NCERT / CBSE Board Exam
1. Electric Charges and Fields
Coulomb’s Law
F = 1 4 π ε 0 ∣ q 1 q 2 ∣ r 2 F=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2} k = 1 4 π ε 0 = 9 × 10 9 N m 2 C − 2 k=\frac{1}{4\pi\varepsilon_0}=9\times10^9\;Nm^2C^{-2}
In a medium:F = 1 4 π ε ∣ q 1 q 2 ∣ r 2 F=\frac{1}{4\pi\varepsilon}\frac{|q_1q_2|}{r^2} ε = K ε 0 \varepsilon=K\varepsilon_0
Electric Field
E ⃗ = F ⃗ q 0 \vec E=\frac{\vec F}{q_0}
For a point charge:E ⃗ = 1 4 π ε 0 q r 2 r ^ \vec E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat r
Superposition Principle
E ⃗ = E ⃗ 1 + E ⃗ 2 + E ⃗ 3 + ⋯ \vec E=\vec E_1+\vec E_2+\vec E_3+\cdots
Electric Dipole
Dipole moment:p ⃗ = q ( 2 a ⃗ ) \vec p=q(2\vec a)
Magnitude:p = q ( 2 a ) p=q(2a)
Direction: from − q -q to + q +q .
Field Due to Dipole
Axial position:E a x i a l = 1 4 π ε 0 2 p r ( r 2 − a 2 ) 2 E_{\rm axial} = \frac{1}{4\pi\varepsilon_0} \frac{2pr}{(r^2-a^2)^2}
For r ≫ a r\gg a :E a x i a l = 1 4 π ε 0 2 p r 3 E_{\rm axial} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}
Equatorial position:E e q u a t o r i a l = 1 4 π ε 0 p ( r 2 + a 2 ) 3 / 2 E_{\rm equatorial} = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2+a^2)^{3/2}}
For r ≫ a r\gg a :E e q u a t o r i a l = 1 4 π ε 0 p r 3 E_{\rm equatorial} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}
Torque on an Electric Dipole
τ ⃗ = p ⃗ × E ⃗ \vec\tau=\vec p\times\vec E τ = p E sin θ \tau=pE\sin\theta
Electric Flux
Φ E = E ⃗ ⋅ A ⃗ \Phi_E=\vec E\cdot\vec A Φ E = E A cos θ \Phi_E=EA\cos\theta
Gauss’s Law
∮ E ⃗ ⋅ d S ⃗ = q e n c l o s e d ε 0 \boxed{\oint\vec E\cdot d\vec S=\frac{q_{\rm enclosed}}{\varepsilon_0}}
Important Results from Gauss’s Law
Infinite straight wire:E = λ 2 π ε 0 r E=\frac{\lambda}{2\pi\varepsilon_0r}
Infinite plane sheet:E = σ 2 ε 0 E=\frac{\sigma}{2\varepsilon_0}
Two oppositely charged infinite sheets:E = σ ε 0 E=\frac{\sigma}{\varepsilon_0}
Spherical shell, outside:E = 1 4 π ε 0 Q r 2 E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}
Inside:E = 0 E=0
2. Electrostatic Potential and Capacitance
Electric Potential
V = W q V=\frac{W}{q}
Point charge:V = 1 4 π ε 0 q r V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}
System of charges:V = 1 4 π ε 0 ∑ i q i r i V=\frac{1}{4\pi\varepsilon_0} \sum_i\frac{q_i}{r_i}
Potential Due to Dipole
V = 1 4 π ε 0 p cos θ r 2 ( r ≫ a ) V=\frac{1}{4\pi\varepsilon_0} \frac{p\cos\theta}{r^2} \qquad(r\gg a)
Potential Difference
