Waves and Optics

Question 1

Mechanical waves:

Answer 1

Pass through a substance and vibrate the particles which continues through the substance.

Question 2

Electromagnetic waves:

Answer 2

Don’t need a substance, electric and magnetic fields vibrating at 90 degrees to each other.

Question 3

Polarisation:

Answer 3

-only transverse waves.
-When a wave passes through a slit in a board the wave is polarised so that only waves that are parallel to the wave can pass through.
-If a second filter is placed in front of polarised waves and aligned at 90 degrees to the other filter then no waves pass through.
-Example is sunglasses (by reducing the intensity of light passes through) or radio waves.

Question 4

Define period of wave:

Answer 4

The amount of time for one wavelength to pass through a fixed point.

Question 5

Define phase difference:

Answer 5

The fraction of a cycle between 2 waves measured in radians.

Phase difference = 2pi*d/wavelength

Question 6

Describe a ripple tank:

Answer 6

Clear tray with sloped sides (to prevent reflection) full of water. This can be used to view wave fronts (lines of constant phase) e.g. crests.

Question 7

Angles in a reflection:

Answer 7

Angle of incidence = angle of reflection.

Question 8

Refraction:

Answer 8

Changes angle when passing into a more or less optically dense substance as they change speed if the angle is not 90.

Question 9

Diffraction:

Answer 9

Waves spread out through a small gap.
-smaller the gap the more the waves spread out.
-longer the wavelength, the more the waves spread out.
Due to each part of the wave front spreading out and creating its own wave.

Question 10

Principle of superposition:

Answer 10

When 2 waves meet, the total displacement at a point is equal to the sum of the individual displacements at that point. e.g. super crest on a boat.

Question 11

Examples of superposition:

Answer 11

-2 people sending progressive waves down a rope which forms nodes and antinodes.
- water waves in ripple tank due to vibrating dipper which makes waves spread out. They start to cancel each other out in certain places.

Question 12

Tests using microwaves:

Answer 12

-place receiver in path of microwave transmitter and then place.
This can be used to show that the intensity is weaker from further away, that the waves diffract when a grating is placed in front of them and that waves cannot pass through metal.

Question 13

Frequency on plucked guitar string:

Answer 13

Plucked gently - a stationary wave of constant frequency is created.
Plucked harshly - contains several different frequencies.

Question 14

Describe first harmonic:

Answer 14

Single loop with 2 nodes and 1 antinode (where amplitude is maximum).

Question 15

What happens if you raise frequency from the first harmonic:

Answer 15

2 loops will appear with 3 nodes and 2 antinodes. This happens when the frequency is double that of the first harmonic. Length of rope = wavelength.

Question 16

How energy works in stationary wave:

Answer 16

Do not transfer energy to their surroundings as there is no energy at the nodes as they are not vibrating. The energy at the antinodes is maximum.

Question 17

How stationary waves are formed:

Answer 17

When they are in phase, their crests / troughs reinforce to create a bigger amplitude.
-quarter of a cycle later, they have moved into antiphase so they cancel each other out.
-after another quarter cycle they move into phase again and produce a larger resultant wave

Question 18

Stationary waves in a pipe and microwaves:

Answer 18

Pipe - closed at one end, the sound resonates when there is an antinode at the open end and a node at the closed end.
Microwave - pointed at a metal plate that reflects it back. Detector signal is 0 along equally placed positions on the line.

Question 19

Practical for stationary waves:

Answer 19

String at one end is attached to mechanical vibrator and the other end passes over a pulley and holds a weight. Frequency of oscillator increases to show different stationary wave patterns.
-First harmonic occurs at lowest possible frequency that shows a stationary wave pattern.
-the next harmonics occur at multiples of the frequency of the first harmonic.

Question 20

Pitch and Frequency/Tension:

Answer 20

Pitch increased - raise tension or shorten the length.

f=1/2l*(T/u)^1/2

Question 21

Define refraction:

Answer 21

Change of direction that occurs when light passes at an angle across a boundary between 2 transparent substances.

Towards normal if more dense. away if less dense.

Partial reflection also occurs when a light ray enters glass.

