Bihar Board Class 12 Physics Full Notes 2025 – हर Chapter Detailed in Hindi

Physics (भौतिक विज्ञान) प्रकृति के नियमों का अध्ययन है। यह हमें शारीरिक घटनाओं और उनके पीछे के सिद्धांतों को समझने में मदद करता है।
Class 12 Physics के 15 chapters हैं, जिन्हें 2 भागों में बांटा गया है। यह नोट्स हर टॉपिक, फॉर्मूला, उदाहरण और व्याख्या के साथ तैयार किया गया है ताकि स्टूडेंट्स परीक्षा में आसानी से तैयारी कर सकें।

Bihar Board Class 12 Physics Full Notes 2025 – हर Chapter Detailed in Hindi

भाग 1 (Part – I)

Chapter 1 – Electric Charges and Fields (विद्युत आवेश और क्षेत्र)

Topics & Explanation:

1. Electric Charge (विद्युत आवेश)

   • दो प्रकार: Positive (+) और Negative (−)

   • Conservation of charge: कुल आवेश हमेशा संरक्षित रहता है।

   • Quantization of charge:

   $$q=ne(n=1,2,3\dots )$$

   Example: यदि 3 electrons का कुल charge निकालना हो:

   q=3×1.6×10−19C=4.8×10−19Cq = 3 \times 1.6 \times 10^{-19} C = 4.8 \times 10^{-19} Cq=3×1.6×10−19C=4.8×10−19C

2. Coulomb’s Law (कुलॉम्ब का नियम)

F=14πε0q1q2r2F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}F=4πε0​1​r2q1​q2​​

• Force (F) प्रतिकर्षण या आकर्षण

• q1,q2q_1, q_2q1​,q2​ – charges

• $r$ – दूरी

Example: q1=2μC,q2=3μC,r=0.5mq_1 = 2\mu C, q_2 = 3\mu C, r = 0.5 mq1​=2μC,q2​=3μC,r=0.5m

F=9×109(2×10−6)(3×10−6)0.52=216NF = 9 \times 10^9 \frac{(2 \times 10^{-6})(3 \times 10^{-6})}{0.5^2} = 216 NF=9×1090.52(2×10−6)(3×10−6)​=216N

3. Electric Field (विद्युत क्षेत्र)

E=Fq=14πε0Qr2E = \frac{F}{q} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}E=qF​=4πε0​1​r2Q​

• Direction: Positive charge से दूर, Negative charge की ओर

• Example: $Q=4\mu C,r=0.2m$

E=9×1094×10−60.22=9×105 N/CE = 9\times10^9 \frac{4\times10^{-6}}{0.2^2} = 9 \times 10^5 \, N/CE=9×1090.224×10−6​=9×105N/C

4. Electric Flux & Gauss’s Law (विद्युत फ्लक्स और गॉस का नियम)

ΦE=∮E⃗⋅dA⃗=qenclosedε0\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}ΦE​=∮E⋅dA=ε0​qenclosed​​

• Example: Sphere में q enclosed = 2µC, ΦE=2×10−68.85×10−12=2.26×105Nm2/C\Phi_E = \frac{2\times10^{-6}}{8.85\times10^{-12}} = 2.26 \times 10^5 N m^2/CΦE​=8.85×10−122×10−6​=2.26×105Nm2/C

Chapter 2 – Electrostatic Potential and Capacitance (विद्युत विभव और धारिता)

1. Electrostatic Potential (V):

V=WqV = \frac{W}{q}V=qW​

• Work done in bringing unit positive charge to a point

Example: W = 0.01 J, q = 2µC

V=0.012×10−6=5000VV = \frac{0.01}{2\times10^{-6}} = 5000 VV=2×10−60.01​=5000V

2. Potential Difference (V):

VAB=VA−VBV_{AB} = V_A – V_BVAB​=VA​−VB​

3. Capacitance (C):

C=QV,C=ε0Ad (Parallel Plate)C = \frac{Q}{V}, \quad C = \varepsilon_0 \frac{A}{d} \text{ (Parallel Plate)}C=VQ​,C=ε0​dA​ (Parallel Plate)

Example: A=0.02m2,d=0.01mA=0.02m^2, d=0.01mA=0.02m2,d=0.01m

C=8.85×10−120.020.01=1.77×10−11FC = 8.85\times10^{-12} \frac{0.02}{0.01} = 1.77 \times 10^{-11} FC=8.85×10−120.010.02​=1.77×10−11F

4. Energy stored in capacitor:

U=12CV2U = \frac{1}{2} C V^2U=21​CV2

Chapter 3 – Current Electricity (विद्युत धारा)