V B − V A = − ∫ A B E ⃗ ⋅ d l ⃗ V_B-V_A=-\int_A^B\vec E\cdot d\vec l
Relation Between E and V
E = − d V d r E=-\frac{dV}{dr}
Potential Energy
Two charges:U = 1 4 π ε 0 q 1 q 2 r U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r}
System of charges:U = 1 4 π ε 0 ∑ i < j q i q j r i j U=\frac{1}{4\pi\varepsilon_0} \sum_{i<j}\frac{q_iq_j}{r_{ij}}
Dipole in External Field
U = − p ⃗ ⋅ E ⃗ U=-\vec p\cdot\vec E U = − p E cos θ U=-pE\cos\theta
Capacitance
C = Q V C=\frac{Q}{V}
Parallel plate capacitor:C = ε 0 A d C=\frac{\varepsilon_0A}{d}
With dielectric:C = K ε 0 A d C=\frac{K\varepsilon_0A}{d}
Capacitors in Series
1 C = 1 C 1 + 1 C 2 + ⋯ \frac1C=\frac1{C_1}+\frac1{C_2}+\cdots
Capacitors in Parallel
C = C 1 + C 2 + ⋯ C=C_1+C_2+\cdots
Energy Stored
U = 1 2 C V 2 U=\frac12CV^2 U = 1 2 Q V U=\frac12QV U = Q 2 2 C U=\frac{Q^2}{2C}
Energy density:u = 1 2 ε 0 E 2 u=\frac12\varepsilon_0E^2
3. Current Electricity
Electric Current
I = d Q d t I=\frac{dQ}{dt}
For steady current:I = Q t I=\frac Qt
Drift Velocity
v d = e E τ m v_d=\frac{eE\tau}{m} I = n e A v d I=neAv_d v d = I n e A v_d=\frac{I}{neA}
Current Density
J = I A J=\frac IA J = n e v d J=nev_d
Ohm’s Law
V = I R V=IR
Resistance
R = ρ l A R=\rho\frac{l}{A}
Conductivity
σ = 1 ρ \sigma=\frac1\rho σ = n e 2 τ m \sigma=\frac{ne^2\tau}{m}
Temperature Dependence
ρ T = ρ 0 [ 1 + α ( T − T 0 ) ] \rho_T=\rho_0[1+\alpha(T-T_0)]
Approximately:R T = R 0 [ 1 + α ( T − T 0 ) ] R_T=R_0[1+\alpha(T-T_0)]
Series Combination
R = R 1 + R 2 + ⋯ R=R_1+R_2+\cdots
Parallel Combination
1 R = 1 R 1 + 1 R 2 + ⋯ \frac1R=\frac1{R_1}+\frac1{R_2}+\cdots
Electrical Power
P = V I P=VI P = I 2 R P=I^2R P = V 2 R P=\frac{V^2}{R}
Electrical Energy
W = V I t W=VIt
Cell
V = E − I r V=\mathcal E-Ir I = E R + r I=\frac{\mathcal E}{R+r}
Terminal voltage during charging:V = E + I r V=\mathcal E+Ir
Cells in Series
For identical cells:E e q = n E \mathcal E_{\rm eq}=n\mathcal E r e q = n r r_{\rm eq}=nr
Cells in Parallel
For n n identical cells:E e q = E \mathcal E_{\rm eq}=\mathcal E r e q = r n r_{\rm eq}=\frac rn
Kirchhoff’s Junction Rule
∑ I i n = ∑ I o u t \boxed{\sum I_{\rm in}=\sum I_{\rm out}}
Kirchhoff’s Loop Rule
∑ Δ V = 0 \boxed{\sum\Delta V=0}
Wheatstone Bridge
At balance:P Q = R S \boxed{\frac{P}{Q}=\frac{R}{S}}
4. Moving Charges and Magnetism
Magnetic Force
F ⃗ = q ( v ⃗ × B ⃗ ) \vec F=q(\vec v\times\vec B) F = q v B sin θ F=qvB\sin\theta
Lorentz Force
F ⃗ = q ( E ⃗ + v ⃗ × B ⃗ ) \vec F=q(\vec E+\vec v\times\vec B)
Force on Current-Carrying Conductor
F ⃗ = I ( l ⃗ × B ⃗ ) \vec F=I(\vec l\times\vec B) F = I l B sin θ F=IlB\sin\theta
Motion of Charged Particle
For v ⊥ B v\perp B :q v B = m v 2 r qvB=\frac{mv^2}{r} r = m v q B r=\frac{mv}{qB} ω = q B m \omega=\frac{qB}{m} T = 2 π m q B T=\frac{2\pi m}{qB} f = q B 2 π m f=\frac{qB}{2\pi m}
Velocity Selector
q E = q v B qE=qvB v = E B v=\frac EB
Biot-Savart Law
d B ⃗ = μ 0 4 π I ( d l ⃗ × r ^ ) r 2 d\vec B= \frac{\mu_0}{4\pi} \frac{I(d\vec l\times\hat r)}{r^2}
Circular Current Loop
At centre:B = μ 0 I 2 R B=\frac{\mu_0I}{2R}
For N N turns:B = μ 0 N I 2 R B=\frac{\mu_0NI}{2R}
On axis:B = μ 0 N I R 2 2 ( R 2 + x 2 ) 3 / 2 B= \frac{\mu_0NIR^2} {2(R^2+x^2)^{3/2}}
Ampere’s Circuital Law
∮ B ⃗ ⋅ d l ⃗ = μ 0 I \oint\vec B\cdot d\vec l=\mu_0I
Long Straight Wire
B = μ 0 I 2 π r B=\frac{\mu_0I}{2\pi r}
Long Solenoid
B = μ 0 n I B=\mu_0nI
wheren = N L n=\frac NL
Force Between Parallel Wires
F l = μ 0 I 1 I 2 2 π d \frac Fl=\frac{\mu_0I_1I_2}{2\pi d}
Magnetic Dipole Moment of Current Loop
m ⃗ = N I A ⃗ \vec m=NI\vec A
Torque on Current Loop
τ ⃗ = m ⃗ × B ⃗ \vec\tau=\vec m\times\vec B τ = N I A B sin θ \tau=NIAB\sin\theta
Moving Coil Galvanometer
N I A B = k θ NIAB=k\theta I = k θ N A B I=\frac{k\theta}{NAB}
Current sensitivity:θ I = N A B k \frac{\theta}{I}=\frac{NAB}{k}
Galvanometer → Ammeter
S = I g G I − I g S=\frac{I_gG}{I-I_g}
Galvanometer → Voltmeter
R = V I g − G R=\frac{V}{I_g}-G
5. Magnetism and Matter
Magnetic Dipole Moment of Bar Magnet
m = ( pole strength ) ( magnetic length ) m=(\text{pole strength})(\text{magnetic length})
Torque
τ ⃗ = m ⃗ × B ⃗ \vec\tau=\vec m\times\vec B τ = m B sin θ \tau=mB\sin\theta
Potential Energy
U = − m ⃗ ⋅ B ⃗ U=-\vec m\cdot\vec B U = − m B cos θ U=-mB\cos\theta
Magnetisation
M = net magnetic moment volume M=\frac{\text{net magnetic moment}}{\text{volume}}
Magnetic Intensity
H = B μ 0 − M H=\frac{B}{\mu_0}-M
Therefore:B = μ 0 ( H + M ) B=\mu_0(H+M)
Magnetic Susceptibility
M = χ m H M=\chi_mH
Relative Permeability
μ r = 1 + χ m \mu_r=1+\chi_m B = μ 0 μ r H B=\mu_0\mu_rH
6. Electromagnetic Induction
Magnetic Flux
Φ B = B ⃗ ⋅ A ⃗ \Phi_B=\vec B\cdot\vec A Φ B = B A cos θ \Phi_B=BA\cos\theta
Faraday’s Law
E = − d Φ B d t \boxed{\mathcal E=-\frac{d\Phi_B}{dt}}
For N N turns:E = − N d Φ B d t \boxed{\mathcal E=-N\frac{d\Phi_B}{dt}}
Motional EMF
E = B l v \mathcal E=Blv
for mutually perpendicular B , l , v B,l,v .
