48 public questions tagged with this topic.
Spreadsheet program correct accounting fundamental; capital revenue classification per accounting principles.
Ref: NCERT Accountancy Class XI Chapter 1 Accounting Fundamentals explains accounting meaning objectives advantages limitations double entry system clearly.
**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Speed in a medium v = (c/n) . Given n = 1.8 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.8) ≈ 1.67 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,
Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle
**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Frequency nu = (c/λ) . λ = 4.6 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/4.6 × 10⁻⁷) ≈ 6.52 × 10¹⁴ Hz . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a
Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation
**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.4 × 10⁸ m/s , n = (3.0 × 10⁸/2.4 × 10⁸) = 1.25 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A =
Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle
**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. Frequency nu = (c/λ) . λ = 5.4 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/5.4 × 10⁻⁷) ≈ 5.56 × 10¹⁴ Hz . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,
Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation
**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Wavelength in a medium λ_m = (λvₐcuuₘ/n) . Given λvₐcuuₘ = 600 nm , n = 1.2 , λ_m = (600/1.2) = 500 nm . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a
Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle
**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Speed in a medium v = (c/n) . Given n = 1.6 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.6) = 1.875 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,
Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle
**Frequency of light** remains unchanged when refracts from air into water because frequency determined by source, energy E= h f conserved, speed decreases v= c/n, wavelength decreases λ'=v/f= λ/n, energy of wave proportional to amplitude² not speed, intensity I =½ c ε₀ E₀², energy not depend on speed directly, when speed decreases amplitude may change but energy conserved, interference redistributes energy, total energy same, bright regions gain from dark. Frequency nu = (c/λ) . λ = 6.3 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/6.3 × 10⁻⁷) ≈ 4.76 × 10¹⁴ Hz . Using Δ = d
Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation
**Nuclear density** nearly constant because R ∝ A^{1/3} so volume ∝ A, mass ∝ A, ratio constant, ~10¹⁷ kg/m³, 10¹⁴ times water density, shows nucleus compact, nuclear force short-range saturated, mass number 16 radius ~3×10⁻¹⁵ m, mass number from radius R=5.4×10⁻¹⁵ m => A=(R/R₀)³=(4.5)³=91. Ebₙ = (E_b/A) . E_b = 425 MeV , A = 50 . Ebₙ = (425/50) = 8.5 MeV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 8.5 MeV, consistent with Bohr model and nuclear binding energy systematics.
Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure
**Nuclear size** order 10⁻¹⁵ m femtometer, atomic size 10⁻¹⁰ m, ratio 10⁵, nucleus contains protons and neutrons bound by strong force, density independent of A indicates incompressibility, R₀ determined from electron scattering experiments. R = R₀ A¹/³ . 3.6 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (3.6/1.2) = 3 . A = 3³ = 27 . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 27, consistent with Bohr model and nuclear binding energy systematics.
Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure
**Aerials produce radio waves** by rapid acceleration/deceleration of electrons in antenna driven by AC, frequency equals driving frequency, for 60 MHz, λ=c/f=3×10⁸/60×10⁶=5 m, half-wave antenna length λ/2=2.5 m, efficient radiation when antenna size comparable to λ. c = (1/√(μ₀ ε₀)) . Substituting values, μ₀ ε₀ = (4 π × 10⁻⁷) × (8.85 × 10⁻¹²) ≈ 1.11 × 10⁻¹⁷ . Thus, c = (1/√(1.11 × 10⁻¹⁷)) ≈ 3 × 10⁸ m/s . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 3 × 10⁸ m/s, illustrating EM wave transverse nature and Maxwell's displacement current concep
Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment
**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. Using v λ = c , we have v = (c/λ) = (3 × 10⁸/20) = 1.5 × 10⁷ Hz . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.5 × 10⁷ Hz, illustrating EM wave transverse
Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves
