant} \)
The net force acting on an object equals the rate of change of its momentum.
Vector Form:
\( \vec{F} = m\vec{a} \)
Force = mass × acceleration
Component Form:
\( F_x = ma_x \), \( F_y = ma_y \), \( F_z = ma_z \)
Key Points:
Forces always occur in pairs - if object A exerts a force on object B, then object B simultaneously exerts an equal and opposite force on object A.
Mathematical Expression:
\( \vec{F}_{AB} = -\vec{F}_{BA} \)
Important Notes:
Examples:
\( W = mg \)
Acts downward
Perpendicular to surface
Contact force
\( f = \mu N \)
Opposes motion
Along string/rope
Pulling force
Free Body Diagrams are essential tools for solving problems using Newton's Laws. They show all forces acting on a single object.
Weight (mg) downward
Normal force perpendicular to surface
Friction parallel to surface
Problem: A 5 kg object experiences a net force of 20 N. What is its acceleration?
Solution:
Using Newton's Second Law: \( F = ma \)
\( 20 = 5 \times a \)
\( a = \frac{20}{5} = 4 \) m/s²
Problem: A 10 kg box is pulled with 50 N force at 30° angle. If friction is 15 N, find acceleration.
Solution:
Horizontal component: \( F_x = 50 \cos 30° = 43.3 \) N
Net horizontal force: \( F_{net} = 43.3 - 15 = 28.3 \) N
Using \( F = ma \): \( 28.3 = 10a \)
\( a = 2.83 \) m/s²
Problem: Two masses 3 kg and 5 kg connected by string over pulley. Find acceleration and tension.
Solution:
For 5 kg mass: \( 5g - T = 5a \)
For 3 kg mass: \( T - 3g = 3a \)
Adding equations: \( 2g = 8a \)
\( a = \frac{2 \times 9.8}{8} = 2.45 \) m/s²
\( T = 3g + 3a = 29.4 + 7.35 = 36.75 \) N
Third law explains rocket propulsion - expelled gases create equal and opposite thrust.
First law explains need for seatbelts - passengers continue moving when car stops suddenly.
Second law determines how much force needed to accelerate balls to desired speeds.
Laws explain orbits - gravitational force provides centripetal acceleration.
1. A 8 kg object accelerates at 3 m/s². What net force acts on it?
[Answer: 24 N]
2. A car mass 1200 kg accelerates from 0 to 60 km/h in 8 seconds. Find average force.
[Answer: 2500 N]
3. Two blocks 2 kg and 4 kg connected by string. If 4 kg block hangs vertically, find acceleration.
[Answer: 6.53 m/s²]
4. A 50 kg box on horizontal surface with μ=0.3. Find minimum force to start moving.
[Answer: 147 N]
Gravitation is a natural phenomenon by which all things with mass or energy are brought toward one another. It's one of the four fundamental forces of nature and governs the motion of celestial bodies.
Gravitation explains everything from falling apples to planetary orbits. This chapter covers Newton's Law, gravitational field, potential energy, satellites, and Kepler's Laws.
\( F = G \frac{m_1 m_2}{r^2} \)
6.674 × 10⁻¹¹ N⋅m²/kg²
First measured by Henry Cavendish in 1798 using torsion balance experiment.
\( g = \frac{GM}{R^2} \)
On Earth's surface: g ≈ 9.8 m/s²
Variation with height: \( g_h = g\left(1 - \frac{2h}{R}\right) \)
\( I = \frac{F}{m} = \frac{GM}{r^2} \)
Force per unit mass
Vector quantity (towards mass)
Units: N/kg or m/s²
\( V = -\frac{GM}{r} \)
Work done to bring unit mass from infinity
Scalar quantity
Units: J/kg
\( U = -\frac{GMm}{r} \)
For two point masses
\( U = mgh \)
Near Earth's surface (approximate)
"All planets move in elliptical orbits with the Sun at one focus."
Eccentricity (e): Measure of ovalness
0 ≤ e < 1 (e=0 for circle)
"A line joining a planet and the Sun sweeps out equal areas in equal intervals of time."
Angular momentum: Conserved
Planets move faster when closer to Sun
"The square of the orbital period is proportional to the cube of the semi-major axis."
\( T^2 \propto a^3 \)
\( \frac{T^2}{a^3} = \text{constant} \)
\( v_o = \sqrt{\frac{GM}{r}} \)
For circular orbit at distance r from center
\( v_o = \sqrt{\frac{gR^2}{r}} \)
Using Earth's surface gravity
Special Cases:
\( T = 2\pi \sqrt{\frac{r^3}{GM}} \)
Geostationary
T = 24 hours
r ≈ 42,164 km
Polar
T = ~100 minutes
r ≈ 7,000 km
GPS
T = 12 hours
r ≈ 26,600 km
\( K = \frac{GMm}{2r} \)
Positive quantity
\( U = -\frac{GMm}{r} \)
Negative quantity
\( E = -\frac{GMm}{2r} \)
Negative (bound orbit)
\( K = -E \)
Kinetic energy = -Total energy
\( U = 2E \)
Potential energy = 2 × Total energy
Virial Theorem:
\( \langle K \rangle = -\frac{1}{2} \langle U \rangle \) for bound systems
Problem: Calculate the gravitational force between two 100 kg masses separated by 1 meter.
Solution:
Using \( F = G \frac{m_1 m_2}{r^2} \), where G = 6.674×10⁻¹¹ N⋅m²/kg²:
\( F = (6.674 \times 10^{-11}) \frac{(100)(100)}{1^2} \)
\( F = (6.674 \times 10^{-11}) \times 10^4 \)
\( F = 6.674 \times 10^{-7} \) N
Problem: Find orbital velocity and period of a satellite 500 km above Earth's surface.
Solution:
Earth radius R = 6371 km, mass M = 5.97×10²⁴ kg
Orbit radius r = R + h = 6371 + 500 = 6871 km = 6.871×10⁶ m
Orbital velocity: \( v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.871\times10^6}} \)
\( v_o = \sqrt{5.798\times10^7} = 7615 \) m/s = 7.615 km/s
Period: \( T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(6.871\times10^6)^3}{(6.674\times10^{-11})(5.97\times10^{24})}} \)
\( T = 5670 \) seconds ≈ 94.5 minutes
Problem: Mars orbits Sun with period 687 days. Earth's orbit radius is 1 AU. Find Mars' orbit radius.
Solution:
Using \( \frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} \)
Earth: T₁ = 365 days, r₁ = 1 AU
Mars: T₂ = 687 days
\( \frac{687^2}{r_2^3} = \frac{365^2}{1^3} \)
\( r_2^3 = \frac{687^2}{365^2} = 3.537 \)
\( r_2 = \sqrt[3]{3.537} = 1.524 \) AU
1. Calculate gravitational force between Earth (5.97×10²⁴ kg) and Moon (7.35×10²² kg) separated by 3.84×10⁸ m.
[Answer: 1.98×10²⁰ N]
2. Find acceleration due to gravity at height 1000 km above Earth's surface.
[Answer: 7.33 m/s²]
3. A geostationary satellite orbits at 42,164 km from Earth's center. Calculate its orbital velocity.
[Answer: 3.07 km/s]
4. Jupiter's moon Io orbits at 421,600 km with period 1.77 days. Find Jupiter's mass.
[Answer: 1.90×10²⁷ kg]
Circular motion describes the movement of an object along a circular path. It can be uniform (constant speed) or non-uniform (changing speed).
Circular motion involves angular displacement, angular velocity, angular acceleration, centripetal force, and their relationships through mathematical equations.
