CBSE Physics — Class 12 Physics Board Complete Chapter Guide

Chapter Overview & Weightage

CBSE Class 12 Physics carries 70 marks in theory (30 marks practical). The board splits the syllabus into 9 units, but for strategy purposes, we group them into 6 major blocks. Here’s the real picture — unit-wise weightage from the last 5 years:

Unit	Topics	2020	2021	2022	2023	2024
Unit 1	Electrostatics	8	8	8	8	8
Unit 2	Current Electricity	7	7	7	7	7
Unit 3	Magnetic Effects + Magnetism	8	8	8	8	8
Unit 4	EMI + AC	8	8	8	8	8
Unit 5	Optics	14	14	14	14	14
Unit 6	Dual Nature + Atoms + Nuclei	12	12	12	12	12
Unit 7	Semiconductors	7	7	7	7	7
Unit 8	Communication Systems	3	3	3	3	3
Unit 9	EM Waves	3	3	3	3	3

Optics (14 marks) and Dual Nature + Atoms + Nuclei (12 marks) together give you 26 marks — that’s 37% of theory. If you’re pressed for time before boards, these two blocks are non-negotiable. Communication Systems (3 marks) is 100% theoretical and takes 2 hours to prepare — do it the night before.

The paper follows a fixed structure: 1-mark MCQs (16Q), 2-mark short answers (5Q), 3-mark short answers (7Q), 5-mark long answers (3Q, internal choice), and case-based MCQs (4Q, 4 marks). Knowing this structure helps you allocate writing time.

Key Concepts You Must Know

Ranked by exam frequency — the topics at the top appear every single year without fail.

Electrostatics (8 marks)

Coulomb’s law and superposition principle
Electric field lines, dipole field, and flux (Gauss’s Law — appears every year)
Capacitors in series/parallel; energy stored; effect of dielectric
Equipotential surfaces and relation between E and V

Current Electricity (7 marks)

Ohm’s law, resistivity, and temperature dependence
Kirchhoff’s laws — both KCL and KVL
Wheatstone bridge and meter bridge
Potentiometer: comparing EMFs and finding internal resistance

Magnetism (8 marks)

Biot-Savart law for circular loop; Ampere’s law for solenoid
Force on a current-carrying conductor; F = BIL sinθ
Moving coil galvanometer — principle, sensitivity, conversion to ammeter/voltmeter
Magnetic properties: dia, para, ferromagnetic substances

EMI and AC (8 marks)

Faraday’s law; Lenz’s law — direction of induced current
Self and mutual inductance; energy stored in inductor
AC: phasors, impedance of LCR circuit, resonance, power factor
Transformer — efficiency, step-up/step-down

Optics (14 marks)

Mirror formula, lens formula, lens maker’s equation
Refraction through prism — angle of minimum deviation
Total Internal Reflection (TIR) — critical angle, optical fibre
Young’s double slit experiment (YDSE) — fringe width, path difference
Diffraction: single slit, condition for minima
Polarisation: Malus’s law, Brewster’s angle

Modern Physics (12 marks)

Photoelectric effect: Einstein’s equation, stopping potential, threshold frequency
de Broglie wavelength
Bohr’s model: energy levels, spectral series of hydrogen
Radioactive decay: activity, half-life, decay law
Nuclear reactions: mass defect, binding energy per nucleon

Semiconductors (7 marks)

p-n junction: forward and reverse bias, I-V characteristics
Rectifier: half-wave and full-wave
Zener diode as voltage regulator
Logic gates: AND, OR, NOT, NAND, NOR — truth tables

Important Formulas

F = \frac{kq_1q_2}{r^2}, \quad k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}

E = \frac{\sigma}{\varepsilon_0} \text{ (infinite plane)}, \quad E = \frac{\sigma}{2\varepsilon_0} \text{ (single sheet)}

C = \frac{\varepsilon_0 A}{d}, \quad U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

When to use: Gauss’s Law works beautifully for high-symmetry charge distributions (sphere, infinite wire, infinite plane). If the problem gives you a non-symmetric shape, go back to direct integration or superposition.

