भौतिकशास्त्र Physics

थोडक्यात सारांश

Polarisation shows that light is a transverse wave. In polarised light, the electric vector vibrates in one particular direction. This topic covers methods of producing polarised light (reflection, refraction, double refraction, selective absorption), Malus's law, Brewster's law, Nicol prism, wave plates and optical activity.

महत्त्वाचे मुद्दे

• In unpolarised light the electric vector vibrates in all directions perpendicular to propagation. In plane-polarised light it vibrates in one plane only.
• Only transverse waves can be polarised; sound (longitudinal) cannot.
• Malus's law: intensity through an analyser varies as cos²θ.
• Brewster's law: at the polarising angle θₚ, the reflected light is completely plane-polarised, and tan θₚ = μ. At this angle the reflected and refracted rays are perpendicular.
• Double refraction: in calcite and quartz a ray splits into the ordinary (O) ray, which obeys Snell's law, and the extraordinary (E) ray, which does not. Both are plane-polarised at right angles to each other.
• Optic axis: direction in the crystal along which there is no double refraction.
• Calcite is a negative crystal (μₒ > μₑ); quartz is a positive crystal (μₑ > μₒ).
• Nicol prism: a calcite crystal cut and joined with Canada balsam; the O-ray is removed by total internal reflection and only the E-ray passes. Used as polariser and analyser.
• Quarter-wave plate introduces a path difference λ/4 (turns plane-polarised light at 45° into circularly polarised light). Half-wave plate introduces λ/2.
• Optical activity: some substances (quartz, sugar solution) rotate the plane of polarisation. Measured with a polarimeter.

महत्त्वाच्या संज्ञा

• Polariser / analyser – device that produces / tests polarised light.
• Polarising angle (ध्रुवण कोन) – angle of incidence giving fully polarised reflected light.
• Dextro-rotatory / laevo-rotatory – rotates the plane clockwise / anticlockwise (looking towards the source).
• Specific rotation – rotation by a 1 dm column of solution of unit concentration (1 g/cc).

सूत्रे

• Malus: I = I₀ cos²θ
• Brewster: μ = tan θₚ; θₚ + r = 90°
• Unpolarised light through a polaroid: I = I₀/2
• Quarter-wave plate thickness: t = λ/(4(μₒ − μₑ)) (use the magnitude of the difference); half-wave: t = λ/(2(μₒ − μₑ))
• Specific rotation: [α] = θ/(l × c) (l in dm, c in g/cc)

५ महत्त्वाचे प्रश्न

Q1. State Brewster's law.

When unpolarised light falls on a transparent surface at the polarising angle θₚ, the reflected light is completely plane-polarised and the refractive index μ = tan θₚ. Then the reflected and refracted rays are at 90°.

Q2. Find the polarising angle for glass of μ = 1.5.

θₚ = tan⁻¹(1.5) ≈ 56.3°. The angle of refraction is 90° − 56.3° = 33.7°.

Q3. Prove that at the polarising angle the reflected and refracted rays are perpendicular.

1. Brewster: μ = tan θₚ = sin θₚ/cos θₚ.
2. Snell: μ = sin θₚ/sin r.
3. Equate: cos θₚ = sin r = cos(90° − r).
4. So θₚ = 90° − r, i.e. θₚ + r = 90°.
5. The angle between reflected and refracted rays = 180° − (θₚ + r) = 90°.

Q4. Calculate the thickness of a quartz quarter-wave plate for λ = 5893 Å (μₒ = 1.544, μₑ = 1.553).

1. t = λ/(4(μₑ − μₒ)) (quartz is positive, so μₑ > μₒ).
2. μₑ − μₒ = 0.009.
3. t = 5893 × 10⁻¹⁰/(4 × 0.009) = 5.893 × 10⁻⁷/0.036.
4. t ≈ 1.64 × 10⁻⁵ m ≈ 0.016 mm.

Q5. A 20 cm tube of sugar solution of concentration 0.1 g/cc rotates the plane of polarisation by 13°. Find the specific rotation. Also find the intensity through two polaroids at 60° if the incident light is unpolarised with intensity I₀.

1. l = 20 cm = 2 dm, c = 0.1 g/cc.
2. [α] = θ/(lc) = 13/(2 × 0.1) = 65° per dm per (g/cc).
3. First polaroid: I₁ = I₀/2.
4. Second (Malus): I₂ = I₁ cos²60° = (I₀/2)(1/4) = I₀/8.

परीक्षेसाठी टीप

In polarimeter numericals, change the tube length to decimetres. Draw the Nicol prism with the Canada balsam layer, O-ray and E-ray labelled — it is a common diagram question.

थोडक्यात सारांश

LASER stands for Light Amplification by Stimulated Emission of Radiation. A laser gives an intense, highly directional, monochromatic and coherent beam. This topic covers absorption, spontaneous and stimulated emission, Einstein's coefficients, population inversion, pumping, the ruby and He–Ne lasers, and applications.

