Particles & Waves Sub Unit Test

Question 1

What is interference?

Answer 1

Occurs when waves from two or more sources meet

Question 2

What is the resultant wave from inference determined by?

Answer 2

The amplitude and relative phase of the waves

Question 3

What happens when two wave meet in phase?

Answer 3

Constructive interference

Question 4

What does ‘in phase’ mean?

Answer 4

When the crests meet the crests and the troughs meet the troughs of another wave

Question 5

What happens when waves are constructive?

Answer 5

Form a singular wave of greater amplitude

Question 6

What happens when two waves out of phase meet?

Answer 6

Destructive interference

Question 7

What does ‘out of phase’ mean?

Answer 7

When crest meets trough

Question 8

What happens when waves are destructive?

Answer 8

The waves cancel each other out

Question 9

What does it mean for waves to be coherent?

Answer 9

Same frequency
Same wavelength
Same velocity

Question 10

What is a maxima?

Answer 10

Maxima are regions where two waves meet in phase and interfere constructively.
In terms of speakers, these are loud regions
In terms of light, these are the bright points

Question 11

What is a minima?

Answer 11

Regions where two waves meet out of phase and interfere destructively.
In terms of speakers, these are the quiet regions
In terms of light, these are the dull/blank regions

Question 12

What is path difference?

Answer 12

The extra distance travelled by one wave source compared to another source

Question 13

How do you work out the path difference for constructive interference?

Answer 13

p.d. = m x wavelength

where m = 0,1,2 (which is what point it is)

Question 14

How do you work out the path difference for destructive waves?

Answer 14

p.d. = (m + 1/2) x wavelength

where m = 0,1,2, 3

Question 15

How would you work out the wavelength of something when given the distance between two speakers and the maxima/minima?

Answer 15

Work out p.d. by distS2p -distS1p
Sub into p.d. = m x wavelength
Rearrange to find wavelength

Question 16

What is Young’s double slit experiment?

Answer 16

Passes light through a narrow slit then using two slits to divide the waves which produces bright and dark fringes on a screen

Question 17

In diffraction, what happens to the fringes when the distance between the two slits (2) increases?

Answer 17

Fringes are closer together

Question 18

What happens when the distance (D) between the slits and the screen increases?

Answer 18

The distance between the fringes increases

Question 19

What is a diffraction grating?

Answer 19

A series of narrow parallel slits etched onto glass

Question 20

What happens when the number of slits on a diffraction grating increase?

Answer 20

A series of bright sharp lines are produced

Question 21

What does m=0 mean?

Answer 21

There is no path difference

Question 22

What does m=1 mean?

Answer 22

There is a path difference is exactly on wavelength

pd = 1 x wavelength

Question 23

If the diffraction grating is given in lines per mm, how would you find d?

Answer 23

Convert into metres by dividing by 1000

d = 1/no. of lines per metre

Question 24

What happens when a wave moves from one medium to another?

Answer 24

Wave changes speed, can lead to a change in direction

Question 25

What is the change of direction when a wave moves from one medium to another referred to as?

Answer 25

Refraction

Question 26

What happens to the velocity, frequency and wavelength of a wave been it enters glass from air?

Answer 26

Velocity decreases Frequency remains unchanged Wavelength decreased

Question 27

When does the direction of a wave change in diffraction?

Answer 27

When it enters a medium at an angle

Question 28

What is the refractive index of a median?

Answer 28

A measure of how much the material changes the speed of light when it passes through it. Also a measure of the direction change

Question 29

What is the symbol for refractive index?

Answer 29

n

Question 30

What is the absolute refractive index in a vacuum and air?

Answer 30

1.00

Question 31

How would you calculate the refractive index?

