All Important Formula — Class 12 Physics Notes

All Important Formula · Class 12 Physics · 4 topics.

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Topics covered in All Important Formula

1.Optics
1. Snell's Law for Refractive Index
μ1μ2=sin⁡isin⁡r\frac{\mu_1}{\mu_2} = \frac{\sin i}{\sin r}
Example:
A light ray travels from air (μ1=1\mu_1 = 1) to water (μ2=1.33\mu_2 = 1.33). If the angle of incidence ii is 30∘30^\circ, find the angle of refraction rr.

Solution:
11.33=sin⁡30∘sin⁡r

\frac{1}{1.33} = \frac{\sin 30^\circ}{\sin r} 0.752=0.5sin⁡r

0.752 = \frac{0.5}{\sin r} sin⁡r=0.50.752=0.665

\sin r = \frac{0.5}{0.752} = 0.665 r=sin⁡−1(0.665)≈41.8∘r = \sin^{-1}(0.665) \approx 41.8^\circ
2. Brewster's Law
μ=tan⁡p(p = Angle of polarization)\mu = \tan p \quad (\text{p = Angle of polarization})

Example:
If the angle of polarization pp is 56∘56^\circ, find the refractive index μ\mu.

Solution:
μ=tan⁡56∘≈1.48\mu = \tan 56^\circ \approx 1.48
3. Relation Between Refractive Indices
n12×n21=1

n_{12} \times n_{21} = 1 n21=1n12n_{21} = \frac{1}{n_{12}}
Example:
If n12=1.5n_{12} = 1.5, find n21n_{21}.

Solution:
n21=11.5=0.67n_{21} = \frac{1}{1.5} = 0.67
4. Refractive Index Formula
μ=sin⁡isin⁡r\mu = \frac{\sin i}{\sin r}
Example:
A light ray travels from air to glass with an incidence angle of 45∘45^\circ and a refraction angle of 28∘28^\circ. Find the refractive index of glass.

Solution:
μ=sin⁡45∘sin⁡28∘=0.7070.469≈1.51\mu = \frac{\sin 45^\circ}{\sin 28^\circ} = \frac{0.707}{0.469} \approx 1.51
5. Mirror Formula
1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
Example:
An object is placed 20 cm in front of a concave mirror with a focal length of 10 cm. Find the image distance vv.

Solution:
1−10=1v+1−20

\frac{1}{-10} = \frac{1}{v} + \frac{1}{-20} 1v=1−10+120=−2+120=−120

\frac{1}{v} = \frac{1}{-10} + \frac{1}{20} = \frac{-2 + 1}{20} = \frac{-1}{20} v=−20 cmv = -20 \, \text{cm}
6. Lens Maker Formula
- In a Medium:
  1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)
- In Air:
  1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)
Example:
For a lens with μ=1.5\mu = 1.5, R1=20 cmR_1 = 20 \, \text{cm} and R2=−30 cmR_2 = -30 \, \text{cm}, find the focal length ff.

Solution:
1f=(1.5−1)(120−1−30)\frac{1}{f} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{-30}\right)
1f=0.5(120+130)\frac{1}{f} = 0.5 \left(\frac{1}{20} + \frac{1}{30}\right)
1f=0.5(3+260)=0.5×560=5120

\frac{1}{f} = 0.5 \left(\frac{3 + 2}{60}\right) = 0.5 \times \frac{5}{60} = \frac{5}{120} f=24 cmf = 24 \, \text{cm}
7. Equivalent Focal Length for Two Focal Lengths f1f_1 and f2f_2
1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}
Example:
Two lenses have focal lengths f1=10 cmf_1 = 10 \, \text{cm} and f2=20 cmf_2 = 20 \, \text{cm}. Find the equivalent focal length FF.

Solution:
1F=110+120=2+120=320

\frac{1}{F} = \frac{1}{10} + \frac{1}{20} = \frac{2 + 1}{20} = \frac{3}{20} F=203≈6.67 cmF = \frac{20}{3} \approx 6.67 \, \text{cm}
8. Power of a Lens
P=100f(in cm)P = \frac{100}{f(\text{in cm})}
Example:
If the focal length of a lens is 25 cm25 \, \text{cm}, find its power.
Solution:
P=10025=4 dioptersP = \frac{100}{25} = 4 \, \text{diopters}
9. Refraction Through Prism
μ=sin⁡(A+Dm2)sin⁡(A2)\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}
Example:
If the refractive angle A=60∘A = 60^\circ and the angle of minimum deviation Dm=40∘D_m = 40^\circ, find the refractive index μ\mu.