Angular Motion: Rotation about a fixed axis
Linear Motion: Straight-line translation
Uniform Circular Motion: Constant angular speed
Non-uniform Circular Motion: Changing angular speed
\( \omega = \omega_0 + \alpha t \)
Final angular velocity
\( \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \)
Angular displacement
\( \omega^2 = \omega_0^2 + 2\alpha\theta \)
Angular velocity-displacement relation
\( \theta = \frac{\omega_0 + \omega}{2}t \)
Average angular velocity
Arc length: \( s = r\theta \)
Tangential displacement: \( \Delta s = r\Delta\theta \)
Tangential velocity: \( v = r\omega \)
Relationship: \( \vec{v} = \vec{\omega} \times \vec{r} \)
Tangential acceleration: \( a_t = r\alpha \)
Centripetal acceleration: \( a_c = \frac{v^2}{r} = r\omega^2 \)
Magnitude: \( a = \sqrt{a_t^2 + a_c^2} \)
Direction: \( \phi = \tan^{-1}\left(\frac{a_t}{a_c}\right) \)
The net force directed toward the center of rotation that keeps an object in circular motion
Formula: \( F_c = \frac{mv^2}{r} = mr\omega^2 \)
A fictitious or pseudo-force that appears in a rotating reference frame
Motion in a circle with constant speed
Motion in a circle with changing speed
The tilting of curved roads to help vehicles navigate turns without relying solely on friction
Ideal Banking Angle
\( \tan\theta = \frac{v^2}{rg} \)
Where θ is the banking angle
Maximum Safe Speed
\( v_{\text{max}} = \sqrt{rg\tan\theta} \)
For frictionless surface
With Friction:
\( v_{\text{max}} = \sqrt{\frac{rg(\mu_s + \tan\theta)}{1 - \mu_s\tan\theta}} \)
Where μₛ is the coefficient of static friction
Circular motion where gravity affects the speed throughout the path
At Highest Point
\( T_{\text{top}} = \frac{mv_{\text{top}}^2}{r} - mg \)
Minimum speed: \( \sqrt{rg} \)
At Lowest Point
\( T_{\text{bottom}} = \frac{mv_{\text{bottom}}^2}{r} + mg \)
Maximum tension
Critical Velocity:
Minimum velocity at top for complete circle: \( v_{\text{critical}} = \sqrt{5rg} \)
When released from horizontal position
Problem: A car moves in a circle of radius 50m at 20 m/s. Find centripetal acceleration and period.
Solution:
Centripetal acceleration: \( a_c = \frac{v^2}{r} = \frac{(20)^2}{50} = 8 \) m/s²
Period: \( T = \frac{2\pi r}{v} = \frac{2\pi(50)}{20} = 15.7 \) s
Problem: A curved road has radius 100m. At what angle should it be banked for vehicles moving at 25 m/s without friction?
Solution:
Using \( \tan\theta = \frac{v^2}{rg} \):
\( \tan\theta = \frac{(25)^2}{(100)(9.8)} = \frac{625}{980} = 0.638 \)
\( \theta = \tan^{-1}(0.638) = 32.5^\circ \)
Problem: A 2kg mass swings in a vertical circle of radius 1m. Find minimum speed at top and tension at bottom if speed at bottom is 7 m/s.
Solution:
Minimum speed at top: \( v_{\text{min}} = \sqrt{rg} = \sqrt{(1)(9.8)} = 3.13 \) m/s
Tension at bottom: \( T = \frac{mv^2}{r} + mg = \frac{(2)(7)^2}{1} + (2)(9.8) \)
\( T = 98 + 19.6 = 117.6 \) N
1. A wheel rotates at 120 rpm. Find its angular velocity in rad/s and period of rotation.
[Answer: ω = 12.57 rad/s, T = 0.5 s]
2. A 1000kg car rounds a curve of radius 80m at 25 m/s. What centripetal force is required?
[Answer: F = 7812.5 N]
3. A stone tied to a string is whirled in a vertical circle of radius 2m. What is the minimum speed at the top for complete circular motion?
[Answer: v = 4.43 m/s]
Thermal physics deals with temperature, heat transfer, and the relationship between heat and other forms of energy.
Thermal physics covers concepts like temperature, heat, thermal expansion, calorimetry, heat transfer mechanisms, and the laws of thermodynamics.
Temperature: Measure of average kinetic energy of molecules
Heat: Energy transferred due to temperature difference
Zeroth Law: If A=B and B=C, then A=C thermally
Equilibrium: No net heat flow between objects
Freezing point: 0°C
Boiling point: 100°C
Freezing point: 32°F
Boiling point: 212°F
Absolute zero: 0 K
Freezing point: 273.15 K
\( K = °C + 273.15 \)
\( °F = \frac{9}{5}°C + 32 \)
\( °C = \frac{5}{9}(°F - 32) \)
−273.15°C or −459.67°F
Theoretical minimum temperature where molecular motion stops
Change in length of solids with temperature
Formula: \( \Delta L = L_0 \alpha \Delta T \)
Where α is the coefficient of linear expansion
Change in area of surfaces with temperature
Formula: \( \Delta A = A_0 \beta \Delta T \)
Where β ≈ 2α is the coefficient of area expansion
Change in volume of solids and liquids with temperature
Formula: \( \Delta V = V_0 \gamma \Delta T \)
Where γ ≈ 3α is the coefficient of volume expansion
\( Q = mc\Delta T \)
Heat required to raise temperature of unit mass by 1°C
Units: J/kg·K or cal/g·°C
\( Q = nC\Delta T \)
Heat required to raise temperature of one mole by 1°C
Units: J/mol·K
Heat lost by hot bodies = Heat gained by cold bodies
\( m_1c_1(T_1 - T_f) = m_2c_2(T_f - T_2) \)
Where T_f is the final equilibrium temperature
Heat required for phase change at constant temperature
Latent Heat of Fusion
\( Q = mL_f \)
Solid ↔ Liquid transition
Latent Heat of Vaporization
\( Q = mL_v \)
Liquid ↔ Gas transition
Solid Phase
\( Q = mc_s\Delta T \)
Temperature increases
Phase Change
\( Q = mL \)
Constant temperature
Gas Phase
\( Q = mc_g\Delta T \)
Temperature increases
Heat transfer through direct molecular contact
\( \frac{Q}{t} = kA\frac{\Delta T}{L} \)
Where k is thermal conductivity, A is area, L is length
Thermal Resistance:
\( R = \frac{L}{kA} \)
For series combination: \( R_{\text{total}} = R_1 + R_2 + \cdots \)
Heat transfer by fluid motion
\( \frac{Q}{t} = hA\Delta T \)
Where h is convection heat transfer coefficient
Heat transfer by electromagnetic waves
\( \frac{Q}{t} = \sigma \epsilon A(T^4 - T_0^4) \)
Where σ = 5.67×10⁻⁸ W/m²K⁴ (Stefan-Boltzmann constant), ε is emissivity
Wien's Displacement Law:
\( \lambda_{\text{max}}T = b \)
Where b = 2.898×10⁻³ m·K
\( PV = \text{constant} \)
At constant temperature
\( \frac{V}{T} = \text{constant} \)
At constant pressure
\( \frac{P}{T} = \text{constant} \)
At constant volume
\( \frac{V}{n} = \text{constant} \)
At constant T and P
\( PV = nRT \)
Where R = 8.314 J/mol·K (Universal gas constant)
Problem: 200g of water at 80°C is mixed with 100g of water at 20°C. Find final temperature.
Solution:
Using calorimetry principle: \( m_1c(T_1 - T_f) = m_2c(T_f - T_2) \)
\( 200(80 - T_f) = 100(T_f - 20) \)
\( 16000 - 200T_f = 100T_f - 2000 \)
\( 300T_f = 18000 \)
\( T_f = 60°C \)
Problem: A steel rail of length 10m at 20°C. If α = 1.2×10⁻⁵/°C, find length at 40°C.
