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#energy conservation

39 public questions tagged with this topic.

Polarization and Malus Law

What causes the intensity of light to remain conserved during interference despite the presence of dark fringes?

**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. Energy is redistributed from dark to bright fringes through interference, conserving total energy as the sum of intensities balances

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

Why does the intensity of light remain unaffected by changes in its speed during refraction?

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. Intensity depends on the square of the amplitude, which remains constant during refraction, while speed changes do not alter the energy per unit area. Using Δ = d sinθ, y = n λ D/d,

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

Why does the intensity of light remain unchanged in terms of energy when it undergoes interference or diffraction?

**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. Interference and diffraction redistribute light energy without loss, as bright and dark regions balance out, conserving total energy

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

Why does the intensity of light not depend on its speed when it enters a denser medium?

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. Intensity depends on the amplitude squared, not speed, which only affects wavelength and not the energy carried per unit area. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ co

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Energy, Intensity and Momentum of EM Waves

In electromagnetic theory, what fundamental principle allows electromagnetic waves to sustain their propagation in vacuu

**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. The principle of mutual induction, where a changing electric field induces a magnetic field and vice versa (via Faraday’s law and Ampere-Maxwell law), ensures self-sustaining propagation in vacuum without energy loss. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Mutual induction of fields, illustrating EM wave transverse nature and Maxwell's displace

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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Energy, Intensity and Momentum of EM Waves

In electromagnetic wave propagation, what ensures energy conservation during wave travel?

**Energy in EM wave** equally divided between electric and magnetic fields, energy density u = ½ ε₀ E² + B²/(2μ₀) = ε₀ E² = B²/μ₀, average u_avg = ½ ε₀ E₀², intensity I = c u_avg = ½ c ε₀ E₀² = E₀ B₀/(2μ₀) = c B₀²/(2μ₀), radiation pressure p = I/c for absorption, 2I/c for reflection. The energy in an electromagnetic wave is carried by both electric and magnetic fields, with the total energy flux described by the Poynting vector, ensuring conservation during propagation. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Poynting

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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Sharing of Charges, Common Potential and Applications

When two identical capacitors, one charged and one uncharged, are connected in parallel, why does the total energy decre

**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. When a charged capacitor ( C , charge Q , voltage V ) is connected in parallel with an uncharged capacitor ( C ), the total capacitance becomes 2C , and the charge redistributes to a final voltage V' = Q/(2C) = V/2 . Initial energy is U_i = (Q²/2C) , while final energy

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Energy Stored in Capacitor and Energy Density

In an isolated system of two capacitors initially charged and then connected in parallel with opposite polarities, what

**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. When two capacitors are connected in parallel with opposite polarities, charges redistribute such that the net charge on the positive plates (and negative plates) adjusts to a new equilibrium. The initial energy stored in the capacitors ( U_i = (1/2) C₁ V₁² + (1/2) C₂ V₂² ) is greater than the final

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Electrical Power, Energy and Heating Effect

In a circuit with resistors, why does the total power dissipated equal the power supplied by the source in steady state?

**Power dissipation** in resistor converts electrical energy to heat, P = V²/R inversely proportional to R for fixed V, directly proportional for fixed I. For battery with internal r, power wasted internally = I² r, useful power = I² R, efficiency η = R/(R+r). In steady state, energy is conserved. The power supplied by the source ( P = V I ) is fully dissipated as heat in the resistors ( P = I² Rtₒtₐl ), with no energy stored or lost elsewhere. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε

Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect

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Forced Oscillations and Resonance

Which condition ensures that the total mechanical energy in an SHM system remains conserved during the motion?

**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. Total mechanical energy (kinetic + potential) is conserved in SHM when no external dissipative forces (e.g., friction) act, allowing energy to transform without loss. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Absence of dissipative forces follows, ref

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Energy in SHM - Kinetic, Potential and Total

In an ideal SHM system, what occurs to the total mechanical energy as the particle moves from the mean position to an ex

**Energy distribution** shows maximum kinetic at equilibrium and maximum potential at extremes, sum constant. This relation enables calculation of amplitude, velocity at any displacement via v = ±√(2(E-U)/m), and understanding of energy storage in oscillating system for NEET problems. Total mechanical energy in ideal SHM (no friction) is conserved, remaining constant as kinetic energy converts to potential energy during the motion. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It remains constant follows, reflecting SHM dependence on amplitude A, ω and

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Heat Transfer Work Distinction and Internal Energy Change

A system in a cyclic process performs 300 J of work and rejects 200 J of heat. What is the heat absorbed?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . Q_absorb - 200 = 300 ⇒ Q_absorb = 500 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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