Induced Current
I = E R I=\frac{\mathcal E}{R}
Self-Inductance
N Φ = L I N\Phi=LI E = − L d I d t \mathcal E=-L\frac{dI}{dt}
Mutual Inductance
N 2 Φ 2 = M I 1 N_2\Phi_2=MI_1 E 2 = − M d I 1 d t \mathcal E_2=-M\frac{dI_1}{dt}
Energy Stored in Inductor
U = 1 2 L I 2 U=\frac12LI^2
7. Alternating Current
AC Current and Voltage
i = i 0 sin ω t i=i_0\sin\omega t v = v 0 sin ω t v=v_0\sin\omega t ω = 2 π f \omega=2\pi f
RMS Values
I r m s = I 0 2 I_{\rm rms}=\frac{I_0}{\sqrt2} V r m s = V 0 2 V_{\rm rms}=\frac{V_0}{\sqrt2}
Inductive Reactance
X L = ω L X_L=\omega L
Capacitive Reactance
X C = 1 ω C X_C=\frac1{\omega C}
Series LCR Circuit
Z = R 2 + ( X L − X C ) 2 Z=\sqrt{R^2+(X_L-X_C)^2} I r m s = V r m s Z I_{\rm rms}=\frac{V_{\rm rms}}Z
Phase Angle
tan ϕ = X L − X C R \tan\phi=\frac{X_L-X_C}{R}
Power Factor
cos ϕ = R Z \cos\phi=\frac RZ
Average Power
P = V r m s I r m s cos ϕ P=V_{\rm rms}I_{\rm rms}\cos\phi
Resonance
X L = X C X_L=X_C ω 0 = 1 L C \omega_0=\frac1{\sqrt{LC}} f 0 = 1 2 π L C f_0=\frac1{2\pi\sqrt{LC}}
At resonance:Z = R Z=R
AC Generator
Φ = N B A cos ω t \Phi=NBA\cos\omega t e = N B A ω sin ω t e=NBA\omega\sin\omega t e 0 = N B A ω e_0=NBA\omega
Transformer
V s V p = N s N p \boxed{\frac{V_s}{V_p}=\frac{N_s}{N_p}}
Ideal transformer:V p I p = V s I s V_pI_p=V_sI_s I s I p = N p N s \frac{I_s}{I_p}=\frac{N_p}{N_s}
8. Electromagnetic Waves
Speed of Electromagnetic Wave
c = 1 μ 0 ε 0 \boxed{c=\frac1{\sqrt{\mu_0\varepsilon_0}}}
Wave Relation
c = ν λ c=\nu\lambda
Relation Between Electric and Magnetic Fields
E 0 B 0 = c \boxed{\frac{E_0}{B_0}=c}
orE 0 = c B 0 E_0=cB_0
Displacement Current
I d = ε 0 d Φ E d t I_d=\varepsilon_0\frac{d\Phi_E}{dt}
Ampere-Maxwell Law
∮ B ⃗ ⋅ d l ⃗ = μ 0 ( I c + I d ) \oint\vec B\cdot d\vec l = \mu_0(I_c+I_d)
9. Ray Optics and Optical Instruments
Refractive Index
n = c v n=\frac cv
Snell’s Law
n 1 sin i = n 2 sin r n_1\sin i=n_2\sin r sin i sin r = n 2 n 1 \frac{\sin i}{\sin r}=\frac{n_2}{n_1}
Critical Angle
For denser to rarer medium:sin C = n r a r e r n d e n s e r \sin C=\frac{n_{\rm rarer}}{n_{\rm denser}}
For air:sin C = 1 n \sin C=\frac1n
Spherical Mirror
1 f = 1 v + 1 u \boxed{\frac1f=\frac1v+\frac1u} f = R 2 f=\frac R2
Magnification:m = h i h o = − v u m=\frac{h_i}{h_o}=-\frac vu
Refraction at Spherical Surface
n 2 v − n 1 u = n 2 − n 1 R \frac{n_2}{v}-\frac{n_1}{u} = \frac{n_2-n_1}{R}
Thin Lens Formula
1 v − 1 u = 1 f \boxed{\frac1v-\frac1u=\frac1f}
Magnification:m = h i h o = v u m=\frac{h_i}{h_o}=\frac vu
Lens Maker’s Formula
In air:1 f = ( n − 1 ) ( 1 R 1 − 1 R 2 ) \frac1f=(n-1) \left( \frac1{R_1}-\frac1{R_2} \right)
Power of Lens
P = 1 f P=\frac1f
f f in metre.