1. Ohm’s Law:

$$V=IR$$

2. Resistances in Series and Parallel:

• Series: Rs=R1+R2+…R_s = R_1 + R_2 + …Rs​=R1​+R2​+…

• Parallel: 1Rp=1R1+1R2+…\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + …Rp​1​=R1​1​+R2​1​+…

3. Power in electrical circuit:

P=VI=I2R=V2RP = VI = I^2 R = \frac{V^2}{R}P=VI=I2R=RV2​

Example: R = 2Ω, I = 3A

V=IR=2×3=6V,P=I2R=9×2=18WV = IR = 2 \times 3 = 6V, \quad P = I^2R = 9\times2 = 18 WV=IR=2×3=6V,P=I2R=9×2=18W

Chapter 4 – Moving Charges & Magnetism (चालक धारा के प्रभाव)

1. Magnetic Force on moving charge:

F⃗=qv⃗×B⃗\vec{F} = q \vec{v} \times \vec{B}F=qv×B

2. Biot-Savart Law (लंबी तार के लिए):

B=μ0I2πrB = \frac{\mu_0 I}{2 \pi r}B=2πrμ0​I​

Example: I=5A, r=0.1m, μ0=4π×10−7\mu_0=4\pi\times10^{-7}μ0​=4π×10−7

B=4π×10−7×52π×0.1=1×10−5TB = \frac{4\pi\times10^{-7}\times5}{2\pi\times0.1} = 1\times10^{-5} TB=2π×0.14π×10−7×5​=1×10−5T

Chapter 5 – Magnetism & Matter (चुंबकत्व और पदार्थ)

1. Magnetic Moment: $M=I\cdot A$

2. Types of Magnetic Material: Diamagnetic, Paramagnetic, Ferromagnetic

Chapter 6 – Electromagnetic Induction (विद्युतचुंबकीय प्रेरण)

1. Faraday’s Law:

ε=−dΦBdt\varepsilon = -\frac{d\Phi_B}{dt}ε=−dtdΦB​​

2. Lenz’s Law: Induced current opposes flux change

Example: Flux decreases 0.02 Wb/sec → $\epsilon =0.02V$

Chapter 7 – Alternating Current (प्रत्यावर्ती धारा)

1. Instantaneous current: i=I0sin⁡ωti = I_0 \sin \omega ti=I0​sinωt

2. RMS value: Irms=I02I_{rms} = \frac{I_0}{\sqrt{2}}Irms​=2​I0​​

Example: I₀=10A → I₍rms₎=7.07A

Chapter 8 – Electromagnetic Waves (विद्युतचुंबकीय तरंगें)

1. Maxwell’s equations: EM waves in vacuum at speed c = 3×10⁸ m/s

2. Spectrum: Radio → γ-rays

भाग 2 (Part II)

Chapter 9 – Ray Optics and Optical Instruments (किरण प्रकाशिकी और ऑप्टिकल यंत्र)

Topics & Explanation

1. Reflection (परावर्तन)

   • नियम: Angle of Incidence = Angle of Reflection

   • Plane Mirror: Image always virtual, erect, laterally inverted

   Example: Object at 10 cm in front of plane mirror → Image 10 cm behind mirror

2. Refraction (अपवर्तन)

   • Snell’s Law:

sin⁡isin⁡r=μ\frac{\sin i}{\sin r} = \musinrsini​=μ

• $i$ = angle of incidence, $r$ = angle of refraction, $\mu$ = refractive index

Example: Light enters glass ($\mu =1.5$) at i=30∘i = 30^\circi=30∘ →

r=arcsin⁡(sin⁡301.5)=19.47∘r = \arcsin(\frac{\sin 30}{1.5}) = 19.47^\circr=arcsin(1.5sin30​)=19.47∘

3. Lens Formula (लेंस सूत्र)

1f=1v−1u\frac{1}{f} = \frac{1}{v} – \frac{1}{u}f1​=v1​−u1​

• $f$ = focal length, $v$ = image distance, $u$ = object distance

Example: Object at 20 cm in front of convex lens, f = 10 cm →

1v=110−1−20=320⇒v=6.67cm\frac{1}{v} = \frac{1}{10} – \frac{1}{-20} = \frac{3}{20} \Rightarrow v = 6.67 cmv1​=101​−−201​=203​⇒v=6.67cm