Solution:
\( \Delta L = L_0 \alpha \Delta T = 10 \times 1.2\times10^{-5} \times (40-20) \)
\( \Delta L = 10 \times 1.2\times10^{-5} \times 20 = 0.0024 \) m
Final length = 10 + 0.0024 = 10.0024 m
Problem: A copper rod (k=400 W/m·K) of length 0.5m and cross-section 0.01m² has 100°C difference. Find heat current.
Solution:
\( \frac{Q}{t} = kA\frac{\Delta T}{L} = 400 \times 0.01 \times \frac{100}{0.5} \)
\( \frac{Q}{t} = 4 \times 200 = 800 \) W
1. Convert 98.6°F to Celsius and Kelvin scales.
[Answer: 37°C, 310.15 K]
2. How much heat is required to melt 500g of ice at 0°C? (L_f = 334 kJ/kg)
[Answer: 167 kJ]
3. A gas occupies 2L at 300K. Find volume at 400K if pressure is constant.
[Answer: 2.67 L]
Thermodynamics is the branch of physics that deals with heat, work, temperature, and their relation to energy, radiation, and physical properties of matter.
The laws of thermodynamics describe how thermal energy is transferred and transformed, governing all natural processes from microscopic molecular interactions to macroscopic systems.
Open System: Can exchange both energy and matter
Closed System: Can exchange energy but not matter
Isolated System: Cannot exchange energy or matter
Intensive: Independent of mass (T, P, ρ)
Extensive: Depend on mass (U, H, S, V)
State Functions: Path independent (U, H, S)
If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
If A = B and B = C, then A = C (thermally)
Energy cannot be created or destroyed, only converted from one form to another.
ΔU = Q - W
Where:
dU = đQ - đW
For infinitesimal processes
Constant Volume
W = 0, ΔU = Q
Constant Pressure
Q = ΔH
No Heat Transfer
Q = 0, ΔU = -W
The total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible.
ΔSuniverse ≥ 0
For any spontaneous process
Heat cannot spontaneously flow from a colder to a hotter body
No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of all this heat into work.
η = 1 - Qc/Qh = 1 - Tc/Th
Carnot efficiency (ideal)
COP = Qc/W = Tc/(Th - Tc)
Coefficient of Performance
The entropy of a perfect crystal approaches zero as the temperature approaches absolute zero.
limT→0 S = 0
For perfect crystals
| Process | Constant | Work (W) | Heat (Q) | Relation |
|---|---|---|---|---|
| Isochoric | Volume | 0 | nCvΔT | P ∝ T |
| Isobaric | Pressure | PΔV | nCpΔT | V ∝ T |
| Isothermal | Temperature | nRT ln(V2/V1) | = W | PV = constant |
| Adiabatic | Q = 0 | (P1V1-P2V2)/(γ-1) | 0 | PVγ = constant |
Problem: A gas expands isothermally at 300K from 2L to 5L. If 2000J of heat is added, find work done and change in internal energy.
Solution:
For isothermal process: ΔU = 0
From first law: Q = W
Work done: W = 2000 J
Change in internal energy: ΔU = 0
Problem: A Carnot engine operates between 500K and 300K. If it absorbs 1000J of heat, find work output and efficiency.
Solution:
Efficiency: η = 1 - Tc/Th = 1 - 300/500 = 0.4 (40%)
Work output: W = η × Qh = 0.4 × 1000 = 400 J
Heat rejected: Qc = Qh - W = 1000 - 400 = 600 J
Problem: A monatomic gas (γ=5/3) at 300K and 1 atm expands adiabatically to twice its volume. Find final temperature.
Solution:
For adiabatic process: TVγ-1 = constant
T1V1γ-1 = T2V2γ-1
300 × V12/3 = T2 × (2V1)2/3
T2 = 300 / 22/3 = 300 / 1.587 = 189 K
1. A system absorbs 500J of heat and does 200J of work. What is the change in internal energy?
[Answer: ΔU = 300 J]
2. Calculate the Carnot efficiency of an engine operating between 400K and 300K.
[Answer: η = 25%]
3. 2 moles of an ideal gas expand isothermally at 27°C from 10L to 20L. Calculate work done.
[Answer: W = 3458 J]
Electrostatics is the branch of physics that studies electric charges at rest, the forces between them, and the electric fields they create.
Electrostatics deals with concepts like electric charge, Coulomb's law, electric field, electric potential, capacitance, and their applications in various systems.
Properties: Quantized, conserved, additive
Types: Positive and negative
Unit: Coulomb (C)
Charge of electron: \( e = 1.6 \times 10^{-19} \) C
Friction: Transfer by rubbing
Conduction: Direct contact
Induction: Without direct contact
\( F = k\frac{q_1 q_2}{r^2} \)
Where \( k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \) Nm²/C²
\( \vec{F}_{12} = k\frac{q_1 q_2}{r^2}\hat{r}_{12} \)
Force on q₁ due to q₂
Electric Field Intensity
\( \vec{E} = \frac{\vec{F}}{q_0} \)
Force per unit charge
Due to Point Charge
\( E = k\frac{q}{r^2} \)
Radially outward/inward
\( \vec{E} = \int k\frac{dq}{r^2}\hat{r} \)
Electric Potential
\( V = \frac{U}{q} = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l} \)
Work done per unit charge
Potential Difference
\( V_{AB} = V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{l} \)
Voltage between points
\( V = k\frac{q}{r} \)
Scalar quantity
\( E_x = -\frac{\partial V}{\partial x} \)
\( \vec{E} = -\nabla V \)
\( \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0} \)
Electric flux through closed surface
Capacitance
\( C = \frac{Q}{V} \)
Charge storing capacity
Energy Stored
\( U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV \)
In electric field
Parallel Plate
\( C = \frac{\epsilon_0 A}{d} \)
With dielectric: \( \kappa\epsilon_0 A/d \)
Spherical
\( C = 4\pi\epsilon_0\frac{ab}{b-a} \)
Concentric spheres
Cylindrical
\( C = \frac{2\pi\epsilon_0 l}{\ln(b/a)} \)
Coaxial cable
\( \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots \)
\( C_{\text{eq}} = C_1 + C_2 + C_3 + \cdots \)
Dielectric Constant
\( \kappa = \frac{C}{C_0} \)
Ratio with and without dielectric
Polarization
\( \vec{P} = \chi_e \epsilon_0 \vec{E} \)
Dipole moment per unit volume
Effects:
Problem: Two point charges +2μC and +3μC are placed 0.5m apart. Find the force between them.
Solution:
Using \( F = k\frac{q_1 q_2}{r^2} \):
\( F = (9 \times 10^9)\frac{(2 \times 10^{-6})(3 \times 10^{-6})}{(0.5)^2} \)
\( F = (9 \times 10^9)\frac{6 \times 10^{-12}}{0.25} = 0.216 \) N
Repulsive force since both are positive
Problem: Calculate electric field at a point 0.3m from a +5μC charge.
Solution:
Using \( E = k\frac{q}{r^2} \):
\( E = (9 \times 10^9)\frac{5 \times 10^{-6}}{(0.3)^2} \)
\( E = (9 \times 10^9)\frac{5 \times 10^{-6}}{0.09} = 5 \times 10^5 \) N/C
Radially outward direction
Problem: Three 6μF capacitors are connected in series to 12V battery. Find charge and energy stored.
Solution:
Equivalent capacitance: \( \frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \)
\( C_{eq} = 2\mu F \)
Charge: \( Q = CV = (2 \times 10^{-6})(12) = 24\mu C \)
Energy: \( U = \frac{1}{2}CV^2 = \frac{1}{2}(2 \times 10^{-6})(12)^2 = 144\mu J \)
1. Two charges +4μC and -6μC are 0.8m apart. Find the force and nature of force.
[Answer: F = 0.3375 N, attractive]
2. A parallel plate capacitor has plates 0.1m² area separated by 1mm. Find capacitance with air and with dielectric (κ=5).