Unit:1 D = 1 m − 1 1D=1m^{-1}
Lenses in Contact
P = P 1 + P 2 + ⋯ P=P_1+P_2+\cdots 1 F = 1 f 1 + 1 f 2 + ⋯ \frac1F=\frac1{f_1}+\frac1{f_2}+\cdots
Prism
δ = i + e − A \delta=i+e-A
At minimum deviation:i = e i=e r 1 = r 2 = A 2 r_1=r_2=\frac A2 n = sin [ ( A + δ m ) / 2 ] sin ( A / 2 ) \boxed{ n= \frac{\sin[(A+\delta_m)/2]} {\sin(A/2)} }
For small-angle prism:δ m = ( n − 1 ) A \delta_m=(n-1)A
Simple Microscope
Final image at infinity:M = D f M=\frac Df
At near point:M = 1 + D f M=1+\frac Df
Compound Microscope
Normal adjustment:M = L f o D f e M=\frac{L}{f_o}\frac{D}{f_e}
Astronomical Telescope
Normal adjustment:M = − f o f e M=-\frac{f_o}{f_e}
Magnitude:∣ M ∣ = f o f e |M|=\frac{f_o}{f_e}
Length:L = f o + f e L=f_o+f_e
10. Wave Optics
Path Difference
Δ = d sin θ \Delta=d\sin\theta
For small angle:Δ ≃ d y D \Delta\simeq\frac{dy}{D}
Constructive Interference
Δ = n λ \Delta=n\lambda
Destructive Interference
Δ = ( 2 n + 1 ) λ 2 \Delta=(2n+1)\frac\lambda2
YDSE Fringe Width
β = λ D d \boxed{\beta=\frac{\lambda D}{d}}
Bright fringe:y n = n λ D d y_n=n\frac{\lambda D}{d}
Dark fringe:y n = ( 2 n + 1 ) λ D 2 d y_n=(2n+1)\frac{\lambda D}{2d}
Resultant Intensity
I = I 1 + I 2 + 2 I 1 I 2 cos ϕ I=I_1+I_2+ 2\sqrt{I_1I_2}\cos\phi
For equal intensities:I = 4 I 0 cos 2 ϕ 2 I=4I_0\cos^2\frac\phi2
Single-Slit Diffraction
Minima:a sin θ = n λ a\sin\theta=n\lambda
For small angle:y n = n λ D a y_n=\frac{n\lambda D}{a}
Width of central maximum:W = 2 λ D a \boxed{W=\frac{2\lambda D}{a}}
11. Dual Nature of Radiation and Matter
Photon Energy
E = h ν E=h\nu E = h c λ E=\frac{hc}{\lambda}
Photon Momentum
p = h λ p=\frac{h}{\lambda} p = E c p=\frac Ec
Einstein’s Photoelectric Equation
h ν = ϕ + K max \boxed{h\nu=\phi+K_{\max}} K max = 1 2 m v max 2 K_{\max}=\frac12mv_{\max}^2 K max = e V 0 K_{\max}=eV_0
Therefore:h ν = ϕ + e V 0 h\nu=\phi+eV_0
Work Function
ϕ = h ν 0 \phi=h\nu_0 ϕ = h c λ 0 \phi=\frac{hc}{\lambda_0}
Stopping Potential
e V 0 = h ( ν − ν 0 ) eV_0=h(\nu-\nu_0)
de Broglie Wavelength
λ = h p \boxed{\lambda=\frac hp} λ = h m v \lambda=\frac{h}{mv}
For particle of kinetic energy K K :λ = h 2 m K \lambda=\frac{h}{\sqrt{2mK}}
For electron accelerated through potential V V :λ = h 2 m e V \lambda=\frac{h}{\sqrt{2meV}}
Approximately:λ ( A ˚ ) = 12.27 V \boxed{\lambda(\mathring A)=\frac{12.27}{\sqrt V}}
12. Atoms
Bohr Quantisation
m v r = n h 2 π mvr=\frac{nh}{2\pi}
Radius of n t h n^{th} Orbit
For hydrogen:r n = n 2 a 0 r_n=n^2a_0
wherea 0 = 0.529 A ˚ a_0=0.529\,\mathring A
For hydrogen-like atom:r n = n 2 a 0 Z r_n=\frac{n^2a_0}{Z}
Velocity
v n = 2.18 × 10 6 Z n m / s v_n=\frac{2.18\times10^6Z}{n}\;m/s
Energy of Electron
E n = − 13.6 Z 2 n 2 e V \boxed{E_n=-\frac{13.6Z^2}{n^2}\;eV}
For hydrogen:E n = − 13.6 n 2 e V E_n=-\frac{13.6}{n^2}\;eV
Photon Emission/Absorption
h ν = ∣ E f − E i ∣ h\nu=|E_f-E_i|
Rydberg Formula
1 λ = R Z 2 ( 1 n 1 2 − 1 n 2 2 ) \boxed{ \frac1\lambda = RZ^2 \left( \frac1{n_1^2}-\frac1{n_2^2} \right) }
where n 2 > n 1 n_2>n_1 .
