Physics Unit 2

• Explain in terms of interference, phase and path difference, how the bright fringes arise.

  At centres of bright fringes:

  1. Path lengths from slits differ by n wl
  2. waves arrive in phase
  3. waves constructively interfere, forming bright fringes
  4. slits act as coherent sources
• Discuss whether you would choose Young’s fringe method, or diffraction grating method, if you wanted an accurate value for wavelength.

  1. More uncertainty with Young’s method
  2. smaller fringe separation therefore more difficult to measure, whereas grating beams are well spaced
• difference between progressive and stationary waves

  progressive: phase changes steadily with distance; transfer energy through medium; constant amplitudes

  stationary: phase reverses at nodes; do not transfer energy through medium; constantly changing amplitudes
• explain how stationary waves can be thought of as arising from progressive waves. [2]

  1. reflections at a fixed point give rise to two waves propagating in opposite directions
  2. same frequency and amplitude
  3. interference between these progressive waves gives stationary waves
  4. nodes + antinodes
• if one of slits covered, dark fringes become brighter. explain this observation. [2]

  1. dark fringes caused by destructive interference
  2. with one slit closed, light from other slit not cancelled
• advantage of choosing 2nd order, rather than 1st order, beam

  uncertainty in measuring angle makes lower uncertainty in d
• principle of superposition

  total displacement at any point is the vector sum of displacements of individual waves
• describe what is seen when a source of polarised light is viewed through a polarising filter which is rotated slowly through 360° [2]

  1. alternates gradually between light and dark
  2. 2 extinctions / minima (no light)
• explain why diffraction at slits is essential to produce interference fringes [1]

  so light from two slits can overlap
• distance from double slits to screen is increased to 7.5m. state two ways in which appearance of fringes on screen is changed. [2]

  1. increased fringe separation
  2. bright fringes become dimmer
• Explain why the reading on the coulombmeter is negative. [1]

  electrons transferred from [polythene] rod to [metal cap]
• State the sign of the charge acquired by the duster. Explain your reasoning. [2]

  positive

  electrons transferred from duster to rod
• Explain carefully, in terms of energy, the meanings of V, E and Ir. [4]

  V is energy delivered per unit charge to external circuit

  E is energy supplied per unit charge from chemical energy of cell

  Ir is energy wasted per coulomb in internal resistance
• Some materials exhibit the property of superconductivity under certain conditions.

  State what is meant by superconductivity.

  Explain the required conditions for the material to become superconducting. [3]

  material with zero resistivity

  resistivity decreases with temperature

  becomes superconducting when transition temperature reached
• State and explain what happens to the resistance of the cable when the embedded filaments of wire are made superconducting. [3]

  resistance decreases to zero

  copper still has resistance

  but this is in parallel with filaments which have 0 resistance

  hence total resistance is zero

  current goes through filaments
• Explain in terms of particles how electrical resistance arises in metal conductors. [3]

  free electrons

  collide

  with metal ions
• Hence suggest an explanation for your results to the experiment in part (a). [2]

  as temperature increases kinetic energy of metal ions increase

  causing more frequent collisions
• Explain why only a small percentage of the total current in the cable passes through the steel wire. [3]

  Al is a better conductor as has less resistance

  6 Al wires therefore 6 times as much CSA

  each Al carries [3x] as much current
• electron diffraction - how it illustrates wave behaviour [3]

  1. concentric
  2. circles
  3. this is a diffraction pattern
• electron diffraction- explain how the pattern arises [2]

  1. [] is crystalline
  2. crystals in [] act as diffraction grating
• why Ek max = hf - ø is not + [3]

  1. Ek max is the maximum KE energy of emitted electron
  2. ø is minimum energy for electron to escape
  3. what is left over of photon’s energy after the escape is its KE
• Ek max is unaffected by increasing light intensity. [2]

  1. Increasing intensity increases number of photons per second
  2. but doesn‘t change energy of individual photons
• what aspect of photo-electric emission is affected by light intensity [1]

  number of electrons emitted per second
• photoelectric experiment [4]

  1. increase p.d from zero
  2. until ammeter reads zero
  3. take voltmeter reading
  4. evaluate Ek max = eV

  
• explain how fringe pattern shows effects of diffraction at individual slits [2]

  1. without diffraction, no fringe pattern
  2. diffraction at single slits cause fading of bright fringes away from centre
• TOTAL number of beams

  n < d / wavelength
• potential divider derivation

  1. Vin = IR1 + IR2 = I (R1 + R2)
  2. Vin / Vout = I (R1 + R2) / IR2
  3. Vout = ( R1 + R2 / R2 ) * Vin
• photoelectric effect [2]

  1. emission of electrons from a metal surface
  2. when light of a short enough wavelength falls on it
• What aspect of photo-electric emission is affected by the light intensity? [1]

  number of emitted electrons per second
• photoelectric effect [4]

  1. increase pd from zero
  2. until ammeter reads zero
  3. take Voltmeter reading
  4. evaluate Ek = eV

  A) variable DC supply

  B) surface

  C) voltmeter

  D) microammeter

  E) light of frequency

  
• Explain why it is an advantage for this angle to be small if data are being transmitted. [2]

  1. smaller differences in distance travelled by light travelling different paths
  2. so less smearing of data

  
• Explain how this arrangement produces interference fringes on the screen. In your answer, explain why slit S should be narrow and why slits S1 and S2 act as coherent sources. [6]

  1. S acts as point source
  2. narrow = wider diffraction + provides coherent light sources at S1 and S2
  3. ensure both S1 and S2 are illuminated by same source giving same wavelength
  4. light diffracted at S1 and S2 + diffracted waves overlap
  5. path difference of n wl + in phase + constructive = bright
  6. path difference of n + 0.5 wl + antiphase + destructive = dark

  
• TIR conditions

  1. incidence > critical
  2. n1 > n2
• Without further calculations, explain how conservation of energy still applies to this collision. [2]

  loss of photon energy = gain in electron kinetic energy

  
• Explain, using the concept of stimulated emission, why more photons arrive back at M2 than are reflected from it. [2]

  1. Stimulated Emission gives 2 photons out for 1 photon in
  2. these photons go on to cause further stimulated emission
• Explain briefly, in terms of electrons, why there is a reading on the microammeter when the variable supply is set to give a pd of zero.

  [2]

  1. some electrons emitted from [metal] surface land on collecting electrode
  2. flow back to emitting surface via microammeter
• grating beams [2]

  1. flat, opaque screen
  2. with many parallel slits
• Explain, in terms of waves, why refraction occurs. Refer to the diagram below in your

  answer. [3]

  1. refraction is a change in direction of travel as waves change medium
  2. AB and CD are wavefronts
  3. AB goes to CD
  4. Waves travel more slowly in 2 so BD < AC
  5. direction of travel is normal to wavefronts

  
• calculate an upper limit to the number of pulses per second that could be sent through 1.2km of the fibre without overlap occurring. State one assumption you are making. [2]

  1. 1 / time taken for light to travel
  2. assume pulse duration is negligible
• path difference

  1. in phase= n wavelength
  2. antiphase = 1/2 wavelength
  3. slit separation = max path difference
  4. wavelength must be less than slit separation but greater than or equal to slit width