[Answer: C₀ = 885pF, C = 4.425nF]
3. Three capacitors 2μF, 3μF, and 6μF are connected in parallel to 24V. Find total charge and energy stored.
[Answer: Q = 264μC, U = 3.168mJ]
Electric current is the flow of electric charge through a conductor. It forms the basis of electrical circuits and is fundamental to understanding electricity.
Electric current involves concepts like current, voltage, resistance, Ohm's law, electrical power, and various circuit configurations including series and parallel combinations.
Definition: Flow of electric charge
Formula: \( I = \frac{Q}{t} \)
Unit: Ampere (A)
DC: Direct Current (constant direction)
AC: Alternating Current (changing direction)
\( I = \frac{Q}{t} \)
Flow rate of charge
Unit: Ampere (A)
\( V = \frac{W}{Q} \)
Electrical potential difference
Unit: Volt (V)
\( R = \frac{V}{I} \)
Opposition to current flow
Unit: Ohm (Ω)
The current through a conductor between two points is directly proportional to the voltage across the two points.
\( V = IR \)
Where V = Voltage, I = Current, R = Resistance
For Current:
\( I = \frac{V}{R} \)
For Voltage:
\( V = IR \)
For Resistance:
\( R = \frac{V}{I} \)
Resistance Formula
\( R = \rho \frac{L}{A} \)
Where:
Temperature Dependence
\( R = R_0[1 + \alpha(T - T_0)] \)
Where:
Basic Power
\( P = VI \)
Unit: Watt (W)
Using Resistance
\( P = I^2R \)
For constant R
Alternative Form
\( P = \frac{V^2}{R} \)
For constant V
\( E = Pt = VIt \)
Unit: Joule (J) or kilowatt-hour (kWh)
1 kWh = 3.6 × 10⁶ J
Equivalent Resistance
\( R_{eq} = R_1 + R_2 + R_3 + \cdots \)
Current
Same through all components
\( I = I_1 = I_2 = I_3 \)
Voltage
Divides across components
\( V = V_1 + V_2 + V_3 \)
Power
\( P = P_1 + P_2 + P_3 \)
Equivalent Resistance
\( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots \)
Voltage
Same across all components
\( V = V_1 = V_2 = V_3 \)
Current
Divides through branches
\( I = I_1 + I_2 + I_3 \)
Power
\( P = P_1 + P_2 + P_3 \)
The algebraic sum of currents meeting at a junction is zero.
\( \sum I_{in} = \sum I_{out} \)
or \( \sum I = 0 \) at any junction
The algebraic sum of all voltages around any closed loop is zero.
\( \sum V = 0 \) around any closed loop
or \( \sum EMF = \sum IR \) in any closed loop
Problem: A resistor has a voltage of 12V across it and current of 0.5A through it. Find its resistance.
Solution:
Using Ohm's Law: \( R = \frac{V}{I} \)
\( R = \frac{12}{0.5} = 24 \) Ω
Problem: Three resistors of 10Ω, 20Ω, and 30Ω are connected in series to a 24V battery. Find total resistance, current, and voltage across each resistor.
Solution:
Total resistance: \( R_{eq} = 10 + 20 + 30 = 60 \) Ω
Current: \( I = \frac{V}{R_{eq}} = \frac{24}{60} = 0.4 \) A
Voltage across 10Ω: \( V_1 = IR_1 = 0.4 \times 10 = 4 \) V
Voltage across 20Ω: \( V_2 = IR_2 = 0.4 \times 20 = 8 \) V
Voltage across 30Ω: \( V_3 = IR_3 = 0.4 \times 30 = 12 \) V
Problem: A 100W bulb operates at 220V. Find its resistance and the current through it.
Solution:
Using \( P = VI \): \( I = \frac{P}{V} = \frac{100}{220} = 0.455 \) A
Resistance: \( R = \frac{V}{I} = \frac{220}{0.455} = 483.5 \) Ω
Alternative: \( R = \frac{V^2}{P} = \frac{(220)^2}{100} = 484 \) Ω
1. A wire carries 2A current for 5 minutes. Calculate the total charge that passes through it.
[Answer: Q = 600 C]
2. Three resistors of 6Ω, 12Ω, and 18Ω are connected in parallel to a 12V battery. Find the total current drawn from the battery.
[Answer: I = 3.67 A]
3. An electric heater rated 1500W at 220V is used for 2 hours daily. Calculate the energy consumed in 30 days and the cost at ₹5 per unit.
[Answer: E = 90 kWh, Cost = ₹450]
Ray optics, also known as geometrical optics, describes light propagation in terms of rays. It deals with reflection, refraction, and the formation of images by optical instruments.
Ray optics studies the behavior of light using the ray model, covering reflection, refraction, lenses, mirrors, and optical instruments based on these principles.
Ray: Straight line path of light
Beam: Collection of rays
Wavefront: Surface of constant phase
Reflection: Bouncing back of light
Refraction: Bending of light
Dispersion: Splitting of light
First Law
Incident ray, reflected ray, and normal all lie in the same plane
Second Law
Angle of incidence = Angle of reflection
\( \angle i = \angle r \)
Regular Reflection
Diffuse Reflection
First Law
Incident ray, refracted ray, and normal all lie in the same plane
Second Law (Snell's Law)
\( n_1 \sin i = n_2 \sin r \)
Where n₁ and n₂ are refractive indices
Absolute Refractive Index
\( n = \frac{c}{v} \)
c = speed in vacuum, v = speed in medium
Relative Refractive Index
\( n_{21} = \frac{n_2}{n_1} = \frac{v_1}{v_2} \)
Critical Angle
\( \sin c = \frac{n_2}{n_1} \)
For total internal reflection
Cartesian Sign Convention
Mirror Formula
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
f = focal length, v = image distance, u = object distance
Linear Magnification
\( m = \frac{h_i}{h_o} = -\frac{v}{u} \)
hᵢ = image height, hₒ = object height
Power of Mirror
\( P = -\frac{1}{f} \)
f in meters
Relation between f and R
\( f = \frac{R}{2} \)
R = radius of curvature
Lens Maker's Formula
\( \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
For thin lens in air
Lens Formula
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Using Cartesian sign convention
Converging Lens (Convex)
Diverging Lens (Concave)
\( \delta = i + e - A \)
δ = angle of deviation, A = angle of prism
\( n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \)
At minimum deviation
\( \omega = \frac{n_V - n_R}{n_Y - 1} \)
ω = dispersive power
Angular dispersion: \( \theta = (n_V - n_R)A \)
Simple Microscope
Magnification: \( M = 1 + \frac{D}{f} \)
For normal adjustment
Compound Microscope
\( M = M_o \times M_e \)
\( M = \frac{v_o}{u_o} \left(1 + \frac{D}{f_e}\right) \)
Astronomical Telescope
Magnification: \( M = -\frac{f_o}{f_e} \)
Length: \( L = f_o + f_e \)
Terrestrial Telescope
Magnification: \( M = \frac{f_o}{f_e} \)
Length: \( L = f_o + 4f + f_e \)
Problem: An object is placed 30cm in front of a concave mirror of focal length 15cm. Find image position and nature.
Solution:
Using mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
\( \frac{1}{-15} = \frac{1}{v} + \frac{1}{-30} \)
\( \frac{1}{v} = -\frac{1}{15} + \frac{1}{30} = -\frac{1}{30} \)
\( v = -30 \) cm (real, inverted, same size)
Problem: A convex lens of focal length 20cm forms an image at 60cm. Find object position.
Solution:
Using lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \frac{1}{20} = \frac{1}{60} - \frac{1}{u} \)
\( \frac{1}{u} = \frac{1}{60} - \frac{1}{20} = -\frac{1}{30} \)
\( u = -30 \) cm (30cm in front of lens)
Problem: Light enters from air to glass (n=1.5) at 30° incidence. Find angle of refraction.