13. Nuclei
Nuclear Radius
R = R 0 A 1 / 3 \boxed{R=R_0A^{1/3}} R 0 ≈ 1.2 × 10 − 15 m R_0\approx1.2\times10^{-15}m
Mass-Energy Equivalence
E = m c 2 E=mc^2
Mass Defect
Δ m = Z m p + ( A − Z ) m n − M n u c l e u s \Delta m= Zm_p+(A-Z)m_n-M_{\rm nucleus}
Using atomic masses:Δ m = Z m H + ( A − Z ) m n − M a t o m \Delta m= Zm_H+(A-Z)m_n-M_{\rm atom}
Binding Energy
B E = Δ m c 2 \boxed{BE=\Delta mc^2}
If Δ m \Delta m is in u:B E = Δ m × 931.5 M e V BE=\Delta m\times931.5\;MeV
Binding Energy per Nucleon
B E A \frac{BE}{A}
14. Semiconductor Electronics: Materials, Devices and Simple Circuits
Energy Gap
Conductor:E g ≈ 0 E_g\approx0
Semiconductor:E g ∼ 1 e V E_g\sim1\,eV
Insulator:E g ≳ 3 e V E_g\gtrsim3\,eV
Current in Semiconductor
I = I e + I h I=I_e+I_h
Conductivity
σ = e ( n e μ e + n h μ h ) \sigma=e(n_e\mu_e+n_h\mu_h)
For intrinsic semiconductor:n e = n h = n i n_e=n_h=n_i
Therefore:σ i = e n i ( μ e + μ h ) \sigma_i=en_i(\mu_e+\mu_h)
Resistivity
ρ = 1 σ \rho=\frac1\sigma
Diode
Forward bias → low resistance
Reverse bias → high resistance
Rectifier
Half-wave rectifier:f o u t p u t = f i n p u t f_{\rm output}=f_{\rm input}
Full-wave rectifier:f o u t p u t = 2 f i n p u t f_{\rm output}=2f_{\rm input}
⭐ Constants to Memorise for CBSE
Constant Value e e 1.6 × 10 − 19 C 1.6\times10^{-19}C m e m_e 9.1 × 10 − 31 k g 9.1\times10^{-31}kg m p m_p 1.67 × 10 − 27 k g 1.67\times10^{-27}kg ε 0 \varepsilon_0 8.85 × 10 − 12 C 2 N − 1 m − 2 8.85\times10^{-12}C^2N^{-1}m^{-2} μ 0 \mu_0 4 π × 10 − 7 T m A − 1 4\pi\times10^{-7}TmA^{-1} k = 1 / 4 π ε 0 k=1/4\pi\varepsilon_0 9 × 10 9 N m 2 C − 2 9\times10^9Nm^2C^{-2} h h 6.626 × 10 − 34 J s 6.626\times10^{-34}Js c c 3 × 10 8 m / s 3\times10^8m/s R R (Rydberg)1.097 × 10 7 m − 1 1.097\times10^7m^{-1} 1 e V 1\,eV 1.6 × 10 − 19 J 1.6\times10^{-19}J 1 u 1\,u 1.66 × 10 − 27 k g 1.66\times10^{-27}kg 1 u c 2 1\,u\,c^2 931.5 M e V 931.5\,MeV
⚠️ Board-exam points to remember
For Ray Optics , always apply the NCERT Cartesian sign convention rather than trying to memorize separate positive/negative versions of the formulas.
For EMI , the negative sign inE = − d Φ d t \mathcal E=-\frac{d\Phi}{dt}
represents Lenz’s law .
For AC , distinguish carefully between peak and RMS values:V r m s = V 0 2 , I r m s = I 0 2 V_{\rm rms}=\frac{V_0}{\sqrt2}, \qquad I_{\rm rms}=\frac{I_0}{\sqrt2}
For Atoms and Nuclei , keep units consistent—especially eV, MeV, joule and atomic mass unit .
For Dual Nature , remember the three equations that solve most CBSE numericals:h ν = ϕ + e V 0 \boxed{h\nu=\phi+eV_0} λ = h p \boxed{\lambda=\frac hp} λ = h 2 m e V \boxed{\lambda=\frac{h}{\sqrt{2meV}}}
And for Wave Optics , the two especially important board formulas are:β = λ D d W c e n t r a l = 2 λ D a \boxed{\beta=\frac{\lambda D}{d}} \qquad \boxed{W_{\rm central}=\frac{2\lambda D}{a}}
This covers the 14 chapters of the current Class 12 NCERT Physics sequence , with the emphasis on formulas that are directly useful for CBSE board preparation.
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