महत्त्वाचे मुद्दे

• Absorption: an atom in lower level E₁ absorbs a photon of energy hν = E₂ − E₁ and goes to E₂.
• Spontaneous emission: an excited atom returns to E₁ on its own, emitting a photon in a random direction and phase (ordinary light).
• Stimulated emission: an incoming photon of energy hν makes an excited atom emit an identical photon — same frequency, phase, direction and polarisation. This is the basis of the laser (idea given by Einstein in 1917).
• Normally lower levels have more atoms (Boltzmann distribution). Population inversion (more atoms in the upper level) is needed for amplification.
• A metastable state has a long lifetime (about 10⁻³ s compared with about 10⁻⁸ s for ordinary excited states); it helps build population inversion.
• Pumping: supplying energy to achieve inversion — optical (flash lamp), electrical discharge, chemical.
• Optical resonator: two parallel mirrors (one fully, one partly reflecting) bounce light back and forth, building up the beam.
• Ruby laser (Maiman, 1960): first laser; Al₂O₃ doped with Cr³⁺ ions; three-level system; optical pumping by flash lamp; pulsed output at 694.3 nm (red).
• He–Ne laser: gas laser; electrical discharge excites He atoms which transfer energy to Ne by collision; four-level system; continuous output at 632.8 nm (red).
• Four-level lasers achieve inversion more easily than three-level ones.
• Applications: optical fibre communication, surgery (eye), cutting and welding, barcode readers, holography, distance measurement, CD/DVD.

महत्त्वाच्या संज्ञा

• Coherence – constant phase relation; laser has high temporal and spatial coherence.
• Monochromaticity – very narrow range of wavelengths.
• Active medium (सक्रिय माध्यम) – material in which inversion is created.
• Einstein coefficients – A₂₁ (spontaneous emission), B₁₂ (absorption), B₂₁ (stimulated emission).

सूत्रे

• Photon energy: E = hν = hc/λ
• Boltzmann: N₂/N₁ = e^(−(E₂ − E₁)/kT)
• Einstein relations: B₁₂ = B₂₁; A₂₁/B₂₁ = 8πhν³/c³
• Ratio of spontaneous to stimulated rate: R = e^(hν/kT) − 1
• Number of photons per second: n = P/(hν) = Pλ/(hc)

५ महत्त्वाचे प्रश्न

Q1. What is population inversion? Why is it necessary?

Population inversion is the state where the number of atoms in a higher energy level exceeds that in a lower level. It is needed because only then do stimulated emissions outnumber absorptions, so light is amplified.

Q2. Distinguish between spontaneous and stimulated emission.

Spontaneous emission happens on its own, giving photons with random phase and direction (incoherent). Stimulated emission is triggered by an incident photon, and the emitted photon is identical to it in frequency, phase and direction (coherent).

Q3. Describe the construction and working of the He–Ne laser.

1. Quartz tube filled with He and Ne (ratio about 10:1 He:Ne); electrodes for discharge; mirrors at both ends (one partly transparent).
2. Discharge excites He atoms to metastable levels.
3. Excited He transfers energy to Ne atoms by collision (levels nearly equal), creating inversion in Ne.
4. Ne atoms drop to a lower level by stimulated emission giving 632.8 nm light; then return to ground state quickly.
5. Mirrors provide feedback; continuous coherent beam comes out.
6. Four-level system, so inversion is easy and output is continuous.

Q4. Find the energy of a photon from a He–Ne laser (λ = 632.8 nm) and the number of photons emitted per second by a 1 mW laser.

1. E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(632.8 × 10⁻⁹).
2. = 1.989 × 10⁻²⁵/6.328 × 10⁻⁷ ≈ 3.14 × 10⁻¹⁹ J.
3. In eV: 3.14 × 10⁻¹⁹/1.6 × 10⁻¹⁹ ≈ 1.96 eV.
4. n = P/E = 10⁻³/3.14 × 10⁻¹⁹ ≈ 3.2 × 10¹⁵ photons per second.

Q5. Explain the ruby laser with an energy level diagram.

1. Ruby rod (Al₂O₃ with about 0.05% Cr³⁺); ends polished and silvered (one partially); helical xenon flash lamp around it.
2. Flash lamp pumps Cr³⁺ ions from ground level E₁ to a broad band E₃.
3. Ions drop quickly (non-radiative) to metastable level E₂.
4. Population inversion between E₂ and E₁ builds up.
5. A spontaneous photon triggers stimulated emission at 694.3 nm; it is amplified by reflections.
6. Output is pulsed, because more than half of the ions must be lifted to E₂ (three-level system).

परीक्षेसाठी टीप

Always draw the energy-level diagram with the pumping, non-radiative and laser transitions labelled. Remember the wavelengths: ruby 694.3 nm, He–Ne 632.8 nm.

थोडक्यात सारांश

Holography is a method of recording and reconstructing the full three-dimensional image of an object using laser light. Unlike a photograph, which records only intensity, a hologram records both the amplitude and the phase of light waves. This topic covers the principle of holography, recording and reconstruction, the properties of holograms, and the main applications of lasers in industry, medicine, communication and science.

महत्त्वाचे मुद्दे

• Holography was invented by Dennis Gabor (1947; Nobel Prize 1971). It became practical only after the laser (1960) gave highly coherent light.
• A hologram is an interference pattern between the light scattered by the object (object beam) and a direct beam (reference beam) from the same laser.
• Recording: a laser beam is split by a beam splitter; one part lights the object, the other goes straight to the photographic plate. The two beams interfere on the plate.
• Reconstruction: the developed plate is lit by the reference beam alone. The fringes act like a diffraction grating and recreate the original object wave, so a 3-D virtual image is seen.
• A hologram looks like a random pattern to the eye; it does not resemble the object.
• Every small piece of a hologram contains information about the whole object (with lower resolution).
• Coherence is essential: the path difference between the two beams must be less than the coherence length of the laser.
• Applications of lasers: cutting, welding and drilling; eye surgery (LASIK), removal of tumours, bloodless surgery; optical-fibre communication; barcode scanners, CD/DVD; LIDAR and range finding; holographic security marks on cards and notes; measuring very small distances (interferometry).

महत्त्वाच्या संज्ञा

• Object beam – laser light reflected or scattered from the object.
• Reference beam – unscattered laser light falling directly on the plate.
• Coherence length – maximum path difference over which light stays coherent.
• Holographic interferometry – comparing two holograms of an object to detect tiny deformations.
• LIDAR – light detection and ranging, radar with laser light.