Solution:
μ=sin⁡(60∘+40∘2)sin⁡(60∘2)\mu = \frac{\sin\left(\frac{60^\circ + 40^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} μ=sin⁡50∘sin⁡30∘\mu = \frac{\sin 50^\circ}{\sin 30^\circ} μ=0.7660.5=1.532
2.Oscillation Formulas
1. Differential Equation for SHM
d2xdt2+ω2x=0\frac{d^2 x}{dt^2} + \omega^2 x = 0
Explanation:
This represents the equation of simple harmonic motion (SHM).

Example:
If ω=3 rad/s\omega = 3 \, \text{rad/s}, the equation becomes:
d2xdt2+32x=0ord2xdt2+9x=0\frac{d^2 x}{dt^2} + 3^2 x = 0 \quad \text{or} \quad \frac{d^2 x}{dt^2} + 9x = 0
2. Velocity in SHM
V=±ωA2−x2V = \pm \omega \sqrt{A^2 - x^2}
Explanation:
This gives the velocity at a displacement xx.

Example:
If ω=4 rad/s\omega = 4 \, \text{rad/s}, A=5 mA = 5 \, \text{m}, and x=3 mx = 3 \, \text{m}:
V=±452−32=±425−9=±416=±16 m/sV = \pm 4 \sqrt{5^2 - 3^2} = \pm 4 \sqrt{25 - 9} = \pm 4 \sqrt{16} = \pm 16 \, \text{m/s}
3. Maximum Velocity (VmaxV_{max})
Vmax=ωAV_{max} = \omega A

Explanation:
This is the highest velocity the particle achieves.

Example:
If ω=6 rad/s\omega = 6 \, \text{rad/s} and A=2 mA = 2 \, \text{m}:
Vmax=6×2=12 m/sV_{max} = 6 \times 2 = 12 \, \text{m/s}
4. Acceleration in SHM
a=−ω2xa = -\omega^2 x
Explanation:
The acceleration is proportional to the displacement but in the opposite direction.

Example:
If ω=5 rad/s\omega = 5 \, \text{rad/s} and x=2 mx = 2 \, \text{m}:
a=−52×2=−25×2=−50 m/s2a = -5^2 \times 2 = -25 \times 2 = -50 \, \text{m/s}^2
5. Displacement in SHM
x=Asin⁡(ωt±ϕ)x = A \sin(\omega t \pm \phi)
Explanation:
The position of the particle as a function of time.

Example:
If A=3 mA = 3 \, \text{m}, ω=2 rad/s\omega = 2 \, \text{rad/s}, and ϕ=0\phi = 0, at t=1 st = 1 \, \text{s}:
x=3sin⁡(2×1)=3sin⁡2≈3×0.909=2.727 mx = 3 \sin(2 \times 1) = 3 \sin 2 \approx 3 \times 0.909 = 2.727 \, \text{m}
6. Time Period for SHM
T=2πωT = \frac{2\pi}{\omega}
Explanation:
Time for one complete oscillation.

