All Important FormulaClass 11 Physics Notes

All Important Formula · Class 11 Physics · 4 topics.

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Topics covered in All Important Formula

  1. 1.Optics

    1. Snell's Law for Refractive Index

    μ1μ2=sin⁡isin⁡r

    Example:

    A light ray travels from air (μ1=1) to water (μ2=1.33). If the angle of incidence i is 30∘, find the angle of refraction r.


    Solution:

    11.33=sin⁡30∘sin⁡r

    0.752=0.5sin⁡r

    sin⁡r=0.50.752=0.665

    r=sin⁡−1(0.665)≈41.8∘


    2. Brewster's Law

    μ=tan⁡p(p = Angle of polarization)


    Example:

    If the angle of polarization p is 56∘, find the refractive index μ.


    Solution:

    μ=tan⁡56∘≈1.48


    3. Relation Between Refractive Indices

    n12×n21=1

    n21=1n12

    Example:

    If n12=1.5, find n21.


    Solution:

    n21=11.5=0.67


    4. Refractive Index Formula

    μ=sin⁡isin⁡r

    Example:

    A light ray travels from air to glass with an incidence angle of 45∘ and a refraction angle of 28∘. Find the refractive index of glass.


    Solution:

    μ=sin⁡45∘sin⁡28∘=0.7070.469≈1.51


    5. Mirror Formula

    1f=1v+1u

    Example:

    An object is placed 20 cm in front of a concave mirror with a focal length of 10 cm. Find the image distance v.


    Solution:

    1−10=1v+1−20

    1v=1−10+120=−2+120=−120

    v=−20 cm


    6. Lens Maker Formula

    • In a Medium:

      1f=(μ−1)(1R1−1R2)
    • In Air:

      1f=(μ−1)(1R1−1R2)

    Example:

    For a lens with μ=1.5, R1=20 cm and R2=−30 cm, find the focal length f.


    Solution:

    1f=(1.5−1)(120−1−30)
    1f=0.5(120+130)
    1f=0.5(3+260)=0.5×560=5120

    f=24 cm


    7. Equivalent Focal Length for Two Focal Lengths f1 and f2

    1F=1f1+1f2

    Example:

    Two lenses have focal lengths f1=10 cm and f2=20 cm. Find the equivalent focal length F.


    Solution:

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

    F=203≈6.67 cm


    8. Power of a Lens

    P=100f(in cm)

    Example:

    If the focal length of a lens is 25 cm, find its power.

    Solution:

    P=10025=4 diopters


    9. Refraction Through Prism

    μ=sin⁡(A+Dm2)sin⁡(A2)

    Example:

    If the refractive angle A=60∘ and the angle of minimum deviation Dm=40∘, find the refractive index μ.


    Solution:

    μ=sin⁡(60∘+40∘2)sin⁡(60∘2)μ=sin⁡50∘sin⁡30∘μ=0.7660.5=1.532

  2. 2.Oscillation Formulas

    1. Differential Equation for SHM

    d2xdt2+ω2x=0

    Explanation:

    This represents the equation of simple harmonic motion (SHM).


    Example:

    If ω=3 rad/s, the equation becomes:

    d2xdt2+32x=0ord2xdt2+9x=0


    2. Velocity in SHM

    V=±ωA2−x2

    Explanation:

    This gives the velocity at a displacement x.


    Example:

    If ω=4 rad/s, A=5 m, and x=3 m:

    V=±452−32=±425−9=±416=±16 m/s


    3. Maximum Velocity (Vmax)

    Vmax=ωA


    Explanation:

    This is the highest velocity the particle achieves.


    Example:

    If ω=6 rad/s and A=2 m:

    Vmax=6×2=12 m/s


    4. Acceleration in SHM

    a=−ω2x

    Explanation:

    The acceleration is proportional to the displacement but in the opposite direction.