सूत्रे

• Intensity on the plate: I = |O + R|² = |O|² + |R|² + O R* + O* R (the cross terms store the phase)
• Fringe spacing for two plane beams at angle θ: d = λ/[2 sin(θ/2)]
• Laser range finding: distance D = ct/2 (t = round-trip time)
• Coherence length L = c × Δt = c/Δν

५ महत्त्वाचे प्रश्न

Q1. How does a hologram differ from an ordinary photograph?

Q2. A laser pulse returns from the Moon after 2.56 s. Find the Earth–Moon distance.

D = ct/2 = 3 × 10⁸ × 2.56/2 = 3.84 × 10⁸ m (about 3.84 lakh km).

Q3. Explain the recording and reconstruction of a hologram with the principle.

1. Principle: the interference of two coherent waves records the phase of one relative to the other as fringe positions.
2. Recording: laser → beam splitter. Beam 1 (object beam) is reflected from the object to the plate; beam 2 (reference beam) reaches the plate via a mirror. The plate records the fringe pattern I = |O + R|².
3. Reconstruction: the developed plate is lit by a beam identical to R. Its transmittance is proportional to I, so the transmitted light contains a term proportional to |R|²O — a copy of the original object wave.
4. An observer looking through the plate sees a virtual 3-D image at the object's position; a real image (conjugate) also forms on the other side.

Q4. Two plane laser beams (λ = 633 nm) meet at 30° on a plate. Find the fringe spacing recorded.

d = λ/[2 sin(θ/2)] = 633 × 10⁻⁹/(2 × sin 15°) = 633 × 10⁻⁹/(2 × 0.2588) = 633 × 10⁻⁹/0.5176 ≈ 1.22 × 10⁻⁶ m. The plate must resolve about 800 lines per mm, so special fine-grain plates are used.

Q5. Describe the applications of lasers in medicine and industry.

Medicine: CO₂ laser as a bloodless scalpel (light seals blood vessels while cutting); excimer laser for LASIK eye correction; argon laser to repair detached retina; lasers for kidney-stone breaking, tattoo and tumour removal; photodynamic therapy for cancer. Industry: Nd:YAG and CO₂ lasers for cutting, welding and drilling hard metals and ceramics with a narrow heat-affected zone; laser marking; alignment in construction; measuring length and flatness by interferometry; non-destructive testing by holographic interferometry.

परीक्षेसाठी टीप

Remember the one-line idea: a hologram stores the phase as fringe positions. Draw the recording set-up with beam splitter, mirror, object and plate — the diagram alone earns marks.

थोडक्यात सारांश

An optical fibre is a thin, flexible strand of glass or plastic that guides light by total internal reflection. Fibres carry huge amounts of information over long distances with low loss. This topic covers structure, acceptance angle and numerical aperture, types of fibres, attenuation and dispersion, and applications.

महत्त्वाचे मुद्दे

• Structure: a central core of higher refractive index n₁, surrounded by cladding of lower index n₂, covered by a protective buffer/jacket.
• Light travels by total internal reflection (TIR) at the core–cladding boundary. Conditions: light goes from denser to rarer medium, and the angle of incidence exceeds the critical angle.
• Acceptance angle: the maximum angle with the fibre axis at which light can enter and still be guided. Rays inside this cone are guided.
• Numerical aperture (NA) = sine of acceptance angle (in air); measures light-gathering ability.
• Step-index fibre: core index is uniform; sudden change at the cladding; rays follow zig-zag paths.
• Graded-index fibre: core index decreases gradually from axis to edge; rays follow curved paths; less dispersion.
• Single-mode fibre: very thin core (a few μm); one mode; no intermodal dispersion; used for long-distance communication.
• Multimode fibre: larger core; many modes; easier to couple light; used for short links.
• Attenuation (loss): due to absorption, scattering (Rayleigh) and bending losses; measured in dB/km.
• Dispersion: spreading of light pulses — intermodal (different paths) and material/chromatic (different wavelengths). It limits the data rate.
• Advantages over copper: huge bandwidth, low loss, no electromagnetic interference, light weight, secure.
• Applications: telecommunication and internet, endoscopy in medicine, sensors, decorative lighting.

महत्त्वाच्या संज्ञा

• Critical angle (क्रांतिक कोन) – angle of incidence in the denser medium for which the refracted ray grazes the surface.
• Mode – an allowed path (pattern) of light in a fibre.
• V-number – fibre parameter that decides the number of modes.
• Fractional index change (Δ) – (n₁ − n₂)/n₁.

सूत्रे

• Critical angle: sin θc = n₂/n₁
• NA = √(n₁² − n₂²) ≈ n₁√(2Δ); sin θₐ = NA/n₀ (n₀ = 1 for air)
• Δ = (n₁ − n₂)/n₁
• V = (2πa/λ) × NA (a = core radius); single-mode if V < 2.405
• Number of modes (step index, large V) ≈ V²/2
• Attenuation: α = (10/L) log₁₀(P_in/P_out) dB/km

५ महत्त्वाचे प्रश्न

Q1. What is total internal reflection? State its conditions.

When light in a denser medium meets a rarer medium at an angle greater than the critical angle, all of it is reflected back. Conditions: light must go from denser to rarer medium, and the angle of incidence must exceed the critical angle.

Q2. Distinguish between step-index and graded-index fibres.

In a step-index fibre the core has a uniform index that drops suddenly at the cladding, and rays move in zig-zag paths. In a graded-index fibre the core index decreases gradually from the axis, rays bend smoothly, and dispersion is less.

