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182 public questions tagged with this topic.

Superposition, Resultant Amplitude and Intensity

In a single-slit diffraction pattern, what happens to the central maximum’s width if the wavelength is reduced to one-th

**Superposition principle** resultant displacement sum of individual, for two coherent waves amplitude a each, resultant amplitude A = √(a² + a² +2a² cosφ)=2a|cos(φ/2)|, phase difference φ, path difference Δ = (φ/2π)λ, for φ=π/2 A=√2 a, for φ=6π cos3π=-1? Actually φ=6π cos3π? A=2a|cos3π|=2a, for φ=4π A=2a, intensity I ∝ A², maximum I_max=4I₀ when φ=0, I=2I₀(1+cosφ)=4I₀ cos²(φ/2). Angular width 2θ = (2λ/a) . If λ is reduced to one-third, 2θ reduces to one-third. 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 cos(φ/2), calculation

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Diffraction - Single-Slit and Central Maximum

What is the angular position of the first secondary maximum in a single-slit diffraction pattern if the slit width is \(

**Single-slit pattern** intensity I = I₀ (sinα/α)², α=π a sinθ/λ, central maximum at α=0, minima at α=nπ, so a sinθ=nλ, width increases with λ and D decreases with a, for a=15 μm λ=750 nm first minimum sinθ=750/15000=0.05 θ≈2.87°, angular width of central maximum 2θ≈5.74°. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the first secondary maximum, n = 1 . λ = 4.5 × 10⁻⁷ m , a = 9.0 × 10⁻⁶ m . sin θ = ((1 + (1/2)) × 4.5 × 10⁻⁷/9.0 × 10⁻⁶) = (1.5 × 4.5 × 10⁻⁷/9.0 × 10⁻⁶) = 0.075 , θ = sin⁻¹(0.075) ≈ 4.3°

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

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Diffraction - Single-Slit and Central Maximum

What is the condition for the fourth secondary maximum in a single-slit diffraction pattern?

**Single-slit pattern** intensity I = I₀ (sinα/α)², α=π a sinθ/λ, central maximum at α=0, minima at α=nπ, so a sinθ=nλ, width increases with λ and D decreases with a, for a=15 μm λ=750 nm first minimum sinθ=750/15000=0.05 θ≈2.87°, angular width of central maximum 2θ≈5.74°. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the fourth secondary maximum, n = 4 , θ ≈ (9λ/2a) . 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 cos(φ/2), calculation gives θ = (9λ/2a),

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

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Interference - Double-Slit and Coherence

What happens to the fringe width in a double-slit experiment if the slit separation is reduced to one-third?

**Interference** occurs when two coherent waves superpose, path difference Δ = d sinθ, bright fringe when Δ = n λ, n integer, seventh bright Δ=7λ, dark fringe Δ = (2n-1)λ/2, second dark Δ=3λ/2, third dark 5λ/2, sixth dark 11λ/2, distance of bright fringe from central y = n λ D/d, D screen distance, d slit separation, for λ=650 nm d=0.5 mm D=1.0 m fifth dark? Actually dark y=(2n-1)λ D/(2d). Fringe width β = (λ D/d) . If d is reduced to (d/3) , β increases to 3β , i.e., triples. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

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Diffraction - Single-Slit and Central Maximum

In a single-slit diffraction pattern, what happens to the intensity of secondary maxima as their order increases?

**Diffraction bending** property of light waves causes bending around corners, width of central maximum inversely proportional to slit width, intensity of secondary maxima decreases with order because less constructive interference, angular position of minima θ_n = n λ/a, n=±1,±2..., second minimum n=2, third n=3, condition for third secondary maximum approx a sinθ = (2n+1)λ/2. The intensity of secondary maxima decreases with increasing order, becoming weaker away from the central maximum. 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 cos(φ

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

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Photoelectric Effect - Work Function and Threshold

Which of the following observations contradicts the wave theory of light in the photoelectric effect?

**Classical wave theory fails** to explain instant emission, threshold existence, K_max dependence on frequency not intensity, saturation current dependence on intensity. Observations: K_max independent of intensity, exists threshold frequency, emission instantaneous, all explained by photon model E = h f, one photon ejects one electron, energy conservation h f = Φ + K_max. The wave theory predicts energy absorption over time, but photoelectric emission is instantaneous (within 10⁻⁹ s ), supporting the photon model. Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√

Ref: NCERT > Physics Book > Dual Nature of Matter > Photoelectric Effect - Work Function and Threshold

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Nuclear Size, Density and Structure

Which factor primarily determines the nuclear radius?

**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. The nuclear radius is given by R = R₀ A¹/³ , where A (mass number) is the key factor determining the size, as the radius scales with the cube root of the number of nucleons. 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

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure

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Mass Defect, Binding Energy and Binding Energy per Nucleon

Which process is responsible for the energy release in an atomic bomb?

**Energy equivalent** E= m c², 0.005 kg matter E=0.005×9×10¹⁶=4.5×10¹⁴ J, 0.01 kg 9×10¹⁴ J, mass defect 0.1 u => BE=0.1×931.5=93.15 MeV, mass defect from BE 149.04 MeV => Δm=149.04/931.5=0.16 u, BE per nucleon 8.5 MeV A=20 total BE=170 MeV. The energy in an atomic bomb comes from uncontrolled nuclear fission, where a heavy nucleus splits into lighter fragments, releasing energy due to increased binding energy per nucleon. 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 Nuclear fission, consistent with Bohr model and nuclear bind

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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Nuclear Size, Density and Structure

A nucleus has a radius of \( 3.6 \times 10^{-15} \, \text{m} \). What is its approximate mass number? (Given \( R_0 = 1.

**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

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the potential energy of an electron in the \( n = 5 \) state of a hydrogen atom? (Use \( E_n = -\frac{13.6}{n^2}

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. E₅ = -0.544 eV , K = 0.544 eV . U = -2K = -2 × 0.544 = -1.088 eV ≈ -1.09 eV . 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 -1.09 eV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Lenz's Law, Eddy Currents and Applications

A coil of self-inductance 0.6 H has its current decreased from 4 A to 1 A in 0.15 s. What is the magnitude of the induce

**Solenoid second coil** experiences emf only when current in solenoid changes because flux linkage changes only then, steady current gives constant Φ, dΦ/dt=0, no emf, when current changes, dΦ/dt ≠0, emf induced, illustrating Faraday's law requirement of changing flux. ε = L (Δ I/Δ t) . Δ I = 1 - 4 = -3 A , Δ t = 0.15 s . ε = 0.6 × (-3/0.15) = 0.6 × (-20) = -12 V , magnitude = 12 V. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L =

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

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Energy Stored in Inductor and Magnetic Energy

A solenoid of 700 turns per meter and area 0.015 m² has a current drop from 8 A to 5 A in 0.3 s. What is the self-induce

**Magnetic energy density** u = B²/(2μ₀), for B=0.5 T, u=0.25/(2×4π×10⁻⁷)=0.25/(2.513×10⁻⁶)=99471 J/m³, large, but volume small, total energy moderate. Inductor stores energy in field, released when current interrupted causing spark. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (700)² × 0.015 × 1 = 0.00923 H . ε = L (Δ I/Δ t) = 0.00923 × (5 - 8/0.3) = 0.00923 × (-10) = 0.0923 V ≈ 0.092 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

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