Answer 31

``` n = sin theta in air / sin theta in material n = V in air / V in material n = wavelength in air / wavelength in material ```

Question 32

What is the critical angle?

Answer 32

The angle of incidence for a material that results in an angle of refraction of 90 degrees

Question 33

What is total internal refraction?

Answer 33

When the angle of incidence is greater than the critical angle and all the light is reflected inside the material.

Question 34

What is a uses of internal reflection?

Answer 34

Optical fibres

Question 35

How would you calculate the critical angle?

Answer 35

sintheta = 1/n | where n=refractive index

Question 36

What is irradiance?

Answer 36

The power of radiation per unit area

Question 37

What is the symbol for Irradiance?

Answer 37

I

Question 38

What are the units for irradiance?

Answer 38

Wm-2 | (Watts per metre squared

Question 39

What is the relationship between irradiance and distance?

Answer 39

As the distance from the source increases, the irradiance decreases

Question 40

What would a graph of distance against irradiance look like?

Answer 40

Same shape as a cooling curve

Question 41

What is the purpose of the inverse square law?

Answer 41

To prove that irradiance varies with the square of the distance from the source i.e. doubling the distance results in a quartered irradiance

Question 42

Describe an experiment to verify the inverse square law of light.

Answer 42

Place a lamp at one end of a metre stick and a light intensity metre at the other end Use the light detector to record intensity levels at different known distances from the lamp Repeat the measurements

Question 43

Describe how apparatus could be used to verify the inverse square law for a point source of light. (With the lamp, metre stick, and the light intensity metre)

Answer 43

Obtain values of irradiance for different distances Plot graph of I against 1/d2 (1 divided by distance squared) Graph of I against 1/d2 is a straight line through the origin to prove that it is constant

Question 44

What are the formulas to find irradiance?

Answer 44

``` I = P/A I = k / d2 I = 1 /d2 I1d1 = I2d2 ```

Question 45

When are line spectra emmited?

Answer 45

When an electric current passes through a low pressure gas

Question 46

Why is spectra described as an atomic fingerprint?

Answer 46

Because each element has its own unique line spectrum

Question 47

How can line emission be observed?

Answer 47

With a spectrometer using a diffraction grating or a prism

Question 48

What can spectra be used for?

Answer 48

Identify elements present in distant stars and galaxies Measure redshift of galaxies Determine the rate of universal expansion

Question 49

What creates line spectra?

Answer 49

The transition of of electrons between energy levels

Question 50

How do absorption lines in sunlight spectrum allow us to determine the elements present in the suns atmosphere?

Answer 50

As light from hotter regions of the sun pass through cooler regions of the upper atmosphere, atoms absorb light at specific frequencies/wavelengths to produce an absorption spectra. These absorption lines allow the elements present to be identified.

Question 51

What is meant by continuous spectra?

Answer 51

Spectrum of light with no visible separate spectra lines

Question 52

How do electrons move between orbits?

Answer 52

Can adopt a new orbit level but cannot stop in between the electron levels

Question 53

What is the lowest energy level of electron orbiting called?

Answer 53

Ground state

Question 54

What is it called when an electron moves to a higher energy level?

Answer 54

It is in an excited state

Question 55

What does ionisation level mean?

Answer 55

That an electron has gained sufficient energy to escape from the atom completely

Question 56

When does an electron have zero energy?

Answer 56

When it has reached the ionisation level

Question 57

Why does an electron being ionised have zero energy?

Answer 57

Because electrons have a negative energy, so to get to zero requires a lot of energy

Question 58

When is delta E used?

Answer 58

When comparing energy levels in atoms

Question 59

When and where to electron transitions take place?

Answer 59

``` Where = between the energy levels of an atom When = an electron either absorbs or emits a photon of electromagnetic radiation ```

Question 60

What does E=hf relate to?

Answer 60

The frequency of an emitted photon | where h is plancks constant

Question 61

What is the relationship between the jump in electron levels and the frequency of the photon emitted?

Answer 61

As the size of the jump increases, the frequency of the photon emitted also increases

Question 62

Describe how an electron can make the transition to a higher energy level.

Answer 62

By absorbing a photon of the electromagnetic radiation that had energy equal to precisely the difference between energy levels

Question 63

Describe the process of an electron moving from a higher energy level to a lower energy level.

Answer 63

The electron loses energy as it moves to a lower energy level and emits a photon in the process. The energy of the photon is equal to the difference between levels