Q3. Derive the expression for numerical aperture.

1. Ray enters from air (n₀) at angle θₐ, refracts at angle θᵣ in core: n₀ sin θₐ = n₁ sin θᵣ.
2. It meets the core–cladding surface at angle (90° − θᵣ), which must be at least θc for TIR.
3. At the limit, sin(90° − θᵣ) = cos θᵣ = n₂/n₁.
4. sin θᵣ = √(1 − n₂²/n₁²) = √(n₁² − n₂²)/n₁.
5. n₀ sin θₐ = n₁ sin θᵣ = √(n₁² − n₂²).
6. NA = n₀ sin θₐ = √(n₁² − n₂²).

Q4. A fibre has n₁ = 1.50 and n₂ = 1.47. Find the critical angle, NA and acceptance angle in air.

1. sin θc = 1.47/1.50 = 0.98 → θc ≈ 78.5°.
2. NA = √(1.50² − 1.47²) = √(2.25 − 2.1609) = √0.0891 ≈ 0.298.
3. θₐ = sin⁻¹(0.298) ≈ 17.4°.

Q5. The power input to a fibre is 10 mW and output after 5 km is 5 mW. Find the attenuation.

1. α = (10/L) log₁₀(P_in/P_out).
2. P_in/P_out = 10/5 = 2; log₁₀ 2 = 0.301.
3. α = (10/5) × 0.301 = 0.602 dB/km.
4. Total loss over 5 km = 3.01 dB.

परीक्षेसाठी टीप

Draw a fibre showing core, cladding, the acceptance cone and a zig-zag ray — it supports both the NA derivation and short answers. Learn NA = √(n₁² − n₂²) and the attenuation formula; numericals use them directly.

थोडक्यात सारांश

Vector calculus describes how fields change in space. Three operations built from the del operator ∇ — gradient, divergence and curl — appear throughout electricity, magnetism, fluids and heat flow. This topic covers scalar and vector fields, the meaning and calculation of grad, div and curl, the Laplacian, and the important identities, with worked examples.

महत्त्वाचे मुद्दे

• A scalar field gives a number at every point (temperature T(x, y, z), potential V). A vector field gives a vector at every point (electric field E, velocity of a fluid).
• Del operator: ∇ = î ∂/∂x + ĵ ∂/∂y + k̂ ∂/∂z. It is a vector differential operator.
• Gradient ∇φ of a scalar field is a vector. It points in the direction of the fastest increase of φ, and its magnitude is that maximum rate. It is perpendicular to the level surfaces φ = constant.
• Directional derivative of φ along unit vector â = ∇φ · â.
• Divergence ∇·A of a vector field is a scalar: the net outward flux per unit volume. Positive → source, negative → sink.
• A field with ∇·A = 0 is solenoidal (e.g. magnetic field B).
• Curl ∇×A is a vector: it measures the rotation (circulation per unit area) of the field.
• A field with ∇×A = 0 is irrotational or conservative, and can be written as A = −∇V (e.g. electrostatic field).
• Laplacian ∇²φ = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z². Laplace's equation ∇²V = 0 holds in charge-free space.
• Key identities: curl of grad is always zero; div of curl is always zero.

महत्त्वाच्या संज्ञा

• Level surface – surface on which a scalar field has a constant value.
• Flux – amount of a vector field passing through a surface.
• Solenoidal field – zero divergence everywhere.
• Irrotational field – zero curl everywhere.
• Scalar potential – scalar function whose (negative) gradient gives the field.

सूत्रे

• ∇φ = (∂φ/∂x) î + (∂φ/∂y) ĵ + (∂φ/∂z) k̂
• ∇·A = ∂Ax/∂x + ∂Ay/∂y + ∂Az/∂z
• ∇×A = (∂Az/∂y − ∂Ay/∂z) î + (∂Ax/∂z − ∂Az/∂x) ĵ + (∂Ay/∂x − ∂Ax/∂y) k̂
• ∇×(∇φ) = 0; ∇·(∇×A) = 0
• ∇·(φA) = φ(∇·A) + A·(∇φ)
• ∇×(∇×A) = ∇(∇·A) − ∇²A
• For r = xî + yĵ + zk̂: ∇·r = 3, ∇×r = 0, ∇r = r/r, ∇(1/r) = −r/r³

५ महत्त्वाचे प्रश्न

Q1. Give the physical meaning of divergence and curl.

Divergence measures how much a field spreads out from a point: the outward flux per unit volume around it. In a fluid, positive divergence means fluid is being created (source). Curl measures how much the field circulates around a point; for fluid flow it equals twice the angular velocity of a small paddle wheel placed there. A uniform flow has zero curl; a whirlpool has non-zero curl.

Q2. Find the gradient of φ = x²y + yz³ at (1, 2, 1) and the directional derivative along (î + 2ĵ + 2k̂)/3.

∇φ = 2xy î + (x² + z³) ĵ + 3yz² k̂. At (1, 2, 1): ∇φ = 4î + 2ĵ + 6k̂.
Directional derivative = (4 × 1 + 2 × 2 + 6 × 2)/3 = 20/3 ≈ 6.67.
Maximum rate = |∇φ| = √(16 + 4 + 36) = √56 ≈ 7.48.

Q3. Find the divergence and curl of A = x²y î + yz ĵ + xz² k̂ at (1, 1, 1).

∇·A = 2xy + z + 2xz. At (1, 1, 1): 2 + 1 + 2 = 5.
∇×A: î(∂(xz²)/∂y − ∂(yz)/∂z) = î(0 − y) = −y î;
ĵ(∂(x²y)/∂z − ∂(xz²)/∂x) = ĵ(0 − z²) = −z² ĵ;
k̂(∂(yz)/∂x − ∂(x²y)/∂y) = k̂(0 − x²) = −x² k̂.
At (1, 1, 1): ∇×A = −î − ĵ − k̂.

