topic 5

Question 1

amplitude

Answer 1

the maximum magnitude of the displacement

Question 2

frequency

Answer 2

number of cycles (vibrations) per second passing a given point

Question 3

period

Answer 3

time taken for whole cycle to complete

Question 4

wavelength

Answer 4

length of one whole wave cycle, e.g. from crest to crest or trough to trough

Question 5

longitudinal waves

Answer 5

waves oscillate parallel to direction of energy transfer.
- made up of compressions and rarefractions

Question 6

pressure variation at rarefraction vs compression stages

Answer 6

pressure: increased at compression and decreased at rarefraction point

Question 7

displacement of particles at rarefraction vs compression stage

Answer 7

rarefraction: neighbouring particles move away from eachother
compressiom: neighbouring particles move towards a point

Question 8

transverse waves

Answer 8

waves oscillate perpendicular to direction of energy transfer.

all EM waves are transverse (speed = 3x10^8)

Question 9

wavefront

Answer 9

surface which is used to represent the points of a wave which have the same phase

Question 10

coherence

Answer 10

when two light sources have the same frequency and wavelength and fixed phase difference

Question 11

path difference

Answer 11

difference in distant travelled by two waves

Question 12

superposition

Answer 12

When two or more waves cross at a point, the displacement at that point is equal to the sum of the displacements of the individual waves

Question 13

interference

Answer 13

when two or more waves superpose with each other

Question 14

phase

Answer 14

measurement of the position of a certain point along the wave cycle

Question 15

in phase

Answer 15

when two waves are at the same point of the wave cycle, hence they have the same frequency and wavelength (coherent) and the phase difference is a multiple of 360° (2π radians)

Question 16

out of phase

Answer 16

when two waves have the same frequency and wavelength (coherent) and their phase difference is an odd integer multiple of 180° (π radians)

Question 17

stationary/standing wave

Answer 17

formed from the superposition of 2 progressive waves, travelling in opposite directions in the same plane, with the same frequency, wavelength and amplitude

Question 18

how standing wave is formed

Answer 18

• in phase: constructive interference occurs so anti-nodes formed (area of max displacement)
• out of phase: destructive interference occurs so nodes formed (regions of no displacement)

Question 19

intensity

Answer 19

power per unit area
I = P/A

Question 20

refractive index

Answer 20

• measures how much it slows down light passing through it
  n=c/v
• a material with a high refractive index can be known as being more optically dense

Question 21

refraction

Answer 21

when a wave enters a medium and causing it to change direction, either towards or away from the normal which causes the wave to either slow down or speed up

Question 22

Snell’s law

Answer 22

n1sinθ1 = n2sinθ2
- n1: refractive index of material 1
- n2: refractive index of material 2
- θ1: angle of incidence of the ray in material 1
- θ2: angle of incidence in material 2

Question 23

critical angle

Answer 23

sinC = 1/n

Question 24

reflection

Answer 24

the wave bounces back when it hits a boundary

Question 25

total internal reflection (TIR)

Answer 25

when the angle of incidence is greater than the critical angle and the incident refractive index (n1) is greater than the refractive index of the material at boundary (n2)

Question 26

measuring refractive index

Question 27

converging lenses

Answer 27

- bring light rays together - cause parallel light rays to move closer together - focal length, f, is the distance between lens axis and focal plane - f is positive for converging lens because it is in front of the lens

Question 28

diverging lens

Answer 28

- cave inwards - cause parallel light rays to move apart - focal length, f, is distance between the lens axis and principal focus - f is negative for diverging lens because it is behind the lens

Question 29

real image

Answer 29

formed when light rays from a point on an object pass through another point in space can be projected onto a screen

Question 30

virtual image

Answer 30

formed when light rays from a point on an object appear to have come from another point in space cannot be projected onto a screen

Question 31

what does converging lens form

Answer 31

converging lens can form both real and virtual images. If the object is further than the focal length, the image is real. If the object is closer, the image is virtual

Question 32

what image does a diverging lens form

Answer 32

always form a virtual image - the position of object in relation to focal length will not affect the type of image formed

Question 33

lens equation

Answer 33

1/f = 1/u + 1/v

Question 34

power

Answer 34

tells you how much a lens bends light. can be calculated by P = 1/f

Question 35

diffraction

Answer 35

the way that waves spread out as they pass through a narrow gap or go around obstacles

Question 36

how does diffraction become more noticable

Answer 36

as the gap decreases, the diffraction becomes more noticeable. if the gap becomes too small the water waves cannot pass through it anymore

Question 37

use Huygen's construction to explain what happens to a wave when it meets a slit or an obstacle

Answer 37

the secondary wavelets that pass through the slit produce the curve of the new wavefront emerging from the slit