R = \frac{\rho L}{A}, \quad \rho = \rho_0[1 + \alpha(T - T_0)]

\text{Wheatstone balance: } \frac{P}{Q} = \frac{R}{S}

\text{Potentiometer EMF: } \frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \text{ (mirror)}, \quad \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \text{ (lens, same form)}

\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \text{ (lens maker's)}

\beta = \frac{\lambda D}{d} \text{ (YDSE fringe width)}

\sin\theta_c = \frac{1}{\mu} \text{ (critical angle for TIR)}

\delta_m = 2i - A, \quad \mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

KE_{max} = h\nu - \phi_0 = eV_s

\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}} \text{ (de Broglie)}

E_n = \frac{-13.6}{n^2} \text{ eV (hydrogen)}

N = N_0 e^{-\lambda t}, \quad T_{1/2} = \frac{0.693}{\lambda}

\Delta m \cdot c^2 = \text{Binding Energy}

Z = \sqrt{R^2 + (X_L - X_C)^2}, \quad X_L = \omega L, \quad X_C = \frac{1}{\omega C}

f_r = \frac{1}{2\pi\sqrt{LC}} \text{ (resonance)}

P = V_{rms} I_{rms} \cos\phi

Solved Previous Year Questions

PYQ 1 — CBSE Board 2023 (5 marks)

Q: A parallel plate capacitor with plate area A and plate separation d is charged to potential V. The battery is then disconnected. A dielectric slab of thickness d and dielectric constant K is inserted. Find the new (i) capacitance, (ii) potential, (iii) energy stored. Comment on the change in energy.

Solution:

Step 1: Find initial quantities.

C_0 = \frac{\varepsilon_0 A}{d}, \quad Q = C_0 V, \quad U_0 = \frac{1}{2}C_0 V^2

Step 2: After disconnecting the battery, charge Q stays constant (no path for charge to escape).

Step 3: Insert dielectric — capacitance changes.

C_{new} = KC_0 = \frac{K\varepsilon_0 A}{d}

Step 4: New potential (charge is fixed, C increased):

V_{new} = \frac{Q}{C_{new}} = \frac{C_0 V}{KC_0} = \frac{V}{K}

Step 5: New energy stored:

U_{new} = \frac{Q^2}{2C_{new}} = \frac{(C_0 V)^2}{2KC_0} = \frac{C_0 V^2}{2K} = \frac{U_0}{K}

Energy decreases by a factor of K. The “lost” energy does work in pulling the dielectric slab into the capacitor — it goes into mechanical energy.

Most students forget whether the battery is connected or disconnected. If connected: V is constant, so Q and C both change. If disconnected: Q is constant, so V and E change. Always identify this first — it decides everything else.

PYQ 2 — CBSE Board 2024 (3 marks)

Q: In Young’s double slit experiment, the slits are 0.5 mm apart and the screen is 1.5 m away. If blue light of wavelength 450 nm is used, find the fringe width and the distance of the 5th bright fringe from the central maximum.

Step 1: Fringe width formula.

\beta = \frac{\lambda D}{d} = \frac{450 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}}

\beta = \frac{675 \times 10^{-9}}{5 \times 10^{-4}} = 1.35 \times 10^{-3} \text{ m} = 1.35 \text{ mm}

Step 2: Position of 5th bright fringe.

y_5 = 5\beta = 5 \times 1.35 = 6.75 \text{ mm}

YDSE numericals in CBSE are almost always straightforward substitution — the marks are mostly for correctly writing the formula with the right variables and showing unit conversion clearly. Write every step.

PYQ 3 — CBSE Board 2023 (3 marks)

Q: The half-life of a radioactive substance is 30 minutes. Starting with 80g of the substance, how much will remain after 2 hours? What is the activity if the molar mass is 180 g/mol?

Step 1: Find number of half-lives.

n = \frac{t}{T_{1/2}} = \frac{120}{30} = 4

Step 2: Amount remaining.

N = N_0 \left(\frac{1}{2}\right)^n = 80 \times \left(\frac{1}{2}\right)^4 = 80 \times \frac{1}{16} = 5 \text{ g}

Step 3: Activity requires $\lambda$ and number of atoms.