Q4. Prove that curl grad φ = 0 and div curl A = 0.

1. ∇×(∇φ): the x-component is ∂/∂y(∂φ/∂z) − ∂/∂z(∂φ/∂y) = ∂²φ/∂y∂z − ∂²φ/∂z∂y = 0, since mixed partial derivatives are equal. The y and z components vanish the same way. So ∇×∇φ = 0.
2. ∇·(∇×A) = ∂/∂x(∂Az/∂y − ∂Ay/∂z) + ∂/∂y(∂Ax/∂z − ∂Az/∂x) + ∂/∂z(∂Ay/∂x − ∂Ax/∂y). The six terms cancel in pairs, so ∇·(∇×A) = 0.

Physical meaning: a gradient (conservative) field has no rotation; a curl field (like B) has no sources.

Q5. Show that the field E = (2xy + z) î + x² ĵ + x k̂ is conservative and find its potential.

∇×E: î(∂x/∂y − ∂x²/∂z) = 0; ĵ(∂(2xy + z)/∂z − ∂x/∂x) = ĵ(1 − 1) = 0; k̂(∂x²/∂x − ∂(2xy + z)/∂y) = k̂(2x − 2x) = 0. So E is irrotational, hence conservative.
Find φ with ∇φ = E: ∂φ/∂x = 2xy + z → φ = x²y + xz + f(y, z). Then ∂φ/∂y = x² + ∂f/∂y = x², so f depends only on z; ∂φ/∂z = x + f′(z) = x, so f is constant. φ = x²y + xz + C (and V = −φ if E = −∇V).

परीक्षेसाठी टीप

Write the curl as a 3×3 determinant with ∂/∂x, ∂/∂y, ∂/∂z in the middle row; it prevents sign errors. Remember: grad acts on a scalar and gives a vector; div gives a scalar; curl gives a vector.

थोडक्यात सारांश

Vector integral theorems connect integrals over a region with integrals over its boundary. Gauss's divergence theorem links a volume integral to a surface integral, Stokes' theorem links a surface integral to a line integral, and Green's theorem is the plane form of Stokes' theorem. This topic covers line, surface and volume integrals, the three theorems with proofs in outline, and worked examples, along with their use in electromagnetism.

महत्त्वाचे मुद्दे

• Line integral ∫A·dl along a curve; for a force field it gives the work done. Around a closed loop, ∮A·dl is called the circulation.
• For a conservative field (∇×A = 0) the line integral depends only on the end points, and ∮A·dl = 0.
• Surface integral ∫A·dS gives the flux of A through a surface; dS = n̂ dS points along the outward normal.
• Volume integral ∫φ dV adds a scalar over a volume (for example total charge = ∫ρ dV).
• Gauss's divergence theorem: ∮_S A·dS = ∫_V (∇·A) dV. Total outward flux = total "source strength" inside.
• Stokes' theorem: ∮_C A·dl = ∫_S (∇×A)·dS. Circulation around a loop = flux of the curl through any surface bounded by it.
• Green's theorem in the plane: ∮(M dx + N dy) = ∬(∂N/∂x − ∂M/∂y) dx dy.
• Physics uses: Gauss's theorem turns Gauss's law into ∇·E = ρ/ε₀; Stokes' theorem turns Ampere's law into ∇×B = μ₀J; the equation of continuity ∇·J + ∂ρ/∂t = 0 follows from charge conservation.

महत्त्वाच्या संज्ञा

• Closed surface – a surface that encloses a volume (sphere, cube).
• Circulation – line integral of a field around a closed path.
• Simply connected region – a region in which every closed curve can be shrunk to a point without leaving it.
• Outward normal – unit vector perpendicular to a surface, pointing out of the enclosed volume.

सूत्रे

• Work W = ∫ F·dl
• Gauss: ∮_S A·dS = ∫_V ∇·A dV
• Stokes: ∮_C A·dl = ∫_S (∇×A)·dS
• Green (plane): ∮_C (M dx + N dy) = ∬_R (∂N/∂x − ∂M/∂y) dx dy
• Area of a plane region: A = ½ ∮ (x dy − y dx)
• Differential form of Gauss's law: ∇·E = ρ/ε₀

५ महत्त्वाचे प्रश्न

Q1. State Gauss's divergence theorem and Stokes' theorem.

Gauss: the surface integral of the normal component of a vector field over a closed surface equals the volume integral of its divergence over the enclosed volume, ∮A·dS = ∫∇·A dV. Stokes: the line integral of the tangential component of a vector field around a closed curve equals the surface integral of the normal component of its curl over any surface bounded by that curve, ∮A·dl = ∫(∇×A)·dS.

Q2. Find the work done by F = 3x² î + (2xz − y) ĵ + z k̂ along the straight line from (0, 0, 0) to (2, 1, 3).

Path: x = 2t, y = t, z = 3t, t from 0 to 1; dx = 2dt, dy = dt, dz = 3dt.
F·dl = 3(4t²)(2dt) + (2·2t·3t − t)dt + 3t(3dt) = (24t² + 12t² − t + 9t)dt = (36t² + 8t)dt.
W = ∫₀¹ (36t² + 8t) dt = 12 + 4 = 16 units.

Q3. Verify Gauss's divergence theorem for A = x î + y ĵ + z k̂ over a cube of side a with one corner at the origin.

