Vector Equations - Higher Physics - Solved Past Paper, Exams of Physics

Download Vector Equations - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! SOLUTION to PHYS1231 FINAL EXAM 82 2010 Total forZ Questions: 60 Marks Question 5. EM Waves (Marks 22) (a) A laser emits sinusoidal electromagnetic (EM) waves that travel in the negative x-direction. The EM waves of wavelength A = 10,600 nm are cmitted from the laser into vacuum with E field parallel to the z-axis; the E field amplitude is 1.5x10° Vm". Write vector equations for E and B as a function of time and position. (12 marks) Solution ‘The wave is travelling in the direction -i. The equations for E and B have the form E(x,t) = KE, cos(kx + at) Bx.t) = JBow cos(kx + at) where the unit vectors k,j give the orientation of E and B. 6 and where E,,., = 1.5x10° Vm" and B,., = Bass 15x10" =5.0x10° T c 30x10 The wave number is k = ox an =5,93x10° radm™ A 10.6x10% m The angular frequency is w =ck = (3x10°)(5 93x10") rads™ =1.78x10"* rads We obtain E(x,t) = k{1 5x10*)cos[5.93x10°x +1.78x10t] and B(x,t) = j{5.0x10")cosf5.93x10°x +1.78x10"t] (b) Ina CD ROM drive light from a semiconductar diode laser having wavelength A = 780nm travels a distance 125 nm in a polycarbonate layer, Polycarbonate is a transparent medium of refractive index 1.58. Calculate, (i) the optical path length (2 marks) (ii) the wavelength of the light in the transparent medium (3 marks) (iii) the phase difference after travelling the distance 125nm with respect to light travelling the same distance in free space. (5 marks) Solution (i) the optical path length (0.p.1) is o.p.] = (physical distance travelled in medium) x (refractive index) = (125x107 m)(1 58) = 1.97x107 Gi) | The wavelength in the polycarbonate medium is A,, given by where A, is the free space (vacuum) wavelength. ii) The physical distance travelled is { =125x10%m The refractive indices are: Dyeaium = 1-58, Myscoum = 1-0 path difference = (Mean! — Myscuune) = (1.58}(125x10°) - (1.0)(125x10°) = 7.25x107 m Phase difference = 27(path difference/wavelength) -¥ = 20 = 0.58 radians Question’? (20 Marks) (a) A sodium atom emits a photon of wavelength 589.0 nm and energy 2.105 eV ina transition from an excited state to the ground state. The atom remains in the excited state for an average ‘lifetime’ 7 = 0.16 ns before the transition to the ground state. Calculate, ti) the uncertainty in the energy of the excited state (4 marks), (ii) the width (i.e. the spread in wavelength) of the line in the observed spectrum associated with this wransition. (5 marks} Solution . y 5 é fole: also conect fo @) AB =— = 66x10 4.110% se ACAT RE a Solution AIxl0* eV 2.105 eV c. Adm (1.95x10™)(589.0 nm) = 11x10 nm (ii) uncertainty in photon energy = =1.95x10* (b) Three materials have the energy band structures shown schematically in the diagram below representing, (1) a metal, (2) an n-type doped semiconductor and (3) an insulator. The shaded areas indicate occupied (by electrons) energy ranges. a) @) G) Bp= SeV [> 5.50V (i) For the metal shown in (1), find the Fermi velocity and the thermal velocity of the electrons at 300K. (4 marks) Gi) Find the wavelength of EM radiation that will cause a sharp increase in the electrical conductivity of material (2). (2 marks) Gii} By comparing the energy gap values for materials (2) and (3) state, with your reasoning, whether material (3) is expected to be transparent or opaque to visible light at room temperature. (The visible region of the EM spectrum spans the wavelength range A=400nm to A=700nm approx.) (5 marks) Solution (i) Fermi velocity x ODA 2k, |25.0x10°°V) m, Y 9.LxlO™ ke = 13x10° ms Vp Thermal velocity: 2(1.38x10 JK~)(300K) 9k =1x10°ms* (ii) hc _ (6.63x10)(3x10°) =1.1x10%m Gi) Solution Material (1) is a metal and will retain metallic conductivity at low temperature and as T tends to 0K. Material (2) is a doped semiconductor and will become an insulator as T approached OK as the free electrons in the conduction band ‘freeze out’.