Waves and Optics ༄

Question 1

Explain how light from the diffraction grating forms a maximum on the screen

Answer 1

(1) Light from slits overlap
(2) Arrive at screen in phase
(3) undergo superposition

Question 2

Explain what is meant by modal dispersion in an optical fibre

Answer 2

(1) Pulse broadening occurs
(2) Because light rays enter the fibre at different angles
(3) Takes a different amount of time to travel along the fibre

Question 3

Explain, with reference to refractive index, why the pulse of red light has a shorter transit
time than the pulse of blue light

Answer 3

(1) The refractive index of core for blue light is greater than the refractive index for
red
(2) The speed of the blue light is less than the speed of the red light and travel the
same distance

Question 4

State why emergent ray does not change direction as it leaves an object

Answer 4

(1) Angle of incidence is 0

Question 5

Progressive wave

Answer 5

Transfers energy without transferring material and is made up of particles of a medium oscillating

Question 6

Amplitude

Answer 6

Wave maximum displacement from equilibrium position

Question 7

Frequency

Answer 7

Number of complete oscillations passing through a point per second

Question 8

Wavelength

Answer 8

Length of one whole oscillation

Question 9

Speed

Answer 9

Distance travelled by the wave per unit time

Question 10

Phase

Answer 10

Position of a certain point on a wave cycle

Question 11

Phase difference

Answer 11

How much a particle/wave lags behind another

Question 12

Period

Answer 12

Time taken for one full oscillation

Question 13

Points in phase

Answer 13

Two points on a wave are in phase if they are both at the same point of the wave cycle, they will
have the same displacement and velocity and their phase difference will be a multiple of 360°
(2π radians), they do not need the same amplitude, only the same frequency and wavelength.

Question 14

Points out of phase

Answer 14

Two points are completely out of phase when they’re an odd integer of half cycles apart e.g. 5
half cycles apart where one half cycle is 180°

Question 15

Speed equation

Answer 15

c = fλ

Question 16

Frequency equation

Answer 16

f = 1/T

Question 17

Transverse waves

Answer 17

Oscillations at perpendicular to direction of energy transfer
All electromagnetic waves are transverse at travel at 3x10^8 ms in a vacuum
Can be demonstrated by shaking a slinky vertically

Question 18

Longitudinal waves

Answer 18

Oscillations are parallel to direction of energy transfer
Made up of compressions and rarefactions
Shown by sound waves or shaking a slinky horizontally

Question 19

Polarising waves

Answer 19

Can only travel in one plane
Only transverse waves can be polarised
Provides evidence for the nature of transverse waves as polarised waves must travel perpendicular to direction of energy transfer

Question 20

Superposition

Answer 20

Where the displacements of two waves are combined as they pass each other,
the resultant displacement is the vector sum of each wave’s displacement

Question 21

Two types of superposition

Answer 21

● Constructive interference occurs when 2 waves have displacement in the same
direction
● Destructive interference occurs when one wave has positive displacement and the other has negative displacement, if the waves have equal but opposite displacements, total destructive interference occurs

Question 22

Stationary wave

Answer 22

A stationary wave is formed from the superposition of 2 progressive waves, travelling in opposite directions in the same plane, with the same frequency, wavelength and amplitude
No energy is transferred by a stationary wave

Question 23

Nodes

Answer 23

● Where the waves meet in phase, constructive interference occurs so antinodes are
formed, which are regions of maximum amplitude
● Where the waves meet completely out of phase, destructive interference occurs and nodes are formed, which are regions of no displacement

Question 24

First harmonic

Answer 24

The lowest frequency at which a stationary wave forms is the first harmonic, which forms a stationary wave with two nodes and a single antinode. The distance between adjacent nodes (or antinodes) is half a wavelength (for any harmonic

Question 25

Frequency of harmonic equation

Answer 25

f = 1/2L root T/mu

Question 26

Example of stationary waves

Answer 26

● Stationary microwaves can be formed by reflecting a microwave beam at a metal plate, to find the nodes and antinodes use a microwave probe
● Stationary sound waves can be formed by placing a speaker at one end of a closed glass tube, lay powder across the bottom of the tube, it will be shaken at the antinodes and settle at the nodes. The distance between each node is half a wavelength, and the frequency of the signal generator to the speaker is known so by c=fλ the speed of sound in air can be found

Question 27

Path difference

Answer 27

The difference in the distance travelled by two waves

Question 28

Coherent light source

Answer 28

Has the same frequency and wavelength and a fixed phase difference

Question 29

Laser nature

Answer 29

Lasers are an example of light which is coherent and monochromatic, meaning they emit a single (or small range of) wavelength(s) of light. Lasers are usually used as sources of light in diffraction experiments as they form clear interference patterns

Question 30

Young’s double slit experiment

Answer 30

Demonstrates interference of light from two-sources. In this experiment you can use two coherent sources of light or you could use one coherent source and a double slit in order to form an interference pattern. If you do not have a coherent source of light for example a light bulb, you could place a single slit before the double slit to make the light have a fixed path difference, and a filter to make the light monochromatic. Below is a brief procedure to describe Young’s double slit experiment:
● Shine a coherent light source through 2 slits about the same size as the wavelength of the laser light so the light diffracts
● Each slit acts as a coherent point source making a pattern of light and dark fringes. Light fringes are formed where the light meets in phase and interferes constructively, this occurs
where the path difference between waves is a whole number of wavelengths
Dark fringes are formed where the light meets completely out of phase and
interferes destructively, this occurs where the path difference is a whole number and a half wavelengths ((n+½)λ).