Volume side: ∇·A = 3, so ∫∇·A dV = 3a³.
Surface side: on the face x = a, A·n̂ = x = a, flux = a × a² = a³; on x = 0, flux = 0. The same holds for y and z faces. Total flux = 3a³.
Both sides equal 3a³, so the theorem is verified.

Q4. Prove Gauss's divergence theorem (outline).

1. Divide the volume V into tiny cubes of volume dx dy dz.
2. For one cube, the net outflow in the x-direction is [Ax(x + dx) − Ax(x)] dy dz = (∂Ax/∂x) dx dy dz. Adding y and z: net outflow = (∇·A) dV.
3. Add over all cubes. Across every inner face the flux leaving one cube enters its neighbour, so these cancel.
4. Only the outer faces remain, giving the flux through the boundary surface S.
5. Hence ∮_S A·dS = ∫_V (∇·A) dV.

Q5. Using Green's theorem, find the area of an ellipse x²/a² + y²/b² = 1.

Area = ½ ∮ (x dy − y dx). Parametrise: x = a cos θ, y = b sin θ, θ from 0 to 2π. Then dx = −a sin θ dθ, dy = b cos θ dθ.
x dy − y dx = ab cos²θ dθ + ab sin²θ dθ = ab dθ.
Area = ½ × ab × 2π = πab.

परीक्षेसाठी टीप

For a "verify" question, always compute both sides separately and show they are equal. Mention the electromagnetic use (Gauss's law, Ampere's law) — it shows the examiner you know why the theorem matters.

थोडक्यात सारांश

Most laws of physics are written as differential equations: radioactive decay, cooling, motion of a spring, current in a circuit and waves. This topic covers the order and degree of a differential equation, first-order equations (variable separable and linear), second-order linear equations with constant coefficients (complementary function and particular integral), and an introduction to partial differential equations and separation of variables, all with physics examples.

महत्त्वाचे मुद्दे

• Order = order of the highest derivative; degree = power of the highest derivative after removing fractions and radicals.
• A linear equation has the dependent variable and its derivatives only to the first power and not multiplied together.
• Variable separable: write f(y) dy = g(x) dx and integrate both sides. Example: radioactive decay dN/dt = −λN gives N = N₀e^(−λt).
• Linear first-order equation dy/dx + P(x)y = Q(x): multiply by the integrating factor e^(∫P dx). Example: an LR circuit.
• Real, different roots m₁, m₂: y = C₁e^(m₁x) + C₂e^(m₂x)
• Equal roots m: y = (C₁ + C₂x)e^(mx)
• Complex roots α ± iβ: y = e^(αx)(C₁ cos βx + C₂ sin βx)
• Second-order homogeneous a y″ + b y′ + c y = 0: try y = e^(mx) to get the auxiliary equation am² + bm + c = 0.
• Non-homogeneous: general solution = complementary function (CF) + particular integral (PI).
• Physical examples: SHM (ẍ + ω²x = 0), damped oscillator, LCR circuit, forced oscillations.
• Partial differential equations contain partial derivatives: wave equation, heat (diffusion) equation, Laplace's equation. They are often solved by separation of variables, u(x, t) = X(x)T(t).

महत्त्वाच्या संज्ञा

• General solution – solution with as many arbitrary constants as the order.
• Particular solution – solution after fixing constants using initial or boundary conditions.
• Integrating factor (IF) – factor that makes the left side an exact derivative.
• Auxiliary equation – algebraic equation obtained by putting y = e^(mx).
• Boundary conditions – values of the solution fixed at the edges of the region.

सूत्रे

• Linear first order: y × IF = ∫Q × IF dx + C, IF = e^(∫P dx)
• Decay: N = N₀ e^(−λt), half-life T½ = 0.693/λ
• SHM: ẍ + ω²x = 0 → x = A cos(ωt + φ)
• LR circuit growth: I = (E/R)(1 − e^(−Rt/L))
• Wave equation: ∂²y/∂t² = v² ∂²y/∂x²; heat equation: ∂u/∂t = α ∂²u/∂x²
• String fixed at x = 0 and x = L: y = Σ A_n sin(nπx/L) cos(nπvt/L)

५ महत्त्वाचे प्रश्न

Q1. Find the order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0.

The highest derivative is d²y/dx², so the order is 2. Its power is 3, so the degree is 3.

Q2. Solve the LR circuit equation L dI/dt + RI = E with I = 0 at t = 0.

Write dI/dt + (R/L)I = E/L. IF = e^(Rt/L).
I e^(Rt/L) = ∫(E/L)e^(Rt/L) dt = (E/R)e^(Rt/L) + C.
At t = 0, I = 0 → C = −E/R. So I = (E/R)(1 − e^(−Rt/L)). The current grows to E/R with time constant L/R.

Q3. Solve y″ − 5y′ + 6y = 0 with y(0) = 2 and y′(0) = 5.

Auxiliary equation m² − 5m + 6 = 0 → (m − 2)(m − 3) = 0 → m = 2, 3.
y = C₁e^(2x) + C₂e^(3x). y(0): C₁ + C₂ = 2. y′(0): 2C₁ + 3C₂ = 5.
Solving: C₂ = 1, C₁ = 1. So y = e^(2x) + e^(3x).
Check: y′(0) = 2 + 3 = 5 ✔.

Q4. Solve the forced equation ẍ + 4x = 8 cos t (find the general solution).

CF: m² + 4 = 0 → m = ±2i → x_c = C₁ cos 2t + C₂ sin 2t.
PI: try x_p = A cos t → −A cos t + 4A cos t = 3A cos t = 8 cos t → A = 8/3.
General solution: x = C₁ cos 2t + C₂ sin 2t + (8/3) cos t. The first part is the natural oscillation, the second follows the driving force.