Solution:
Using Snell's Law: \( n_1 \sin i = n_2 \sin r \)
\( 1 \times \sin 30^\circ = 1.5 \times \sin r \)
\( 0.5 = 1.5 \sin r \)
\( \sin r = \frac{1}{3} = 0.333 \)
\( r = \sin^{-1}(0.333) = 19.47^\circ \)
1. A concave mirror has focal length 20cm. Where should an object be placed to get a real image three times the size?
[Answer: u = -26.67 cm]
2. A convex lens forms an image at 30cm when object is at 15cm. Find focal length and magnification.
[Answer: f = 10 cm, m = -2]
3. Calculate critical angle for glass-air interface (n_glass = 1.5).
[Answer: c = 41.81°]
Reflection is the phenomenon where light bounces back when it strikes a surface. The laws of reflection govern how light behaves when it encounters different types of surfaces.
Reflection involves the study of how light interacts with surfaces, including plane mirrors, spherical mirrors, and the formation of images through reflection.
Regular/Specular: Smooth surfaces, parallel reflected rays
Diffused/Irregular: Rough surfaces, scattered reflected rays
Incident Ray: Ray approaching the surface
Reflected Ray: Ray leaving the surface
The incident ray, reflected ray, and normal all lie in the same plane.
Angle of incidence equals angle of reflection:
\( \angle i = \angle r \)
Center point of the mirror surface
Center of the sphere of which mirror is part
Distance between pole and center of curvature
Line joining pole and center of curvature
Point where parallel rays converge/ appear to diverge
Distance between pole and focus
Characteristics:
Image Properties:
\( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
Where:
\( f \) = focal length
\( u \) = object distance
\( v \) = image distance
Sign Convention:
• Object towards left: \( u \) = -ve
• Real image: \( v \) = -ve
• Virtual image: \( v \) = +ve
Focal Length:
• Concave mirror: \( f \) = -ve
• Convex mirror: \( f \) = +ve
\( m = -\frac{v}{u} = \frac{h_i}{h_o} \)
Where:
\( h_i \) = image height
\( h_o \) = object height
Interpretation:
• \( m > 0 \): Erect image
• \( m < 0 \): Inverted image
• \( |m| > 1 \): Magnified
• \( |m| < 1 \): Diminished
Principal Rays:
Image Cases:
Principal Rays:
Image Properties:
Problem: A man 1.8m tall stands 2m from a plane mirror. What is the minimum height of mirror required to see his full image?
Solution:
For plane mirror, minimum height = \( \frac{\text{Height of object}}{2} \)
Minimum height = \( \frac{1.8}{2} = 0.9 \) m
The mirror should be positioned such that its top is at eye level
Problem: An object is placed 30cm from a concave mirror of focal length 15cm. Find image position and nature.
Solution:
Using mirror formula: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
\( \frac{1}{-15} = \frac{1}{-30} + \frac{1}{v} \)
\( \frac{1}{v} = \frac{1}{-15} - \frac{1}{-30} = -\frac{1}{15} + \frac{1}{30} = -\frac{1}{30} \)
\( v = -30 \) cm
Magnification: \( m = -\frac{v}{u} = -\frac{(-30)}{(-30)} = -1 \)
Image is real, inverted, same size, located at 30cm
Problem: An object is placed 20cm from a convex mirror of focal length 10cm. Find image position and magnification.
Solution:
Using mirror formula: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
\( \frac{1}{10} = \frac{1}{-20} + \frac{1}{v} \)
\( \frac{1}{v} = \frac{1}{10} - \frac{1}{-20} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20} \)
\( v = \frac{20}{3} = 6.67 \) cm
Magnification: \( m = -\frac{v}{u} = -\frac{6.67}{-20} = +0.33 \)
Image is virtual, erect, diminished, located at 6.67cm behind mirror
1. A concave mirror has focal length 20cm. Where should an object be placed to get a real image three times the size of object?
[Answer: u = -26.67cm for diminished, u = -13.33cm for magnified]
2. A convex mirror used in a store has radius of curvature 3m. If a customer is 4m away, where is his image formed?
[Answer: v = +0.86m behind mirror]
3. An object 2cm high is placed 15cm from a concave mirror of focal length 10cm. Find image height and nature.
[Answer: hᵢ = -4cm, real and inverted]
Interference is a phenomenon in which two or more waves superpose to form a resultant wave of greater, lower, or the same amplitude.
Interference deals with the superposition of waves, conditions for interference, types of interference, and applications in various optical phenomena.
Definition: When two or more waves overlap, the resultant displacement equals the vector sum of individual displacements
\( y = y_1 + y_2 + y_3 + \cdots \)
Temporal Coherence: Constant phase relationship over time
Spatial Coherence: Constant phase relationship across space
Two Waves with Phase Difference δ
\( I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\delta \)
Equal Amplitudes (I₁ = I₂ = I₀)
\( I = 4I_0\cos^2\left(\frac{\delta}{2}\right) \)
Relation
\( \delta = \frac{2\pi}{\lambda}\Delta \)
Where Δ is path difference
Conditions
Constructive: \( \Delta = n\lambda \), \( \delta = 2n\pi \)
Destructive: \( \Delta = (2n+1)\frac{\lambda}{2} \), \( \delta = (2n+1)\pi \)
Two coherent sources created by illuminating two slits with a single light source
Fringe Width
\( \beta = \frac{\lambda D}{d} \)
Distance between consecutive bright/dark fringes
Position of nth Maximum
\( y_n = \frac{n\lambda D}{d} \)
From central maximum
Phase Difference
\( \delta = \frac{2\pi d\sin\theta}{\lambda} \approx \frac{2\pi d y}{\lambda D} \)
Resultant Intensity
\( I = 4I_0\cos^2\left(\frac{\pi d y}{\lambda D}\right) \)
For a film of thickness t and refractive index μ:
Normal Incidence
Path difference: \( \Delta = 2\mu t \)
Due to reflection and double path
Additional Phase Change
π phase change when reflecting from denser medium
Affects constructive/destructive conditions
Reflection from Denser Medium
Constructive: \( 2\mu t = n\lambda \)
Destructive: \( 2\mu t = (2n+1)\frac{\lambda}{2} \)
Transmitted Light
Constructive: \( 2\mu t = (2n+1)\frac{\lambda}{2} \)
Destructive: \( 2\mu t = n\lambda \)
Circular interference pattern formed between a plano-convex lens and a flat glass surface
Radius of nth Dark Ring
\( r_n = \sqrt{n\lambda R} \)
For air film, reflected light
Radius of nth Bright Ring
\( r_n = \sqrt{\frac{(2n-1)\lambda R}{2}} \)
For air film, reflected light
Applications:
Splits light beam into two paths and recombines them to create interference pattern
Path Difference
\( \Delta = 2d\cos\theta \)
Where d is mirror displacement
Fringe Shift
\( \Delta N = \frac{2d}{\lambda} \)
Number of fringes shifted
Applications:
Problem: In Young's experiment, slits are 0.5mm apart and screen is 1m away. If fringe width is 1mm, find wavelength of light.
Solution:
Using \( \beta = \frac{\lambda D}{d} \):
\( \lambda = \frac{\beta d}{D} = \frac{(1\times10^{-3})(0.5\times10^{-3})}{1} \)
\( \lambda = 5 \times 10^{-7} \) m = 500 nm
Problem: A soap film (μ=1.33) appears bright for λ=600nm at normal incidence. Find minimum thickness.
Solution:
For constructive interference with phase change:
\( 2\mu t = n\lambda \) for minimum n=1
\( t = \frac{\lambda}{2\mu} = \frac{600}{2 \times 1.33} = 225.6 \) nm
Problem: The diameter of 10th dark ring in Newton's rings apparatus is 0.5cm. If radius of curvature is 1m, find wavelength.
