28 11 Light from a laser pointer was passed through a diffraction grating. The light was perpendicular to the diffraction grating as shown. A diffraction pattern was produced on a screen. laser pointer screen x θ1 D diffraction grating The distance between the first order maximum and the central maximum of the diffraction pattern was x. The distance between the diffraction grating and the screen was D. (a) Distance x was measured to be 0.500 m with a metre rule. The wavelength of light λ1 from the laser pointer was 650 nm. The laser pointer was replaced with one that produced light of a different wavelength. The new distance x was measured to be 0.400 m. D = 1.45 m Calculate the wavelength λ2 of the light emitted by the replacement laser pointer. (5) ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................... λ2 = ....................................................... 29 (b) Explain one modification to this method that would decrease the uncertainty in the calculated value of λ2. 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(c) In another experiment, the light from the laser pointer was not quite perpendicular to the screen. Explain how this would change the diffraction pattern produced on the screen. 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(Total for Question 11 = 10 marks) TOTAL FOR PAPER = 120 MARKS 30 List of data, formulae and relationships Acceleration of free fall g = 9.81 m s−2 (close to Earth’s surface) Boltzmann constant k = 1.38 × 10−23 J K−1 Coulomb law constant k = 1 4πε0 = 8.99 × 109 N m2 C−2 Electron charge e = −1.60 × 10−19 C Electron mass me = 9.11 × 10−31 kg Electronvolt 1 eV = 1.60 × 10−19 J Gravitational constant G = 6.67 × 10−11 N m2 kg−2 Gravitational field strength g = 9.81 N kg−1 (close to Earth’s surface) Permittivity of free space ε0 = 8.85 × 10−12 F m−1 Planck constant h = 6.63 × 10−34 J s Proton mass mp = 1.67 × 10−27 kg Speed of light in a vacuum c = 3.00 × 108 m s−1 Stefan-Boltzmann constant σ = 5.67 × 10−8 W m−2 K−4 Unified atomic mass unit u = 1.66 × 10−27 kg Mechanics Kinematic equations of motion s =  (u + v)t 2 v = u + at s = ut + 1 2 at2 v2 = u2 + 2as Forces ∑F = ma g =  F m W = mg moment of force = Fx Momentum p = mv Work, energy and power ΔW = FΔs Ek = 1 2 mv2 ΔEgrav = mgΔh P = E t P =  W t efficiency = useful energy output total energy input efficiency =  useful power output total power input 31 Electric circuits Potential difference V =  W Q Resistance R = V I Electrical power and energy P = VI P = I 2R P = V 2 R W = VIt Resistivity R =  ρl A Current I = ΔQ Δt I = nqvA Materials Density ρ =  m V Stokes’ law F = 6πηrv Hooke’s law ΔF = kΔ x Young modulus Stress σ =  F A Strain ε = Δ x x E =  σ ε Elastic strain energy ΔEel = 1 2 FΔ x Waves and particle nature of light Wave speed v = f λ Speed of a transverse wave on a string v = T μ Intensity of radiation I =  P A Power of a lens P =  1 f P = P1 + P2 + P3 + … Thin lens equation 1 u  +  1 v  =  1 f Magnification for a lens m =  image height object height =  v u Diffraction grating nλ = d sin θ Refractive index n1 sin θ1 = n2 sin θ2 n = c v Critical angle sin C = 1 n Photon model E = h f Einstein’s photoelectric equation hf = ϕ + 1 2 mv2 max de Broglie wavelength λ = h p 32 Further mechanics Impulse FΔt = Δp Kinetic energy of a non-relativistic particle Ek =  p2 2m Motion in a circle v = ωr T = 2π ω F = ma =  mv2 r a = v2 r a = rω2 F = mrω2 Fields Coulomb’s law F =  Q1Q2 4πε0r2 Electric field strength E =  F Q E = Q 4πε0r2 E =  V d Electric potential V = Q 4πε0r Capacitance C = Q V Energy stored in a capacitor W = 1 2 QV W = 1 2 CV 2 W = 1 2 Q 2 C Capacitor discharge Q = Q0e−t/RC I = I0e−t/RC V = V0e−t/RC ln Q = ln Q0 −  t RC ln I = ln I0 −  t RC ln V = ln V0 −  t RC In a magnetic field F = BIl sin θ F = Bqv sin θ Faraday’s and Lenz’s laws E =  −d(Nϕ) dt Root-mean-square values Vr ms =  V0 √2 Ir ms =  I0 √2 33 Nuclear and particle physics In a magnetic field r =  p BQ Thermodynamics Heating ΔE = mcΔθ ΔE = LΔm Molecular kinetic theory 1 2 mác2ñ = 3 2 kT pV = 1 3 Nmác2ñ Ideal gas equation pV = NkT Stefan-Boltzmann law L = σAT 4 L = 4πr2σT 4 Wien’s law λmaxT = 2.898 × 10−3 m K Space Intensity I =  L 4πd 2 Redshift of electromagnetic radiation z = Δλ λ  ≈ Δf f  ≈ v c Cosmological expansion v = H0d Nuclear radiation Mass-energy ΔE = c2Δm Radioactive decay A = λN dN dt  = −λN λ =  ln 2 t½ N = N0 e−λt A = A0 e−λt Gravitational fields Gravitational force F = Gm1m2 r 2 Gravitational field strength g = Gm r 2 Gravitational potential Vgrav = −Gm r Oscillations Simple harmonic motion F = −k x a = −ω2x x = A cos ωt v = −Aω sin ωt a = ‒Aω2 cos ωt T = 1 f  =  2π ω ω = 2π f Simple harmonic oscillator T = 2π m k T = 2π l g 34 BLANK PAGE 35 BLANK PAGE 36 BLANK PAGE