Q5. Solve the one-dimensional wave equation for a string fixed at both ends by separation of variables.

1. Put y(x, t) = X(x)T(t): X T″ = v² X″ T → X″/X = T″/(v²T) = −k² (a constant).
2. X″ + k²X = 0 → X = A sin kx + B cos kx. y(0, t) = 0 → B = 0. y(L, t) = 0 → sin kL = 0 → k = nπ/L, n = 1, 2, 3 …
3. T″ + k²v²T = 0 → T = C cos(nπvt/L) + D sin(nπvt/L).
4. y = Σ sin(nπx/L)[C_n cos(nπvt/L) + D_n sin(nπvt/L)]: the normal modes with frequencies f_n = nv/(2L).

The constants C_n and D_n come from the initial shape and velocity using Fourier series.

परीक्षेसाठी टीप

In second-order problems, write the auxiliary equation first and name the case (real, equal or complex roots). Always put the initial conditions back to check your constants.

थोडक्यात सारांश

Any periodic function that meets certain conditions can be written as a sum of sines and cosines — its Fourier series. In physics this lets us break complex waveforms (square wave, sawtooth, rectified AC) into a fundamental and its harmonics. This topic covers Dirichlet conditions, Fourier coefficients, even and odd functions and standard waveforms.

महत्त्वाचे मुद्दे

• A function is periodic with period T if f(t + T) = f(t).
• Fourier's theorem: a periodic function can be expressed as a constant plus a series of sine and cosine terms of frequencies that are whole multiples of the fundamental.
• Dirichlet conditions: in one period, f must be single-valued, finite, have a finite number of discontinuities, and a finite number of maxima and minima.
• At a jump discontinuity, the series converges to the average of left and right values.
• Even function (f(−x) = f(x)): only cosine terms (bₙ = 0).
• Odd function (f(−x) = −f(x)): only sine terms (a₀ = aₙ = 0).
• Square wave and sawtooth wave are odd (if centred suitably), so they contain only sine terms.
• A square wave contains only odd harmonics with amplitudes falling as 1/n.
• Uses: analysing musical sounds (quality/timbre depends on harmonics), electrical signals and filters, heat conduction, vibrating strings.
• Fourier analysis: finding coefficients from a waveform; synthesis: building the waveform from harmonics.

महत्त्वाच्या संज्ञा

• Fundamental – the lowest frequency term (n = 1).
• Harmonics (संनादी) – terms with frequencies 2, 3, 4 … times the fundamental.
• Fourier coefficients – a₀, aₙ, bₙ.
• Orthogonality – ∫ sin mx sin nx dx over a period = 0 for m ≠ n (basis for finding coefficients).

सूत्रे

For period 2π (−π to π):

• f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx)
• a₀ = (1/π) ∫₋π^π f(x) dx; aₙ = (1/π) ∫₋π^π f(x) cos nx dx; bₙ = (1/π) ∫₋π^π f(x) sin nx dx
• Odd function: bₙ = (2/π) ∫₀^π f(x) sin nx dx
• Period T in time (ω = 2π/T): f(t) = a₀/2 + Σ(aₙ cos nωt + bₙ sin nωt), aₙ = (2/T) ∫₀^T f(t) cos nωt dt
• Full-wave rectified wave: |sin x| = 2/π − (4/π)[cos 2x/3 + cos 4x/15 + cos 6x/35 + …]

५ महत्त्वाचे प्रश्न

Q1. State Dirichlet's conditions for a Fourier series.

In one period the function must be single-valued and finite, have a finite number of discontinuities and a finite number of maxima and minima. Then its Fourier series converges to f(x), and to the mean value at a discontinuity.

Q2. Why does the Fourier series of an odd function contain only sine terms?

For odd f, f(x) cos nx is odd, so its integral over (−π, π) is zero; hence a₀ = aₙ = 0. Only bₙ survives, giving a pure sine series.

Q3. Find the Fourier series of the square wave f(x) = −1 for −π < x < 0 and +1 for 0 < x < π.

1. f is odd, so a₀ = aₙ = 0.
2. bₙ = (2/π) ∫₀^π sin nx dx = (2/π)[−cos nx/n]₀^π = (2/(nπ))(1 − cos nπ).
3. cos nπ = (−1)ⁿ: for even n, bₙ = 0; for odd n, bₙ = 4/(nπ).
4. f(x) = (4/π)[sin x + (sin 3x)/3 + (sin 5x)/5 + …].
5. Only odd harmonics are present, with amplitudes 1, 1/3, 1/5 … of the fundamental.

Q4. Find the Fourier series of the sawtooth wave f(x) = x, −π < x < π.

1. f is odd: a₀ = aₙ = 0.
2. bₙ = (2/π) ∫₀^π x sin nx dx.
3. Integrate by parts: ∫₀^π x sin nx dx = [−x cos nx/n]₀^π + (1/n)∫₀^π cos nx dx = −π cos nπ/n + 0 = π(−1)ⁿ⁺¹/n.
4. bₙ = (2/π) × π(−1)ⁿ⁺¹/n = 2(−1)ⁿ⁺¹/n.
5. f(x) = 2[sin x − (sin 2x)/2 + (sin 3x)/3 − …]. All harmonics are present.

Q5. Explain the importance of Fourier series in physics.

1. Breaks a complex periodic signal into simple harmonic components.
2. Explains the quality (timbre) of musical notes: same pitch, different harmonics.
3. Used in electronics: rectifier output, filters, signal analysis and communication.
4. Solves physical problems like vibrating strings and heat flow.
5. Leads to the Fourier transform used for non-periodic signals and spectroscopy.