Solution:
Using \( r_n = \sqrt{n\lambda R} \):
\( \lambda = \frac{r_n^2}{nR} = \frac{(0.25\times10^{-2})^2}{10 \times 1} \)
\( \lambda = 6.25 \times 10^{-7} \) m = 625 nm
1. In Young's experiment, the distance between slits is 0.1mm and screen is 2m away. If 5th bright fringe is 6cm from central maximum, find wavelength.
[Answer: λ = 600 nm]
2. A soap bubble (μ=1.33) of thickness 300nm is illuminated by white light. What color does it appear at normal incidence?
[Answer: Green (λ ≈ 532 nm)]
3. In Michelson interferometer, moving mirror by 0.1mm causes 400 fringes to cross. Find wavelength of light used.
[Answer: λ = 500 nm]
This chapter explores the wave-particle duality of both matter and radiation, bridging classical and quantum physics through revolutionary concepts like photons, matter waves, and quantum mechanics.
The dual nature principle states that both matter and radiation exhibit both wave-like and particle-like properties depending on the experimental context, fundamentally changing our understanding of physics at atomic scales.
Huygens, Young, Fresnel: Light as waves
Evidence: Interference, diffraction, polarization
Explained wave phenomena but failed with blackbody radiation
Einstein, Planck: Light as particles (photons)
Evidence: Photoelectric effect, Compton effect
Explained quantum phenomena but failed with wave effects
Energy
\( E = h\nu = \frac{hc}{\lambda} \)
h = Planck's constant = 6.626 × 10⁻³⁴ Js
Momentum
\( p = \frac{h}{\lambda} = \frac{h\nu}{c} \)
Massless but carries momentum
Rest Mass
\( m_0 = 0 \)
Photons have zero rest mass
Velocity
\( c = 3 \times 10^8 \) m/s
Constant in vacuum
Energy Conservation
\( h\nu = \phi + K_{\text{max}} \)
Photon energy = Work function + Max KE
Maximum Kinetic Energy
\( K_{\text{max}} = h\nu - \phi = h(\nu - \nu_0) \)
ν₀ = threshold frequency
Stopping Potential
\( eV_0 = K_{\text{max}} = h\nu - \phi \)
V₀ = stopping potential
Work Function
\( \phi = h\nu_0 = \frac{hc}{\lambda_0} \)
Minimum energy to eject electron
Every moving particle has wave properties associated with it
de Broglie Wavelength
\( \lambda = \frac{h}{p} = \frac{h}{mv} \)
For non-relativistic particles
For Charged Particles
\( \lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2meV}} \)
Accelerated through potential V
For Electrons
\( \lambda = \frac{12.27}{\sqrt{V}} \) Å
V in volts, λ in angstroms
For Gas Molecules
\( \lambda = \frac{h}{\sqrt{3mkT}} \)
At temperature T
Davisson-Germer Experiment
G.P. Thomson Experiment
Scattering of X-rays by electrons, demonstrating particle nature of radiation
Compton Shift
\( \Delta\lambda = \lambda' - \lambda = \frac{h}{m_0c}(1 - \cos\phi) \)
Change in wavelength
Compton Wavelength
\( \lambda_c = \frac{h}{m_0c} = 0.0243 \) Å
For electron scattering
Key Points:
Localized waves formed by superposition of many waves, representing particles
Position-Momentum
\( \Delta x \cdot \Delta p_x \geq \frac{\hbar}{2} \)
\( \hbar = \frac{h}{2\pi} \)
Energy-Time
\( \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \)
For excited states
Physical Significance:
| Property | Wave Nature | Particle Nature |
|---|---|---|
| Evidence | Interference, diffraction | Photoelectric effect, Compton effect |
| Energy | Depends on amplitude | E = hν (quantized) |
| Momentum | Not defined classically | p = h/λ |
Electron Microscope
Photocells
Problem: Light of wavelength 2000Å falls on aluminum surface (work function 4.2eV). Find maximum KE of photoelectrons and stopping potential.
Solution:
Photon energy: \( E = \frac{hc}{\lambda} = \frac{12400}{2000} = 6.2 \) eV
Maximum KE: \( K_{\text{max}} = E - \phi = 6.2 - 4.2 = 2.0 \) eV
Stopping potential: \( V_0 = \frac{K_{\text{max}}}{e} = 2.0 \) V
Problem: Find de Broglie wavelength of electron accelerated through 100V potential.
Solution:
Using \( \lambda = \frac{12.27}{\sqrt{V}} \) Å
\( \lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \) Å
Problem: X-rays of wavelength 1Å are scattered at 90°. Find wavelength of scattered X-rays.
Solution:
Compton shift: \( \Delta\lambda = \lambda_c(1 - \cos 90^\circ) = 0.0243(1 - 0) = 0.0243 \) Å
Scattered wavelength: \( \lambda' = \lambda + \Delta\lambda = 1 + 0.0243 = 1.0243 \) Å
1. Light of wavelength 3000Å falls on a surface with work function 2.3eV. Calculate the maximum kinetic energy of emitted photoelectrons.
[Answer: 1.83 eV]
2. Find the de Broglie wavelength of a proton accelerated through 400V potential. (mass of proton = 1.67×10⁻²⁷ kg)
[Answer: 1.43×10⁻¹² m]
3. X-rays of wavelength 0.5Å undergo Compton scattering at 60°. Find the wavelength of scattered X-rays.
[Answer: 0.51215 Å]
The photoelectric effect is the emission of electrons from a material when light of sufficient frequency strikes it. This phenomenon provided crucial evidence for the particle nature of light.
The photoelectric effect demonstrates the quantum nature of light, where light behaves as discrete packets of energy called photons. It challenged classical wave theory and laid the foundation for quantum mechanics.
Definition: Discrete packets of electromagnetic energy
Energy: \( E = h\nu = \frac{hc}{\lambda} \)
Momentum: \( p = \frac{h}{\lambda} \)
Definition: Minimum energy needed to eject an electron
Symbol: \( \phi \) (phi)
Typical values: 2-6 eV for metals
\( K_{\text{max}} \)
Maximum kinetic energy of ejected electrons
\( h\nu \)
Photon energy
\( \phi \)
Work function of material
\( \nu_0 = \frac{\phi}{h} \)
Threshold frequency
\( \lambda_0 = \frac{hc}{\phi} \)
Threshold wavelength
\( V_0 = \frac{h\nu - \phi}{e} \)
Stopping potential
\( K_{\text{max}} = eV_0 \)
Kinetic energy from stopping potential
Light Source → Photocathode
↓ Electrons
Anode ← Ammeter
[Experimental setup visualization]
Intensity Dependence
Classical: Higher intensity → Higher kinetic energy
Experimental: Higher intensity → More electrons, same Kmax
Frequency Threshold
Classical: No minimum frequency
Experimental: Threshold frequency exists
Classical Prediction
Time needed for energy accumulation
Experimental Result
Instantaneous emission (≤ 10-9 s)
Quantum Explanation
One photon → One electron (if ν ≥ ν₀)
Slope = h (Planck's constant)
Intercept = -φ
Slope = h
-φ
Linear relationship: Kmax = hν - φ
Slope = h/e
Intercept = -φ/e
Slope = h/e
-φ/e
V₀ = (h/e)ν - φ/e
Saturation current depends on intensity
Different intensities, same stopping potential
Linear relationship
I ∝ Intensity (for ν > ν₀)
| Material | Work Function (eV) | Threshold Frequency (×10¹⁴ Hz) | Threshold Wavelength (nm) |
|---|---|---|---|
| Cesium (Cs) | 1.95 | 4.71 | 637 |
| Potassium (K) | 2.30 | 5.56 | 540 |
| Sodium (Na) | 2.75 | 6.65 | 452 |
| Calcium (Ca) | 2.87 | 6.94 | 433 |
| Copper (Cu) | 4.65 | 11.2 | 267 |
| Silver (Ag) | 4.73 | 11.4 | 263 |
| Platinum (Pt) | 6.35 | 15.3 | 196 |
Problem: Light of wavelength 200 nm falls on a metal surface with work function 4.2 eV. Calculate the maximum kinetic energy of emitted electrons.
Solution:
Photon energy: \( E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{200 \text{ nm}} = 6.2 \text{ eV} \)
Using Einstein's equation: \( K_{\text{max}} = h\nu - \phi = 6.2 - 4.2 = 2.0 \text{ eV} \)
Note: Used hc = 1240 eV·nm for convenience
Problem: A metal has work function 2.3 eV. Find its threshold frequency and wavelength.
Solution:
Threshold frequency: \( \nu_0 = \frac{\phi}{h} = \frac{2.3 \text{ eV}}{4.14 \times 10^{-15} \text{ eV·s}} = 5.56 \times 10^{14} \text{ Hz} \)
Threshold wavelength: \( \lambda_0 = \frac{hc}{\phi} = \frac{1240 \text{ eV·nm}}{2.3 \text{ eV}} = 539 \text{ nm} \)
Problem: When light of wavelength 300 nm falls on a metal, the stopping potential is 1.5 V. Find the work function.
Solution:
Photon energy: \( E = \frac{1240}{300} = 4.13 \text{ eV} \)
Maximum kinetic energy: \( K_{\text{max}} = eV_0 = 1.5 \text{ eV} \)
Work function: \( \phi = E - K_{\text{max}} = 4.13 - 1.5 = 2.63 \text{ eV} \)
1. Light of wavelength 250 nm falls on a metal with work function 3.5 eV. Calculate the maximum kinetic energy of photoelectrons.
[Answer: 1.46 eV]
2. The threshold wavelength for potassium is 558 nm. Calculate its work function in eV.
[Answer: 2.22 eV]
3. When light of frequency 8×10¹⁴ Hz falls on a metal, photoelectrons with maximum kinetic energy 0.8 eV are emitted. Find the stopping potential and work function.
[Answer: V₀ = 0.8 V, φ = 2.51 eV]
Modern communication involves the transmission, reception, and processing of information using electronic systems and electromagnetic waves.
Modern communication systems cover modulation techniques, propagation methods, communication channels, and digital communication technologies.
Transmitter: Converts information into signals
Channel: Medium for signal transmission
Receiver: Recovers information from signals
Analog: Continuous time-varying signals
Digital: Discrete signals with binary values
Baseband: Original information signal
Amplitude Modulation (AM)
\( s(t) = A_c[1 + m_a\cos(2\pi f_m t)]\cos(2\pi f_c t) \)
Modulation index: \( m_a \)
Frequency Modulation (FM)
\( s(t) = A_c\cos[2\pi f_c t + \beta\sin(2\pi f_m t)] \)
Modulation index: \( \beta \)
Phase Modulation (PM)
\( s(t) = A_c\cos[2\pi f_c t + k_p m(t)] \)
Phase sensitivity: \( k_p \)
Amplitude Shift Keying (ASK)
\( s(t) = A\cos(2\pi f_c t) \) for binary '1'
\( s(t) = 0 \) for binary '0'
Frequency Shift Keying (FSK)
\( s(t) = A\cos(2\pi f_1 t) \) for binary '1'
\( s(t) = A\cos(2\pi f_2 t) \) for binary '0'
Phase Shift Keying (PSK)
\( s(t) = A\cos(2\pi f_c t) \) for binary '0'
\( s(t) = A\cos(2\pi f_c t + \pi) \) for binary '1'
Quadrature Phase Shift Keying (QPSK)
Uses four phases: 45°, 135°, 225°, 315°
2 bits per symbol
| Band | Frequency Range | Wavelength | Applications |
|---|---|---|---|
| VLF | 3-30 kHz | 10-100 km | Submarine communication |
| LF | 30-300 kHz | 1-10 km | Radio navigation |
| MF | 300 kHz-3 MHz | 100-1000 m | AM broadcasting |
| HF | 3-30 MHz | 10-100 m | Shortwave radio |
| VHF | 30-300 MHz | 1-10 m | FM radio, TV |
| UHF | 300 MHz-3 GHz | 10 cm-1 m | TV, mobile phones |
| SHF | 3-30 GHz | 1-10 cm | Satellite communication |
Waves that travel along the Earth's surface
Waves reflected by ionosphere
Line-of-sight propagation
Cellular Concept
Generations
Orbit Types
Applications
Components
Advantages
\( f_s \geq 2f_m \)
Where \( f_s \) is sampling frequency and \( f_m \) is maximum signal frequency
Step size: \( \Delta = \frac{V_{max} - V_{min}}{L} \)
Where L is number of quantization levels
Bits required: \( n = \log_2 L \)
For L quantization levels
\( R_b = n \times f_s \)
Where n is bits per sample
Problem: A carrier wave of 1 MHz with amplitude 10V is amplitude modulated by a 5 kHz audio signal. If modulation index is 0.8, find frequencies present in AM wave.
Solution:
Carrier frequency: \( f_c = 1 \) MHz
Modulating frequency: \( f_m = 5 \) kHz
Frequencies present:
\( f_c = 1000 \) kHz
\( f_c + f_m = 1005 \) kHz
\( f_c - f_m = 995 \) kHz
Problem: An audio signal has maximum frequency 8 kHz. What is minimum sampling rate? If 128 quantization levels are used, find bit rate.
Solution:
Minimum sampling rate: \( f_s = 2 \times f_m = 2 \times 8 = 16 \) kHz
Bits per sample: \( n = \log_2 128 = 7 \) bits
Bit rate: \( R_b = n \times f_s = 7 \times 16000 = 112 \) kbps
Problem: A TV tower has height 160m. Find maximum coverage distance. (Radius of Earth = 6400 km)
Solution:
Using \( d = \sqrt{2Rh} \):
\( d = \sqrt{2 \times 6400 \times 10^3 \times 160} \)
\( d = \sqrt{2048 \times 10^6} = 45,254 \) m ≈ 45.25 km
1. A carrier wave of 800 kHz is modulated by 10 kHz signal. Find the frequencies present in AM wave.
[Answer: 790 kHz, 800 kHz, 810 kHz]
2. Calculate minimum sampling rate for signal with maximum frequency 12 kHz and find bit rate if 8 bits per sample are used.
[Answer: f_s = 24 kHz, R_b = 192 kbps]
3. A transmitting antenna has height 100m. Calculate maximum line-of-sight distance. (R = 6400 km)
[Answer: d = 35.78 km]
Waves are disturbances that transfer energy through matter or space without transferring matter. Oscillations are repetitive variations about a central value.
This chapter covers simple harmonic motion, wave properties, wave types, wave equations, interference, standing waves, and sound waves.