Example:
If ω=2 rad/s\omega = 2 \, \text{rad/s}:
T=2π2=π≈3.14 sT = \frac{2\pi}{2} = \pi \approx 3.14 \, \text{s}
7. Angular Frequency (ω\omega)
ω=2πT\omega = \frac{2\pi}{T}
Example:
If T=4 sT = 4 \, \text{s}:
ω=2π4=π2≈1.57 rad/s\omega = \frac{2\pi}{4} = \frac{\pi}{2} \approx 1.57 \, \text{rad/s}
8. Potential Energy (P.E.)
P.E.=12mω2x2=12kx2P.E. = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} k x^2
Example:
If m=1 kgm = 1 \, \text{kg}, ω=3 rad/s\omega = 3 \, \text{rad/s}, and x=2 mx = 2 \, \text{m}:
P.E.=12×1×32×22=12×9×4=18 JP.E. = \frac{1}{2} \times 1 \times 3^2 \times 2^2 = \frac{1}{2} \times 9 \times 4 = 18 \, \text{J}
9. Kinetic Energy (K.E.)
K.E.=12mω2(A2−x2)K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)
Example:
If m=2 kgm = 2 \, \text{kg}, ω=4 rad/s\omega = 4 \, \text{rad/s}, A=3 mA = 3 \, \text{m}, x=1 mx = 1 \, \text{m}:
K.E.=12×2×42×(32−12)=1×16×(9−1)=128 JK.E. = \frac{1}{2} \times 2 \times 4^2 \times (3^2 - 1^2) = 1 \times 16 \times (9 - 1) = 128 \, \text{J}
10. Total Energy (T.E.)
T.E.=12mω2A2=12kA2T.E. = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2
Example:
If m=1 kgm = 1 \, \text{kg}, ω=5 rad/s\omega = 5 \, \text{rad/s}, and A=2 mA = 2 \, \text{m}:
T.E.=12×1×52×22=12×25×4=50 JT.E. = \frac{1}{2} \times 1 \times 5^2 \times 2^2 = \frac{1}{2} \times 25 \times 4 = 50 \, \text{J}
11. Time Period of a Simple Pendulum
T=2πlgT = 2\pi \sqrt{\frac{l}{g}}
Example:
If l=1 ml = 1 \, \text{m} and g=9.8 m/s2g = 9.8 \, \text{m/s}^2:
T=2π19.8≈2π×0.319=2.006 sT = 2\pi \sqrt{\frac{1}{9.8}} \approx 2\pi \times 0.319 = 2.006 \, \text{s}
12. Time Period for a Mass-Spring System
T=2πmkT = 2\pi \sqrt{\frac{m}{k}}
Example:
If m=2 kgm = 2 \, \text{kg} and k=8 N/mk = 8 \, \text{N/m}:
T=2π28=2π0.25=2π×0.5≈3.14 sT = 2\pi \sqrt{\frac{2}{8}} = 2\pi \sqrt{0.25} = 2\pi \times 0.5 \approx 3.14 \, \text{s}
13. Frequency (ff)
f=1Tf = \frac{1}{T}
Example:
If T=2 sT = 2 \, \text{s}:
f=12=0.5 Hz
3.Gravitation Formulas and Constant Values of Physical Quantities
Gravitation Formulas

1. Newton’s Law of Gravitation
F=GMmr2F = \frac{GMm}{r^2}
- Explanation:
  
  The gravitational force between two masses MM and mm separated by distance rr.
- Example:
  If M=10 kgM = 10 \, \text{kg}, m=5 kgm = 5 \, \text{kg}, r=2 mr = 2 \, \text{m}, and G=6.67×10−11 Nm2/kg2G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2:
  
  F=6.67×10−11×10×522=6.67×10−11×504=8.34×10−10 NF = \frac{6.67 \times 10^{-11} \times 10 \times 5}{2^2} = \frac{6.67 \times 10^{-11} \times 50}{4} = 8.34 \times 10^{-10} \, \text{N}
2. Gravitational Constant
G=6.67×10−11 Nm2/kg2G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2
- Explanation:
  
  Universal constant used in the calculation of gravitational force.
3. Acceleration Due to Gravity (gg)
g=GMR2g = \frac{GM}{R^2}
- Explanation:
  
  The acceleration due to gravity on the surface of a planet of mass MM and radius RR.
- Example:
  
  If M=5.98×1024 kgM = 5.98 \times 10^{24} \, \text{kg} and R=6.37×106 mR = 6.37 \times 10^6 \, \text{m}:
  
  g=6.67×10−11×5.98×1024(6.37×106)2≈9.8 m/s2g = \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{(6.37 \times 10^6)^2} \approx 9.8 \, \text{m/s}^2
4. Gravitational Potential Energy
U=−GMmrU = -\frac{GMm}{r}
- Explanation:
  
  The potential energy between two masses MM and mm separated by distance rr.
- Example:
  
  If M=10 kgM = 10 \, \text{kg}, m=5 kgm = 5 \, \text{kg}, r=2 mr = 2 \, \text{m}:
  U=−6.67×10−11×10×52=−1.67×10−9 JU = -\frac{6.67 \times 10^{-11} \times 10 \times 5}{2} = -1.67 \times 10^{-9} \, \text{J}
5. Orbital Velocity (v0v_0)
v0=GMrv_0 = \sqrt{\frac{GM}{r}}
- Explanation:
  
  The velocity required to keep a body in a circular orbit around a planet.
- Example:
  
  If M=5.98×1024 kgM = 5.98 \times 10^{24} \, \text{kg} and r=6.37×106 mr = 6.37 \times 10^6 \, \text{m}:
  
  v0=6.67×10−11×5.98×10246.37×106≈7.9 km/sv_0 = \sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.37 \times 10^6}} \approx 7.9 \, \text{km/s}
6. Escape Velocity (vev_e)
ve=2GMRv_e = \sqrt{\frac{2GM}{R}}
- Explanation:
  
  The minimum velocity needed to escape the gravitational pull of a planet.
- Example:
  