परीक्षेसाठी टीप

Check whether the function is even or odd before integrating — it halves the work. Use (−1)ⁿ for cos nπ and state clearly which harmonics vanish.

थोडक्यात सारांश

A Fourier series breaks a periodic function into sine and cosine terms with discrete frequencies. The Fourier transform extends this idea to non-periodic functions, giving a continuous spread (spectrum) of frequencies. This topic covers the Fourier integral, the Fourier transform and its inverse, sine and cosine transforms, important properties, transforms of standard functions such as the rectangular pulse and Gaussian, and uses in optics, signals and quantum mechanics.

महत्त्वाचे मुद्दे

• A non-periodic function can be treated as a periodic one whose period L → ∞. The sum over harmonics then becomes an integral (the Fourier integral).
• Fourier transform: F(k) = (1/√2π) ∫ f(x) e^(−ikx) dx over all x. Inverse: f(x) = (1/√2π) ∫ F(k) e^(ikx) dk over all k.
• f(x) and F(k) form a transform pair: one describes the function in space (or time), the other in wave number (or frequency).
• For an even function the transform reduces to the cosine transform; for an odd function to the sine transform.
• Properties: linearity; shifting f(x − a) multiplies F(k) by e^(−ika); scaling f(ax) gives (1/|a|)F(k/a); the transform of df/dx is ik F(k).
• Convolution theorem: the transform of a convolution is the product of transforms (times √2π).
• Parseval's theorem: ∫|f(x)|² dx = ∫|F(k)|² dk — total energy is the same in both pictures.
• A narrow pulse has a wide spectrum and a wide pulse has a narrow spectrum: Δx·Δk ≈ constant. This is the mathematical root of the Heisenberg uncertainty principle.
• Uses: Fraunhofer diffraction pattern is the Fourier transform of the aperture; spectrum analysers; image processing; solving differential equations; momentum-space wavefunctions.

महत्त्वाच्या संज्ञा

• Spectrum – distribution of amplitude (or energy) over frequency.
• Transform pair – f(x) and its Fourier transform F(k).
• Rectangular pulse (top-hat) – f = 1 for |x| ≤ a, 0 otherwise.
• Sinc function – sin(u)/u, the transform of a rectangular pulse.
• Bandwidth – spread of frequencies present in a signal.

सूत्रे

• F(k) = (1/√2π) ∫ f(x) e^(−ikx) dx; f(x) = (1/√2π) ∫ F(k) e^(ikx) dk (limits −∞ to +∞)
• Cosine transform: F_c(k) = √(2/π) ∫₀^∞ f(x) cos kx dx
• Sine transform: F_s(k) = √(2/π) ∫₀^∞ f(x) sin kx dx
• Rectangular pulse of half-width a: F(k) = √(2/π) sin(ka)/k
• Gaussian: f(x) = e^(−x²/2σ²) → F(k) = σ e^(−σ²k²/2)
• Exponential: f(x) = e^(−a|x|) → F(k) = √(2/π) a/(a² + k²)
• Parseval: ∫|f|² dx = ∫|F|² dk

५ महत्त्वाचे प्रश्न

Q1. Distinguish between a Fourier series and a Fourier transform.

Q2. Find the Fourier transform of a rectangular pulse f(x) = 1 for |x| ≤ a and 0 elsewhere.

F(k) = (1/√2π) ∫ from −a to a of e^(−ikx) dx = (1/√2π) [e^(−ikx)/(−ik)] from −a to a = (1/√2π)(e^(ika) − e^(−ika))/(ik) = (1/√2π)(2 sin ka)/k.
So F(k) = √(2/π) sin(ka)/k. Its first zeros are at k = ±π/a. A wider pulse (bigger a) gives a narrower central peak: this is the single-slit diffraction pattern.

Q3. Find the Fourier transform of f(x) = e^(−a|x|), a positive.

F(k) = (1/√2π)[∫ from −∞ to 0 of e^(ax)e^(−ikx) dx + ∫ from 0 to ∞ of e^(−ax)e^(−ikx) dx]
= (1/√2π)[1/(a − ik) + 1/(a + ik)] = (1/√2π) × 2a/(a² + k²).
So F(k) = √(2/π) a/(a² + k²), a Lorentzian shape. Natural spectral lines have this shape because excited states decay exponentially.

Q4. Prove the shifting property and the derivative property.

1. Shifting: transform of f(x − a) = (1/√2π)∫f(x − a)e^(−ikx) dx. Put u = x − a: = e^(−ika)(1/√2π)∫f(u)e^(−iku) du = e^(−ika) F(k). A shift changes only the phase, not the amplitude spectrum.
2. Derivative: transform of f′(x) = (1/√2π)∫f′(x)e^(−ikx) dx. Integrate by parts; the boundary term vanishes when f → 0 at ±∞: = (1/√2π)(ik)∫f e^(−ikx) dx = ik F(k). This turns differential equations into algebraic ones.

Q5. Show how the Fourier transform explains the uncertainty principle.

Take a Gaussian wave packet of width σ in x: f(x) = e^(−x²/2σ²). Its transform is F(k) = σe^(−σ²k²/2), a Gaussian of width 1/σ in k. So Δx ≈ σ and Δk ≈ 1/σ, giving Δx·Δk ≈ 1. With p = ħk (de Broglie), Δx·Δp ≈ ħ. A particle confined to a small region must contain a wide range of momenta. (With standard deviations the exact limit is Δx·Δp ≥ ħ/2.)

परीक्षेसाठी टीप

Learn three standard pairs by heart — rectangular pulse → sinc, exponential → Lorentzian, Gaussian → Gaussian. Most numerical questions are one of these with different constants.