    Example:

    If ω=5 rad/s and x=2 m:

    a=−52×2=−25×2=−50 m/s2


    5. Displacement in SHM

    x=Asin⁡(ωt±ϕ)

    Explanation:

    The position of the particle as a function of time.


    Example:

    If A=3 m, ω=2 rad/s, and ϕ=0, at t=1 s:

    x=3sin⁡(2×1)=3sin⁡2≈3×0.909=2.727 m


    6. Time Period for SHM

    T=2πω

    Explanation:

    Time for one complete oscillation.


    Example:

    If ω=2 rad/s:

    T=2π2=π≈3.14 s


    7. Angular Frequency (ω)

    ω=2πT

    Example:

    If T=4 s:

    ω=2π4=π2≈1.57 rad/s


    8. Potential Energy (P.E.)

    P.E.=12mω2x2=12kx2

    Example:

    If m=1 kg, ω=3 rad/s, and x=2 m:

    P.E.=12×1×32×22=12×9×4=18 J


    9. Kinetic Energy (K.E.)

    K.E.=12mω2(A2−x2)

    Example:

    If m=2 kg, ω=4 rad/s, A=3 m, x=1 m:

    K.E.=12×2×42×(32−12)=1×16×(9−1)=128 J


    10. Total Energy (T.E.)

    T.E.=12mω2A2=12kA2

    Example:

    If m=1 kg, ω=5 rad/s, and A=2 m:

    T.E.=12×1×52×22=12×25×4=50 J


    11. Time Period of a Simple Pendulum

    T=2πlg

    Example:

    If l=1 m and g=9.8 m/s2:

    T=2π19.8≈2π×0.319=2.006 s


    12. Time Period for a Mass-Spring System

    T=2πmk

    Example:

    If m=2 kg and k=8 N/m:

    T=2π28=2π0.25=2π×0.5≈3.14 s


    13. Frequency (f)

    f=1T

    Example:

    If T=2 s:

    f=12=0.5 Hz

  3. 3.Gravitation Formulas and Constant Values of Physical Quantities

    Gravitation Formulas

    1. Newton’s Law of Gravitation

    F=GMmr2

    • Explanation:

      The gravitational force between two masses M and m separated by distance r.


    • Example:
      If M=10 kg, m=5 kg, r=2 m, and G=6.67×10−11 Nm2/kg2:

      F=6.67×10−11×10×522=6.67×10−11×504=8.34×10−10 N

    2. Gravitational Constant

    G=6.67×10−11 Nm2/kg2

    • Explanation:

      Universal constant used in the calculation of gravitational force.

    3. Acceleration Due to Gravity (g)

    g=GMR2

    • Explanation:

      The acceleration due to gravity on the surface of a planet of mass M and radius R.

    • Example:

      If M=5.98×1024 kg and R=6.37×106 m:

      g=6.67×10−11×5.98×1024(6.37×106)2≈9.8 m/s2

    4. Gravitational Potential Energy

    U=−GMmr

    • Explanation:

      The potential energy between two masses M and m separated by distance r.

    • Example:

      If M=10 kg, m=5 kg, r=2 m:

      U=−6.67×10−11×10×52=−1.67×10−9 J

    5. Orbital Velocity (v0)

    v0=GMr

    • Explanation:

      The velocity required to keep a body in a circular orbit around a planet.

    • Example:

      If M=5.98×1024 kg and r=6.37×106 m:

      v0=6.67×10−11×5.98×10246.37×106≈7.9 km/s

    6. Escape Velocity (ve)

    ve=2GMR

    • Explanation:

      The minimum velocity needed to escape the gravitational pull of a planet.

    • Example:

      For Earth, if M=5.98×1024 kg and R=6.37×106 m:

      ve=2×6.67×10−11×5.98×10246.37×106≈11.2 km/s

    7. Time Period of a Satellite (T)

    T=2πr3GM

    • Explanation:

      The time taken for one complete revolution of a satellite around a planet.