  For Earth, if M=5.98×1024 kgM = 5.98 \times 10^{24} \, \text{kg} and R=6.37×106 mR = 6.37 \times 10^6 \, \text{m}:
  
  ve=2×6.67×10−11×5.98×10246.37×106≈11.2 km/sv_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.37 \times 10^6}} \approx 11.2 \, \text{km/s}
7. Time Period of a Satellite (TT)
T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}
- Explanation:
  
  The time taken for one complete revolution of a satellite around a planet.
8. Kepler’s Third Law
T2∝r3T^2 \propto r^3
- Explanation:
  
  The square of the time period of a planet’s orbit is proportional to the cube of the semi-major axis of the orbit.
9. Weight on a Planet (WW)
W=mgW = mg
- Explanation:
  
  The weight of an object is the force due to gravity acting on it.
Constant Values of Physical Quantities
1. Velocity of Light (cc)
  3×108 m/s
  
  3 \times 10^8 \, \text{m/s}
2. Gravitational Constant (GG)
  6.67×10−11 Nm2/kg
  
  26.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2
3. Acceleration Due to Gravity on Earth (gg)
  9.8 m/s
  
  29.8 \, \text{m/s}^2
4. Planck’s Constant (hh)
  6.63×10−34 Js
  
  6.63 \times 10^{-34} \, \text{Js}
5. Avogadro’s Number (NAN_A)
  6.022×1023 mol
  
  −16.022 \times 10^{23} \, \text{mol}^{-1}
6. Boltzmann Constant (kk)
  1.38×10−23 J/K
  
  1.38 \times 10^{-23} \, \text{J/K}
7. Universal Gas Constant (RR)
  8.314 J/mol\K
  
  8.314 \, \text{J/mol·K}
8. Electron Charge (ee)
  1.6×10−19 C
  
  1.6 \times 10^{-19} \, \text{C}
9. Mass of Electron
  9.11×10−31 kg
  
  9.11 \times 10^{-31} \, \text{kg}
10. Mass of Proton
  1.67×10−27 kg
  
  1.67 \times 10^{-27} \, \text{kg}
11. Permittivity of Free Space (ϵ0\epsilon_0)
  8.85×10−12 C2/Nm28.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2
12. Permeability of Free Space (μ0\mu_0)
  4π×10−7 Tm/A
  
  4\pi \times 10^{-7} \, \text{Tm/A}
13. Stefan-Boltzmann Constant (σ\sigma)
  5.67×10−8 W/m2K45.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4
14. Gas Density of Air
  1.29 kg/m31.29 \, \text{kg/m}^3
15. Speed of Sound in Air
  343 m/s343 \, \text{m/s}
4.Electromagnetic Induction Formula
1. Faraday's Law of Induction
Formula:

ε=−dϕdt\varepsilon = -\frac{d\phi}{dt}

Explanation:
The induced EMF (ε\varepsilon) is proportional to the rate of change of magnetic flux (ϕ\phi) through a circuit. The negative sign follows Lenz's Law.

Example:
If the magnetic flux changes by 0.2 Wb0.2 \, \text{Wb} in 0.1 s0.1 \, \text{s}, the induced EMF is:

ε=−0.20.1=−2 V\varepsilon = -\frac{0.2}{0.1} = -2 \, \text{V}
2. Magnetic Flux

Formula:
ϕ=B⋅A⋅cos⁡θ

\phi = B \cdot A \cdot \cos\theta
Explanation:
Magnetic flux (ϕ\phi) depends on the magnetic field (BB), area (AA), and the angle (θ\theta) between the field lines and the normal to the surface.

Example:
If B=5 TB = 5 \, \text{T}, A=0.1 m2A = 0.1 \, \text{m}^2, and θ=30∘\theta = 30^\circ:
ϕ=5×0.1×cos⁡30∘=0.5×0.866=0.433 Wb\phi = 5 \times 0.1 \times \cos 30^\circ = 0.5 \times 0.866 = 0.433 \, \text{Wb}
3. Induced EMF in a Loop

Formula:
ε=−Ndϕdt\varepsilon = -N \frac{d\phi}{dt}

Explanation:
If a coil with NN turns experiences a changing magnetic flux, the induced EMF is proportional to NN and the rate of change of flux.
4. Induced EMF for a Moving Conductor

Formula:
ε=Bℓvsin⁡θ\varepsilon = B \ell v \sin\theta

Explanation:
The EMF induced in a conductor of length ℓ\ell moving with velocity vv through a magnetic field BB.