Question 31

Young’s double slit equation

Answer 31

w = λD/s

Question 32

White light instead of mono

Answer 32

Using white light instead of monochromatic laser light gives wider maxima and a less intense diffraction pattern with a central white fringe with alternating bright fringes which are spectra, violet is closest to the central maximum and red furthest

Question 33

Diffraction

Answer 33

The spreading out of waves when they pass through or around a gap

Question 34

Diffraction sizes

Answer 34

The greatest diffraction occurs when the gap is the same size as the wavelength. When the gap is smaller than the wavelength most waves are reflected, whereas when it is larger there is less noticeable diffraction. When a wave meets an obstacle you get diffraction round the edges, the wider the obstacle compared to the wavelength, the less diffraction

Question 35

Monochromatic light on a screen

Answer 35

Can be diffracted through a single slit onto a screen, which forms an
interference pattern of light and dark fringes. The pattern has a bright central fringe, which is double the width of all other fringes, with alternating dark and bright fringes on either side, the bright fringes are caused by constructive interference where the waves meet in phase and the dark fringes are caused by destructive interference where waves arrive completely out of phase

Question 36

How to vary slit width

Answer 36

● Increasing the slit width decreases the amount of diffraction so the central maximum becomes narrower and its intensity increases.
● Increasing the light wavelength increases the amount diffraction as the slit is closer in size to the light’s wavelength, therefore the central maximum becomes wider and its intensity decreases

Question 37

Diffraction grating

Answer 37

A diffraction grating is a slide containing many equally spaced slits very close together. When monochromatic light is passed through a diffraction grating the interference pattern is much sharper and brighter than it would be after being passed through a double slit like in Young’s double slit, this is because there are many more rays of light reinforcing the pattern. This means measurements of slit widths are much more accurate as they are easier to take

Question 38

Order lines

Answer 38

The ray of light passing through the centre of a diffraction grating is called the zero order line, lines either side of the zero order are the first order lines, then the lines outside the two first order lines are the second order lines

Question 39

Diffraction equation

Answer 39

d sinθ = nλ
(Where d is the distance between the slits, θ is the angle to the normal made by the maximum, n is the order and λ is the wavelength. As λ increases (by for example changing the laser light from blue to red), the distance between the orders will increase because θ is larger due to the increase in diffraction as the slit spacing is closer in size to the wavelength, this means the pattern will spread out.
The maximum value of sin θ is 1, therefore any values of n, which give sin θ as greater than 1 are impossible)

Question 40

Refractive index (n)

Answer 40

A property of a material which measures how much it slows down light passing through it. This is calculated by dividing the speed of light in a vacuum (c) by the speed of light in that substance (cs)
n = c/cs

Question 41

Refraction

Answer 41

Occurs when a wave enters a different medium, causing it to change direction, either
towards or away from the normal depending on the material’s refractive index

Question 42

Snell’s law

Answer 42

n1sinθ1=n2sinθ2

Question 43

Total internal reflection

Answer 43

can occur 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 the boundary (n2)

Question 44

TIR use

Answer 44

A useful application of total internal reflection are optical fibres; these are flexible, thin tubes of plastic or glass which carry information in the form of light signals. They have an optically dense
core surrounded by cladding with a lower optical density allowing TIR to occur, this cladding also protects the core from damage and prevents signal degradation through light escaping the core, which can cause information to be lost

Question 45

Signal degradation causes

Answer 45

● Absorption - where part of the signal’s energy is absorbed by the fibre, reducing the amplitude of the signal, which could lead to a loss of information
● Dispersion - this causes pulse broadening, which is where the received signal is broader than the original transmitted signal. Broadened signals can overlap causing loss of information

Question 46

Types of dispersion

Answer 46

➔ Modal - caused by light rays entering the fibre at different angles, therefore they take different paths along the fibre, (for example some may travel down the middle
of the fibre, while others are reflected repeatedly,) this leads to the rays taking a different amount of time to travel along the fibre, causing pulse broadening.
➢ This can be reduced by making the core very narrow, therefore making the possible difference in path lengths smaller.
➔ Material - caused by using light consisting of different wavelengths, meaning light rays will travel at different speeds along the fibre, which leads to pulse broadening
➢ This can be prevented by using monochromatic light