\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{30 \times 60} = 3.85 \times 10^{-4} \text{ s}^{-1}

N_{atoms} = \frac{5}{180} \times 6.022 \times 10^{23} = 1.67 \times 10^{22}

A = \lambda N = 3.85 \times 10^{-4} \times 1.67 \times 10^{22} = 6.43 \times 10^{18} \text{ Bq}

Difficulty Distribution

For CBSE Class 12 Physics board exams, the difficulty typically breaks down as:

Level	Percentage	What it means
Easy	~40%	Direct formula substitution, definitions, 1-2 mark questions
Medium	~45%	Derivations, PYQ-style numericals, circuit problems
Hard	~15%	Multi-step problems, unfamiliar contexts, case-based questions

The “hard” 15% in CBSE is not JEE-hard — it’s unfamiliar application of known concepts. A student who has seen enough PYQs will find most of it manageable. The real differentiator in CBSE is not intelligence — it’s how completely you write your answers.

The 5-mark questions (3 total, with internal choice) are almost always: one derivation, one numerical with multiple parts, and one diagram-heavy question (ray optics or circuit diagram). These three carry 15 marks — prepare them with full written steps.

Expert Strategy

Month-wise breakdown (assuming 3 months to boards):

Month 1: Cover Electrostatics, Current Electricity, and Magnetism. These are conceptually dense but numerically straightforward. Do derivations from NCERT — CBSE lifts them verbatim.

Month 2: Optics (spend extra time here — 14 marks), EMI, and AC. YDSE, lens formula, and the prism deviation derivation appear every year. The AC chapter has beautiful internal connections — understanding phasors makes everything else fall into place.

Month 3: Modern Physics, Semiconductors, Communication Systems. This is also revision month. Do the last 5 years’ PYQs timed.

NCERT is non-negotiable for CBSE. Every derivation asked in the board exam — Gauss’s law for sphere, energy stored in capacitor, force between parallel wires, lens maker’s equation — appears in NCERT with the exact same structure. If you can reproduce NCERT derivations with diagrams, you’re securing 80% of your marks.

For numericals: CBSE rarely invents new problem types. The last 10 years of PYQs cover ~85% of what will appear. After solving a PYQ, don’t just check your answer — check if your presentation matches what the marking scheme expects. CBSE gives step marks, and a wrong final answer with correct steps can still fetch 3/5.

Diagrams: Every ray optics question, every circuit diagram, and every magnetic field line diagram needs to be drawn with labels. Examiners are instructed to deduct marks for missing labels. Practice drawing these quickly but neatly.

Common Traps

Trap 1 — Sign convention in mirrors and lenses. CBSE follows the New Cartesian Convention universally. Distances in the direction of incident light are positive. For a concave mirror with object in front: u is negative, f is negative. Students who haven’t drilled this lose marks on straightforward problems.

Trap 2 — Galvanometer conversion. To convert to ammeter: shunt in parallel (low resistance). To convert to voltmeter: high resistance in series. Students flip these under exam pressure. Remember: ammeter must have low resistance (doesn’t disturb current), voltmeter must have high resistance (doesn’t draw current).

Trap 3 — Photoelectric effect with frequency vs wavelength. $KE_{max}$ depends on frequency, not intensity. Increasing intensity increases the number of photoelectrons, not their energy. In case-based questions, CBSE often gives a graph where intensity changes — the stopping potential remains unchanged. This catches students who haven’t thought about it.

Trap 4 — Lenz’s law direction. When asked for the direction of induced current, always ask: “What change in flux is happening?” If flux is increasing into the page, induced current opposes that — so it flows counterclockwise (by right-hand rule, its field points out of the page). Many students apply the right-hand rule without thinking about opposition first.

Trap 5 — Logic gates: NAND and NOR. These are “universal gates” — any logic function can be built from NAND alone or NOR alone. CBSE asks for this fact and also for truth tables. Students who memorise NAND = NOT(AND) and NOR = NOT(OR) can quickly derive any truth table on the spot rather than memorising it separately.

Final week strategy: Don’t start new topics in the last 7 days. Revise formulas chapter by chapter, redraw all standard diagrams from memory, and solve one full past paper timed. Physics boards reward systematic, clearly laid-out answers over clever shortcuts — write more, not less.