PHYSICS PAPER 2 6 MARKERS
QUESTIONS WITH COMPLETE
SOLUTIONS 2025.
Polarisation - ANSWER: - Light from the source is unpolarised Or light from source has
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
oscillations in all planes.
rr rr rr rr
- Intensity is reduced to 1⁄2 by a filter
rr rr rr rr rr rr rr rr
- By transmitting the parallel components and absorbing the perpendicular components.
rr rr rr rr rr rr rr rr rr rr
- At 0° / 180° filter 2 aligned with filter 1 so all light through filter 1 passes through filter 2
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
- As filter 2 is rotated only the component of the light from filter 1 in the plane of filter 2 is
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
allowed through, so the intensity reduces.
rr rr rr rr rr rr
- At 90°, all light is absorbed because their planes (of polarisation) are at right angles.
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
photoelectric effect - ANSWER: - photon energy E = hf rr rr rr rr rr rr rr rr rr
- photon energy must be greater than work function (of metal) for photon to provide
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
enough energy for photoemission
rr rr rr rr
- Higher frequency photons have more energy
rr rr rr rr rr rr
- one photon interacts with one electron
rr rr rr rr rr rr
- more electrons are emitted in a given time (so the charge is lost more quickly)
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
- Photoelectrons emitted instantaneously when radiation incident on surface
rr rr rr rr rr rr rr rr
- There is no photoemission below the threshold frequency
rr rr rr rr rr rr rr rr
- The maximum ke of the photoelectrons is independent of the intensity of the incident
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
radiation
rr
- The rate of photoemission is proportional to the intensity of the incident radiation
rr rr rr rr rr rr rr rr rr rr rr rr rr
- With waves, energy can be supplied to the electron
rr rr rr rr rr rr rr rr rr
continuously or with waves, energy can 'build up' rr rr rr rr rr rr rr
Oscilations - ANSWER: - The pendulums have the same length, so they have the same
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
time period/frequency
rr rr
- The first pendulum causes forced oscillations of the second pendulum
rr rr rr rr rr rr rr rr rr rr
- The driving frequency equals the natural frequency
rr rr rr rr rr rr rr
- Resonance occurs, so there is maximum transfer of energy so the amplitude increases
rr rr rr rr rr rr rr rr rr rr rr rr rr
until all energy is transferred
rr rr rr rr rr
- The second pendulum then acts as a driver for the first pendulum; or the process
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
repeats with energy transfer from B to A
rr rr rr rr rr rr rr rr
- When the lengths differ the driving frequency is not the natural frequency of the second
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
pendulum so little energy transfer occurs
rr rr rr rr rr rr
- 720 Hz is the natural frequency of the bowl
rr rr rr rr rr rr rr rr rr
- The generator/hand causes forced/driven oscillations
rr rr rr rr rr
QUESTIONS WITH COMPLETE
SOLUTIONS 2025.
Polarisation - ANSWER: - Light from the source is unpolarised Or light from source has
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
oscillations in all planes.
rr rr rr rr
- Intensity is reduced to 1⁄2 by a filter
rr rr rr rr rr rr rr rr
- By transmitting the parallel components and absorbing the perpendicular components.
rr rr rr rr rr rr rr rr rr rr
- At 0° / 180° filter 2 aligned with filter 1 so all light through filter 1 passes through filter 2
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
- As filter 2 is rotated only the component of the light from filter 1 in the plane of filter 2 is
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
allowed through, so the intensity reduces.
rr rr rr rr rr rr
- At 90°, all light is absorbed because their planes (of polarisation) are at right angles.
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
photoelectric effect - ANSWER: - photon energy E = hf rr rr rr rr rr rr rr rr rr
- photon energy must be greater than work function (of metal) for photon to provide
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
enough energy for photoemission
rr rr rr rr
- Higher frequency photons have more energy
rr rr rr rr rr rr
- one photon interacts with one electron
rr rr rr rr rr rr
- more electrons are emitted in a given time (so the charge is lost more quickly)
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
- Photoelectrons emitted instantaneously when radiation incident on surface
rr rr rr rr rr rr rr rr
- There is no photoemission below the threshold frequency
rr rr rr rr rr rr rr rr
- The maximum ke of the photoelectrons is independent of the intensity of the incident
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
radiation
rr
- The rate of photoemission is proportional to the intensity of the incident radiation
rr rr rr rr rr rr rr rr rr rr rr rr rr
- With waves, energy can be supplied to the electron
rr rr rr rr rr rr rr rr rr
continuously or with waves, energy can 'build up' rr rr rr rr rr rr rr
Oscilations - ANSWER: - The pendulums have the same length, so they have the same
rr rr rr rr rr rr rr rr rr rr rr rr rr rr
time period/frequency
rr rr
- The first pendulum causes forced oscillations of the second pendulum
rr rr rr rr rr rr rr rr rr rr
- The driving frequency equals the natural frequency
rr rr rr rr rr rr rr
- Resonance occurs, so there is maximum transfer of energy so the amplitude increases
rr rr rr rr rr rr rr rr rr rr rr rr rr
until all energy is transferred
rr rr rr rr rr
- The second pendulum then acts as a driver for the first pendulum; or the process
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
repeats with energy transfer from B to A
rr rr rr rr rr rr rr rr
- When the lengths differ the driving frequency is not the natural frequency of the second
rr rr rr rr rr rr rr rr rr rr rr rr rr rr rr
pendulum so little energy transfer occurs
rr rr rr rr rr rr
- 720 Hz is the natural frequency of the bowl
rr rr rr rr rr rr rr rr rr
- The generator/hand causes forced/driven oscillations
rr rr rr rr rr