Oscillation: Repetitive motion about equilibrium
Wave: Disturbance propagating through medium
Mechanical: Requires medium (sound, water)
Electromagnetic: No medium required (light, radio)
Displacement
\( x = A\cos(\omega t + \phi) \)
or \( x = A\sin(\omega t + \phi) \)
Velocity
\( v = -A\omega\sin(\omega t + \phi) \)
\( v_{\text{max}} = A\omega \)
Acceleration
\( a = -A\omega^2\cos(\omega t + \phi) \)
\( a_{\text{max}} = A\omega^2 \)
Force
\( F = -kx \)
Hooke's Law
Angular Frequency
\( \omega = 2\pi f = \sqrt{\frac{k}{m}} \)
Period and Frequency
\( T = \frac{1}{f} = 2\pi\sqrt{\frac{m}{k}} \)
Energy in SHM
Total Energy: \( E = \frac{1}{2}kA^2 \)
Constant
Energy Distribution
PE: \( \frac{1}{2}kx^2 \), KE: \( \frac{1}{2}mv^2 \)
Sum is constant
Amplitude (A): Maximum displacement
Wavelength (λ): Distance between crests
Frequency (f): Oscillations per second
Period (T): Time for one oscillation
Wave Speed (v): Speed of wave propagation
Phase: Position in wave cycle
Wave Speed
\( v = f\lambda = \frac{\omega}{k} \)
Wave Number
\( k = \frac{2\pi}{\lambda} \)
Wave Equation
\( y = A\sin(kx - \omega t + \phi) \)
Traveling wave
Phase Difference
\( \Delta\phi = \frac{2\pi}{\lambda}\Delta x \)
Particles oscillate perpendicular to wave direction
Particles oscillate parallel to wave direction
String/Wire
\( v = \sqrt{\frac{T}{\mu}} \)
T = tension, μ = mass/length
Sound in Air
\( v = \sqrt{\frac{\gamma RT}{M}} \)
~343 m/s at 20°C
When two waves meet, their displacements add algebraically
Constructive Interference
Phase difference: \( \Delta\phi = 2n\pi \)
Amplitude increases
Destructive Interference
Phase difference: \( \Delta\phi = (2n+1)\pi \)
Amplitude decreases
Result from interference of two identical waves traveling in opposite directions
Equation
\( y = 2A\sin(kx)\cos(\omega t) \)
Nodes and Antinodes
Nodes: points of zero amplitude
Antinodes: points of maximum amplitude
Harmonics:
String fixed at both ends: \( f_n = \frac{nv}{2L} \), n = 1,2,3...
Open pipe: \( f_n = \frac{nv}{2L} \), Closed pipe: \( f_n = \frac{nv}{4L} \) (n odd)
Change in frequency due to relative motion between source and observer
Source Moving
\( f' = f\frac{v}{v \pm v_s} \)
+ when moving away, - when approaching
Observer Moving
\( f' = f\frac{v \pm v_o}{v} \)
+ when approaching, - when moving away
Pitch: Related to frequency
Loudness: Related to intensity
Quality: Related to waveform
\( I = \frac{P}{A} = 2\pi^2\rho v f^2 A^2 \)
Measured in W/m²
\( \beta = 10\log_{10}\left(\frac{I}{I_0}\right) \) dB
I₀ = 10⁻¹² W/m² (threshold)
\( f_{\text{beat}} = |f_1 - f_2| \)
Result of two close frequencies
Problem: A 0.5kg mass on a spring (k=200 N/m) oscillates with amplitude 0.1m. Find period, max speed, and total energy.
Solution:
Angular frequency: \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = 20 \) rad/s
Period: \( T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = 0.314 \) s
Max speed: \( v_{\text{max}} = A\omega = 0.1 \times 20 = 2 \) m/s
Total energy: \( E = \frac{1}{2}kA^2 = \frac{1}{2}(200)(0.1)^2 = 1 \) J
Problem: A wave has wavelength 2m and frequency 5Hz. Find wave speed. If tension doubles, find new speed.
Solution:
Wave speed: \( v = f\lambda = 5 \times 2 = 10 \) m/s
For string: \( v \propto \sqrt{T} \)
New speed: \( v_{\text{new}} = 10 \times \sqrt{2} = 14.14 \) m/s
Problem: A string of length 2m fixed at both ends has wave speed 40 m/s. Find fundamental frequency and first two overtones.
Solution:
Fundamental: \( f_1 = \frac{v}{2L} = \frac{40}{2 \times 2} = 10 \) Hz
First overtone: \( f_2 = 2f_1 = 20 \) Hz
Second overtone: \( f_3 = 3f_1 = 30 \) Hz
1. A spring-mass system has m=2kg, k=50 N/m. Find period and frequency of oscillation.
[Answer: T = 1.26 s, f = 0.796 Hz]
2. A wave has frequency 440Hz and speed 340 m/s. Calculate wavelength.
[Answer: λ = 0.773 m]
3. A string 1.5m long fixed at both ends vibrates at 300Hz in its fundamental mode. Find wave speed.
[Answer: v = 900 m/s]
4. Calculate sound level for intensity 10⁻⁵ W/m².
[Answer: β = 70 dB]
The Doppler Effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.
The Doppler Effect explains why a siren's pitch appears higher as it approaches and lower as it moves away. It applies to sound waves, light waves, and other wave phenomena.
Source Moving: Frequency changes due to relative motion
Observer Moving: Different formulas for approach vs recession
Sound Waves: Change in perceived pitch
Light Waves: Redshift and blueshift in astronomy
\( f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) \)
Where:
\( f' \) = observed frequency
\( f \) = source frequency
\( v \) = speed of sound
Sign Convention:
\( v_o \) = observer's speed
\( v_s \) = source's speed
+ when moving toward, - when moving away
Source Moving, Observer Stationary
Approaching: \( f' = f \left( \frac{v}{v - v_s} \right) \)
Receding: \( f' = f \left( \frac{v}{v + v_s} \right) \)
Observer Moving, Source Stationary
Approaching: \( f' = f \left( \frac{v + v_o}{v} \right) \)
Receding: \( f' = f \left( \frac{v - v_o}{v} \right) \)
For light waves, we must use relativistic formulas due to the constant speed of light
Source Approaching
\( f' = f \sqrt{\frac{1 + \beta}{1 - \beta}} \)
Blueshift (higher frequency)
Source Receding
\( f' = f \sqrt{\frac{1 - \beta}{1 + \beta}} \)
Redshift (lower frequency)
Where \( \beta = \frac{v}{c} \), with \( v \) = relative speed and \( c \) = speed of light
Galaxies moving away
\( z = \frac{\lambda' - \lambda}{\lambda} \)
Objects approaching
Negative z values
\( v = H_0 d \)
Universe expansion
Concentric circular wavefronts
Equal wavelength in all directions
Compressed waves ahead
Stretched waves behind
Mach Number
\( M = \frac{v_s}{v} \)
Ratio of source speed to sound speed
Mach Cone Angle
\( \sin\theta = \frac{v}{v_s} = \frac{1}{M} \)
Angle of shock wave cone
Sonic Boom Characteristics:
Problem: A police siren emits sound at 1000 Hz. If the police car moves toward you at 30 m/s, what frequency do you hear? (Speed of sound = 340 m/s)
Solution:
Using \( f' = f \left( \frac{v}{v - v_s} \right) \):
\( f' = 1000 \left( \frac{340}{340 - 30} \right) = 1000 \left( \frac{340}{310} \right) \)
\( f' = 1000 \times 1.097 = 1097 \) Hz
Problem: You drive toward a stationary siren at 25 m/s. The siren emits 800 Hz sound. What frequency do you hear? (Speed of sound = 340 m/s)
Solution:
Using \( f' = f \left( \frac{v + v_o}{v} \right) \):
\( f' = 800 \left( \frac{340 + 25}{340} \right) = 800 \left( \frac{365}{340} \right) \)
\( f' = 800 \times 1.074 = 859 \) Hz
Problem: A jet flies at Mach 2.0. What is the angle of its shock wave cone?
Solution:
Using \( \sin\theta = \frac{1}{M} \):
\( \sin\theta = \frac{1}{2.0} = 0.5 \)
\( \theta = \sin^{-1}(0.5) = 30^\circ \)
1. A train whistle emits 440 Hz sound. If the train moves away from you at 20 m/s, what frequency do you hear? (v = 340 m/s)
[Answer: 415.6 Hz]
2. You run toward a stationary sound source at 10 m/s. The source emits 500 Hz sound. What frequency do you hear? (v = 340 m/s)
[Answer: 514.7 Hz]
3. A galaxy shows redshift of z = 0.1. What fraction of light speed is it receding?
[Answer: v/c ≈ 0.095]
4. Calculate the Mach cone angle for an aircraft flying at 680 m/s. (v = 340 m/s)
[Answer: θ = 30°]
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