light and wave particle duality - turning points

Question 1

summarise how newton explained refraction using his corpuscular theory

Answer 1

• corpuscles of light travel in straight lines and are attracted by a force into the gas
• so they travel at a faster speed in the denser medium (the glass)
• the attraction only affects motion at the boundary
• the component of velocity perpendicular to the boundary increases, but the component parallel to the boundary stays unchanged

Question 2

what is newtons corpuscular theory of light

Answer 2

• he imagined a light ray as a stream of tiny particles
• when light is reflected by a plane mirror, Newton said the corpuscles bounce off the mirror without the loss of the speed
• the normal component of velocity is reversed and the parallel component is unchanged
• since the magnitude of the normal and parallel components of velocity are unchanged on reflection, it can be shown that the angle of reflection is equal to the angle of incidence

Question 3

what did newton say happened to the corpuscles as they entered a more dense medium

Answer 3

• the stronger the attraction the denser medium produces on the corpuscles, a net force and accelerates towards it

Question 4

how did newton explain reflection

Answer 4

• they bounce off the boundary and reflect the incident corpuscles path
• the reflection was due to a force that pushed the particles away from the surface

Question 5

if newtons theory was true, describe what the edges of the shadows would be like

Answer 5

shadows formed would be sharp (clear edges as no diffraction

Question 6

what was huygens wave theory or light

Answer 6

• every point on a wavefront may be considered to be a point source of secondary wavelets that spread out in the forward direction at the speed of the wave
• the new wavefront is the surface that is tangential to all these secondary wavelet
• every point on the wavefront acts as a secondary source of wavelets
• (light waves travelled slower in a transparent substance than in air)

Question 7

what did huygens wave theory predict about the speed of light in air

Answer 7

• speed in glass travelled slower than the speed in air
• diffraction and interference fringes due to constructive and destructive interference

Question 8

why was newtons theory preferred over huygens

Answer 8

• at the time, the speed of light could not be measure in air or glass, so there was no experimental evidence
• newton had a much stronger scientific reputation than huygens
• the wave theory of light was considered in terms of longitudinal waves so could not explain polarisation of light

Question 9

what was observed in youngs double slit

Answer 9

• light is a wave
• many bright and dark fringes observed, due to different distances (path difference = phase difference) - LIGHT IS A WAVE AND COHERENT
• every point on a wavefront acts as a source of secondary wavelets
• when the wave reaches slit, each point at slit produces secondary wavelets
• wavelets diffract and overlap on screen (superpose)
• path difference due to different due to different distances between a point on the screen and the two slits
• path difference introduces phase differences
• bright fringes from where path difference is whole number of wavelengths/ waves arrive in phase
• dark fringes where path difference is odd number of half wavelengths/ waves arrive in antiphase

Question 10

what would have been seen in youngs double slit if light was a particle

Answer 10

• predicts 2 bright lines in the screen - corpuscles travel in straight lines and the edges of the shadows would be sharp

Question 11

how can the spacing of each fringe be changed

Answer 11

• move the slits further from each other, the smaller the fringe spacing is
• narrower slits, the greater amount of diffraction therefore, more fringes observed using widely spaced slits compared using widely spaced slits compared with closely spaced slits of the same width

Question 12

how did fizeau measure the speed of light

Answer 12

• passed a beam of light through the gap between two cog teeth to a reflector about 9km away, the cog was rotated at exactly the right speed so that the reflected beam was blocked by the next cog tooth
• using the frequency of rotation and the number of gaps Fizeau was able to calculate the time taken for the light to travel to the reflector and back to the cog
• so fizeau could calculate the time taken and the distance travelled to calculate the speed of light

Question 13

key points from fizeaus experiment

Answer 13

• s = 2D/t (travelled there are back)
• goes through the gap, comes back and gets blocked by the adjacent tooth
• frequency of the cog gets increased from 0 until the reflected light could not been seen
• black teeth stops unwanted reflections
• monochromatic light ensures parallel beams
• showed the speed of light in water is less then in air