Question 38

measuring the wavelength of light using a diffraction grating

Answer 38

1) position a laser in front of a diffraction grating so that the light travels through the grating and creates an interference pattern on a flat wall or screen a few metres away 2) measure the distance, D, between the diffraction grating and the wall 3) measure the distance, x, between the zero order maximum and the 1st order maximum for both sides and take an avg of the two readings 4) repeat the measurements for more order lines to find an avg wavelength

Question 39

wavelength of light

Answer 39

nλ = dsinθ

Question 40

general conclusions from nλ = dsinθ

Answer 40

1) if λ is bigger, sinθ is bigger, so θ is bigger. This means the larger the wavelength, the more the pattern will spread out 2) if d is bigger, sinθ is smaller. This means that the coarser the grating, the less the pattern will spread out 3) values of sinθ greater than 1 are impossible

Question 41

plane polarisation

Answer 41

wave only oscillates in one direction

Question 42

polarising filter

Answer 42

- when filters are aligned all of the light that passes through 1st filter also passes through the 2nd - rotating the 2nd filter the amount of light that passes through varies - as 2nd filter rotates, less light will pass - at 90 degrees no light will pass through

Question 43

light - wave or photon? why?

Answer 43

- both - light produces interference and diffraction patterns - alternating bands of dark and light. this can only be explained using waves. - photoelectric effect is only explainable if light behaved as a particle - known as a photon

Question 44

what is a photon

Answer 44

a quantum of EM radiation - Einstein suggested EM waves can only exist in discrete packets - photons - photons have no charge

Question 45

equation that relates photon energy to wave frequency

Answer 45

E = hf = hc λ

Question 46

how do electrons move down energy levels

Answer 46

by emitting a photon

Question 47

electronvolt - eV

Answer 47

1.60 x 10^-19 J

Question 48

how do electrons move up energy levels

Answer 48

by absorbing a photon with the exact energy difference between the two levels

Question 49

hot gases - atomic line spectra

Answer 49

produce line emission spectra - atoms become excited when heated so move to higher energy levels - as electrons within the atom fall back down, they emit photons - if you split light from a hot gas within a diffraction grating, a line emission spectra is produced (black one with thin coloured lines) -

Question 50

cool gases - atomic line spectra

Answer 50

produce line absorption spectra - at low temps, most electrons in the gas will be in their ground states - when white light passes through a cool gas and then a diffraction grating, a line absorption spectra is produced (colour one with thin black lines) - photons of correct wavelength are absorbed by electrons to excite them to higher energy levels

Question 51

how does the photoelectric effect provide evidence for the particle nature of EM radiation - photoelectric effect

Answer 51

- free electrons on the surface of a metal absorb energy from light - if an electron absorbs enough energy the bonds holding it to the metal break and electron is released

Question 52

first conclusion drawn from the photoelectric effect

Answer 52

1. for a given metal no photoelectrons are emitted if the radiation has a frequency below a certain value - threshold frequency

Question 53

second conclusion drawn from the photoelectric effect

Answer 53

the photoelectrons are emitted with a variety of kinetic energies ranging from zero to a max. value. This value of max KE increases with the frequency of the radiation and is unaffected by the intensity of the radiation

Question 54

third conclusion drawn from the photoelectric effect

Answer 54

the number of photoelectrons emitted per second is proportional to the intensity of the radiation

Question 55

threshold frequency

Answer 55

the minimum frequency of incident radiation below which the photoelectric emission is not possible completely

Question 56

equation linking threshold frequency and work function

Answer 56

hf = φ + ½mv^2

Question 57

light as a wave - wave model

Answer 57

newton's theory suggested light could interfere and diffract (both wave like properties)

Question 58

light as a particle - photon model

Answer 58

after 100 years, Einstein explained that beam of light was a series of particle-like photons if a photon is a discrete bundle of energy, then it can interact with an electron all the energy in the photon is given to one electron

Question 59

De Broglie equation

Answer 59

λ = h / p

Question 60

how does a pulse-echo technique provide information about the position of an object

Answer 60

you send out a sound wave to an object an unknown distance awa

Question 61

using ultrasound waves in scans

Answer 61

- frequency of ultrasound waves is too high for humans to hear - ultrasound is directed into body via a transducer - gel is applied to skin so there is no air between transducer and skin - when the waves reach an interface inside body, some of them are reflected - computer detects how long it takes for reflected waves to return

Question 62

why are shorter pulses used

Answer 62

produces clearer images - pulses of ultrasound transmitted must be very short so reflections from nearby interfaces don't reach transducer - gap between pulses must be long so all reflected waves from one pulse return to the transducer before the next pulse is transmitted

Question 63

resonant frequency

Answer 63

minimum