Example:
If B=2 TB = 2 \, \text{T}, ℓ=0.5 m\ell = 0.5 \, \text{m}, v=3 m/sv = 3 \, \text{m/s}, and θ=90∘\theta = 90^\circ:
ε=2×0.5×3=3 V\varepsilon = 2 \times 0.5 \times 3 = 3 \, \text{V}
5. Lenz's Law

Formula:
ε=−dϕdt\varepsilon = -\frac{d\phi}{dt}

Explanation:
The induced EMF opposes the change in magnetic flux that caused it.
6. Self-Inductance (L)

Formula:
L=NϕIL = \frac{N\phi}{I}

Explanation:
Self-inductance (LL) is the ratio of the magnetic flux (ϕ\phi) linked with the coil to the current (II) producing it.
7. Induced EMF due to Self-Inductance

Formula:
ε=−LdIdt\varepsilon = -L \frac{dI}{dt}

Explanation:
The EMF induced due to a changing current in the same coil.
8. Energy Stored in an Inductor

Formula:
U=12LI2U = \frac{1}{2} L I^2

Explanation:
The energy stored in an inductor due to the current II flowing through it.
9. Mutual Inductance (M)

Formula:
M=N2ϕ21I1M = \frac{N_2 \phi_{21}}{I_1}

Explanation:
Mutual inductance (MM) is the ratio of the magnetic flux linked with the second coil to the current in the first coil.
10. Induced EMF due to Mutual Inductance

Formula:
ε2=−MdI1dt\varepsilon_2 = -M \frac{dI_1}{dt}

Explanation:
The EMF induced in one coil due to a changing current in another coil.
11. Angular Frequency (ω\omega)

Formula:
ω=2πf\omega = 2\pi f

Explanation:
Angular frequency ω\omega is related to frequency ff.
12. Power in AC Circuit

Formula:
P=VIcos⁡ϕP = VI \cos \phi

Explanation:
Power (PP) in an AC circuit depends on voltage (VV), current (II), and phase angle (ϕ\phi).
13. Impedance in Series R-L Circuit

Formula:
Z=R2+(ωL)2Z = \sqrt{R^2 + (\omega L)^2}

Explanation:
Impedance (ZZ) of a series circuit with resistance (RR) and inductance (LL).
14. Resonant Frequency

Formula:
f0=12πLCf_0 = \frac{1}{2\pi \sqrt{LC}}

Explanation:
The frequency at which a circuit naturally oscillates.

All Important Formula — Class 12 Physics Notes

Topics covered in All Important Formula

1.Optics

1. Snell's Law for Refractive Index

Example:

2. Brewster's Law

Example:

3. Relation Between Refractive Indices

Example:

4. Refractive Index Formula

Example:

5. Mirror Formula

Example:

6. Lens Maker Formula

Example:

7. Equivalent Focal Length for Two Focal Lengths f1f_1 and f2f_2

Example:

8. Power of a Lens

Example:

9. Refraction Through Prism

Example:

2.Oscillation Formulas

1. Differential Equation for SHM

Explanation:

Example:

2. Velocity in SHM

Explanation:

Example:

3. Maximum Velocity (VmaxV_{max})

Explanation:

Example:

4. Acceleration in SHM

Explanation:

Example:

5. Displacement in SHM

Explanation:

Example:

6. Time Period for SHM

Explanation:

Example:

7. Angular Frequency (ω\omega)

Example:

8. Potential Energy (P.E.)

Example:

9. Kinetic Energy (K.E.)

Example:

10. Total Energy (T.E.)

Example:

11. Time Period of a Simple Pendulum

Example:

12. Time Period for a Mass-Spring System

Example:

13. Frequency (ff)

Example:

3.Gravitation Formulas and Constant Values of Physical Quantities

Gravitation Formulas

1. Newton’s Law of Gravitation

2. Gravitational Constant

3. Acceleration Due to Gravity (gg)

4. Gravitational Potential Energy

5. Orbital Velocity (v0v_0)

6. Escape Velocity (vev_e)

7. Time Period of a Satellite (TT)

8. Kepler’s Third Law

9. Weight on a Planet (WW)

Constant Values of Physical Quantities

4.Electromagnetic Induction Formula

1. Faraday's Law of Induction

2. Magnetic Flux

3. Induced EMF in a Loop

4. Induced EMF for a Moving Conductor

5. Lenz's Law

6. Self-Inductance (L)

7. Induced EMF due to Self-Inductance

8. Energy Stored in an Inductor

9. Mutual Inductance (M)

10. Induced EMF due to Mutual Inductance

11. Angular Frequency (ω\omega)

12. Power in AC Circuit

13. Impedance in Series R-L Circuit