Question 14

speed of light equation

Answer 14

speed = 4DNf

D = distance between cog and mirror
N = number of teeth
f = lowest frequency that blocks the light

Question 15

describe Maxwell’s theory of em waves

Answer 15

a changing current in a wire creates waves of changing electric and magnetic fields that radiate from the wire

Question 16

how are the oscillations of the electric and magnetic fields related

Answer 16

• both fields create each other
• therefore, they oscillate and create electromagnetic waves

Question 17

what did maxwell predict

Answer 17

• that a change to electric and magnetic fields would create EM waves, radiating out from the source of disturbance
• he predicted there would be a spectrum of EM waves, travelling at the same speed with different frequencies

Question 18

how was what maxwell predicted and fizeaus experiment related

Answer 18

maxwells value of c was close to the value measure by fizeau, so this provided strong evidence that light, as well as uv and infrared radiation beyond the visible spectrum, is an em wave

Question 19

properties of em waves

Answer 19

• in phase
• perpendicular to each other
• perpendicular to the direction of energy propagation

Question 20

what is e0

Answer 20

related to the electric field strength due to a charged object in free space

Question 21

what is u0

Answer 21

relates to the magnetic flux density due to a current carrying wire in free space

Question 22

what did hertz use to produce and detect radio waves

Answer 22

• the radio waves produced by the high voltage spark gap transmitter spread out from the spark gap pass through the detector loop
• the waves passing through the detector loop cause a voltage to be induced in the detector which makes spars jump across the detector gap

Question 23

how could the waves be reflected in hertz experiment

Answer 23

• by placing a metal sheet between the transmitter and the detector
• a concave metal sheet placed behind the transmitter made the detector sparks stronger because it reflected radio waves travelling away from the detector

Question 24

in the hertz experiment, what was the evidence that these waves had a magnetic nature

Answer 24

• the induced voltage in the detector loop is due to the oscillating magnetic field of the radio waves
• as the waves travel across the loop, the oscillating magnetic field causes oscillating changes in the magnetic flux passing through the loop, which causes an alternating pd to be induced in the loop

Question 25

describe how the speed of the radio waves can be found

Answer 25

• hertz was able to calculate the frequency of the radio waves from the electrical characteristics of the circuit in which the spark gap transmitter was
• the charge created by a spark oscillates back and forth across the gap, causing radio waves to be emitted at the same frequency as the spark oscillations
• c = f(wavelength)

Question 26

how could a spark be induced in a dipole detector and what happened when the detector was rotated

Answer 26

• the reflector focuses the radio waves onto the rods such that the oscillating electric field of the radio waves creates an alternating pd between the 2 rods, causing sparks at the spark gap
• when the detector was rotated, the pd was decreased and became 0 at an angle of rotation at 90 degrees from the initial position, this was because the radio waves were polarised, therefore, could not produce a pd between the rods

Question 27

what is the uv catastrophe and black body radiation

Answer 27

black body:
- a body that is a perfect absorber of radiation (100% of all wavelength that hit it are absorbed)

eg. star and a small hole in the side of a furnace

black body radiation:
- em radiation emitted from a black body
- it consists of a continuous range of wavelengths
- the position of the spectrum’s peak and the intensity at each wavelength depends on the temp of the black body

uv catastrophe (classical theory: radiation as a wave):
- predicted intensity is similar to measurements as long wavelengths

• predicted the intensity would tend to infinity as the wavelength got smaller and smaller (and hence disagree with what is observed for short wavelengths)
• this is impossible - infinite energy would be needed

Question 28

what was plancks solution to the uv catastrophe

Answer 28

em radiation is emitted in quanta

energy of a quantum is related to the frequency of the radiation by E = hf

he suggested that the energy of the vibrating atoms that give rise to the radiation is quantised and is proportional to its frequency

Question 29

what are the key observations of the photoelectric effects

Answer 29

• no photoelectric emission if incident light below threshold frequency
• emission is instantaneous (as soon as light is incident on the metal)
• have a range of kinetic energies from 0 to max (depending on metal type and freq of light)
• number of photoelctric electrons per second is proportional to intensity