4. Magnification

m=vu=hihom = \frac{v}{u} = \frac{h_i}{h_o}m=uv​=ho​hi​​

5. Optical Instruments

   • Microscope: Total magnification = mo×mem_o \times m_emo​×me​

   • Telescope: Magnification = fofe\frac{f_o}{f_e}fe​fo​​

Chapter 10 – Wave Optics (तरंग प्रकाशिकी)

1. Huygens Principle: Every point on wavefront acts as secondary source

2. Interference

   • Two coherent sources produce bright & dark fringes

Fringe width β=λDd\text{Fringe width } \beta = \frac{\lambda D}{d}Fringe width β=dλD​

• $\lambda$ = wavelength, D = screen distance, d = slit separation

Example: λ=600 nm, D=1 m, d=0.2 mm →

β=600×10−9×10.2×10−3=3mm\beta = \frac{600\times10^{-9}\times1}{0.2\times10^{-3}} = 3 mmβ=0.2×10−3600×10−9×1​=3mm

3. Diffraction: Bending of light at edges

4. Polarisation: Restriction of light vibration in one plane

Chapter 11 – Dual Nature of Radiation and Matter (विकिरण और पदार्थ का द्वैत स्वभाव)

1. Photoelectric Effect

$$E=h\nu -W$$

• h = Planck’s constant, W = work function

Example: ν=10¹⁵ Hz, W=2 eV →

Ek=6.626×10−34×1015−2×1.6×10−19≈4.26×10−19JE_{k} = 6.626\times10^{-34} \times 10^{15} – 2\times1.6\times10^{-19} \approx 4.26\times10^{-19} JEk​=6.626×10−34×1015−2×1.6×10−19≈4.26×10−19J

2. De Broglie Wavelength

λ=hmv\lambda = \frac{h}{mv}λ=mvh​

Example: Electron m=9.1×10⁻³¹ kg, v=10⁶ m/s →

λ=6.626×10−349.1×10−31×106≈7.27×10−10m\lambda = \frac{6.626\times10^{-34}}{9.1\times10^{-31}\times10^6} \approx 7.27\times10^{-10} mλ=9.1×10−31×1066.626×10−34​≈7.27×10−10m

Chapter 12 – Atoms (परमाणु)

1. Bohr Model of Hydrogen Atom

En=−13.6n2eVE_n = -\frac{13.6}{n^2} eVEn​=−n213.6​eV

• n = energy level

Example: n=2 → E2=−3.4eVE_2 = -3.4 eVE2​=−3.4eV

2. Radius of orbit

rn=n2h2ϵ0πme2r_n = n^2 \frac{h^2 \epsilon_0}{\pi m e^2}rn​=n2πme2h2ϵ0​​

3. Spectral Lines (Balmer, Lyman Series)

Chapter 13 – Nuclei (नाभिक)

1. Mass Defect & Binding Energy

Δm=Zmp+(A−Z)mn−mnucleus\Delta m = Zm_p + (A-Z)m_n – m_{\text{nucleus}}Δm=Zmp​+(A−Z)mn​−mnucleus​ EB=Δm⋅931MeVE_B = \Delta m \cdot 931 MeVEB​=Δm⋅931MeV

Example: Z=2, A=4, m_p=1.0073, m_n=1.0087, m_nucleus=4.0026 →

Δm=0.0302u⇒EB=28.14MeV\Delta m = 0.0302 u \Rightarrow E_B = 28.14 MeVΔm=0.0302u⇒EB​=28.14MeV

2. Radioactivity: α, β, γ decay

3. Half-life: N=N0e−λtN = N_0 e^{-\lambda t}N=N0​e−λt

Chapter 14 – Semiconductor Devices (अर्धचालक युक्तियाँ)

1. Types of Semiconductors: P-type (holes), N-type (electrons)

2. Diode: I-V characteristics

I=I0(eeV/kT−1)I = I_0 (e^{eV/kT} – 1)I=I0​(eeV/kT−1)

3. Transistor: Amplification, Switching

4. Logic Gates: AND, OR, NOT (basic electronics)

5. LED & Photodiode: Applications

Chapter 15 – Communication Systems (संचार प्रणाली)

1. Elements of Communication:

   • Transmitter → Channel → Receiver

2. Modulation:

   • AM: Amplitude modulation

   • FM: Frequency modulation

3. Signal Propagation: Radio waves, Microwaves

4. Bandwidth & Noise:

S/N ratio =Signal PowerNoise PowerS/N \text{ ratio } = \frac{\text{Signal Power}}{\text{Noise Power}}S/N ratio =Noise PowerSignal Power​

निष्कर्ष (Conclusion)

• Physics केवल सूत्रों का विषय नहीं है, यह तर्क और विश्लेषण की कला है।

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