    8. Kepler’s Third Law

    T2∝r3

    • 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 (W)

    W=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 (c)
      3×108 m/s

    2. Gravitational Constant (G)
      6.67×10−11 Nm2/kg

      2

    3. Acceleration Due to Gravity on Earth (g)
      9.8 m/s

      2

    4. Planck’s Constant (h)
      6.63×10−34 Js

    5. Avogadro’s Number (NA)
      6.022×1023 mol

      −1

    6. Boltzmann Constant (k)
      1.38×10−23 J/K

    7. Universal Gas Constant (R)
      8.314 J/mol\K

    8. Electron Charge (e)
      1.6×10−19 C

    9. Mass of Electron
      9.11×10−31 kg

    10. Mass of Proton
      1.67×10−27 kg

    11. Permittivity of Free Space (ϵ0)
      8.85×10−12 C2/Nm2

    12. Permeability of Free Space (μ0)
      4π×10−7 Tm/A

    13. Stefan-Boltzmann Constant (σ)
      5.67×10−8 W/m2K4

    14. Gas Density of Air
      1.29 kg/m3

    15. Speed of Sound in Air
      343 m/s

  4. 4.Electromagnetic Induction Formula

    1. Faraday's Law of Induction

    Formula:

    ε=−dϕdt


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


    Example:
    If the magnetic flux changes by 0.2 Wb in 0.1 s, the induced EMF is:

    ε=−0.20.1=−2 V


    2. Magnetic Flux

    Formula:
    ϕ=B⋅A⋅cos⁡θ

    Explanation:
    Magnetic flux (ϕ) depends on the magnetic field (B), area (A), and the angle (θ) between the field lines and the normal to the surface.


    Example:
    If B=5 T, A=0.1 m2, and θ=30∘:
    ϕ=5×0.1×cos⁡30∘=0.5×0.866=0.433 Wb


    3. Induced EMF in a Loop


    Formula:
    ε=−Ndϕdt


    Explanation:
    If a coil with N turns experiences a changing magnetic flux, the induced EMF is proportional to N and the rate of change of flux.


    4. Induced EMF for a Moving Conductor


    Formula:
    ε=Bℓvsin⁡θ


    Explanation:
    The EMF induced in a conductor of length ℓ moving with velocity v through a magnetic field B.


    Example:
    If B=2 T, ℓ=0.5 m, v=3 m/s, and θ=90∘:
    ε=2×0.5×3=3 V


    5. Lenz's Law

    Formula:
    ε=−dϕdt


    Explanation:
    The induced EMF opposes the change in magnetic flux that caused it.


    6. Self-Inductance (L)


    Formula:
    L=NϕI


    Explanation:
    Self-inductance (L) is the ratio of the magnetic flux (ϕ) linked with the coil to the current (I) producing it.


    7. Induced EMF due to Self-Inductance


    Formula:
    ε=−LdIdt


    Explanation:
    The EMF induced due to a changing current in the same coil.


    8. Energy Stored in an Inductor


    Formula:
    U=12LI2


    Explanation:
    The energy stored in an inductor due to the current I flowing through it.


    9. Mutual Inductance (M)


    Formula:
    M=N2ϕ21I1


    Explanation:
    Mutual inductance (M) 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


    Explanation:
    The EMF induced in one coil due to a changing current in another coil.


    11. Angular Frequency (ω)


    Formula:
    ω=2πf


    Explanation:
    Angular frequency ω is related to frequency f.


    12. Power in AC Circuit


    Formula:
    P=VIcos⁡ϕ


    Explanation:
    Power (P) in an AC circuit depends on voltage (V), current (I), and phase angle (ϕ).


    13. Impedance in Series R-L Circuit


    Formula:
    Z=R2+(ωL)2


    Explanation:
    Impedance (Z) of a series circuit with resistance (R) and inductance (L).


    14. Resonant Frequency


    Formula:
    f0=12πLC


    Explanation:
    The frequency at which a circuit naturally oscillates.

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