Question 22

What is refractive index: Formula:

Answer 22

Ratio of sin of incident angle to the sin of refracted angle.

n = sin1/sin2

Question 23

Refraction angles in a rectangular block:

Answer 23

Angle of incidence is equal to the angle that the ray leaves the box with.

Question 24

Why refraction happens:

Answer 24

Speed of light is is different in each substance.

Question 25

Give refractive index formula with c and the one with wavelength:

Answer 25

n=c/c in the substance

smaller the speed of light in a substance the higher the refractive index.

n=wavelength/wavelength in the substance

Question 26

prove snells law:

Answer 26

sini/sinr = c1/c2

therefore 1/c1sini = 1/c2sinr

mutliply both sides by c

n1sini=n2sinr

(also when light passes from vacuum into a transparent subtance sini/sinr = n

Question 27

Give refractive index of air / a vacuum

Answer 27

air is very close to a vacuum
-very close to 1 so it is treated as 1.

Question 28

Give approximate wavelengths of red to violet: what does this show:

Answer 28

red - 650nm
violet- 350nm

the refractive index of violet is more than refractive index of red light per n=wavelength over wavelength in substance.

Question 29

Critical angle

Answer 29

i = c then it refracts along the boundary
i > c then it reflects inside.

Question 30

Name 2 requirements for TIR:

Answer 30

-Incident substance must have a larger refractive index
-Angle of incidence must exceed the critical angle.

Question 31

Prove critical angle formula

Answer 31

n1sini=n2sinr (since r, refracted angle is 90 degrees) then n1sini = n2

therefore sinc = n2/n1

Question 32

partial reflection:

Answer 32

In TIR, part of the light is reflected inside and part of it is refracted through the substance. The energy is split.

This always occurs when the angle of incidence is less than or equal to the critical angle.

Question 33

why diamonds sparkle:

Answer 33

Very high n value so it splits the colours more than anything else. This also gives it a low critical angle so TIR happens many times before it emerges so the colours spread out more.

Question 34

Why optical fibres are efficient:

Answer 34

-TIR occurs at core-cladding boundary. Cladding has low refractive index so the amount of TIR is increased.
-Core is very narrow to prevent modal dispersion. This happens in wide cores when light travelling straight down the tube travels short distance per metre than one undergoing TIR. This would make the wavelength too long and it could merge with the next pulse.

Question 35

More on modal dispersion:

Answer 35

Also happens if white light is used as this is full of different wavelengths so they travel at different speeds in the optical fibre. The violet light would travel slower than red so the white light beam becomes longer so modal dispersion can happen.

Question 36

Youngs double slit:

Answer 36

Double slit is illuminated from the light from the single slit. Bright and dark fringes can be seen on the white screen in front of the double slit. They are evenly spaced and parallel to the double slits.

Bright fringes are formed when light from 1 slit reinforces the other as they arrive in phase.
Dark fringes are formed when light from one slit cancels the other out as they are pi out of phase.

Question 37

Why the single slit must be narrow:

Answer 37

If wide, each part of the slit produces its own fringe pattern which makes the dark fringes narrower than the bright fringes.

Question 38

Fringe seperation:

Answer 38

Distance from centre of 1 bright fringe to the next.

w =lamda*D/s

Question 39

Path difference

Answer 39

Difference in distance travelled by the light in the double slit experiment for bright fringes = n*wavelength

Question 40

Notes when carrying out young’s double slit:

Answer 40

1) Measure from centre of dark fringes and divide by number of fringes you measured across.
2) This process also occurs with sound waves. You would hear varying levels of sound intensity across the board.

Question 41

Coherence and examples/non examples:

Answer 41

Coherent: Double slit as it produces light waves of the same frequency with a constant phase difference.

e.g. double slit experiment and lasers or vapour lamps and discharge tubes (yellow/orange)

non examples: the sun or filament bulb as they produce light with varying wavelengths/diff colours.

Question 42

Describe white light fringes:

Answer 42

central white maxima, inner fringes tinted blue with red on the outer side.

outer fringes merge into white light as the diff. colours reinforce and overlap.

Question 43

Single slit experiment:

Answer 43

parallel beam of light directed at single slit.

-Shows central maxima that is twice the width of the other fringes either side.
-The fringes get weaker the further out it goes.
-Each outer fringe is equal in width.
-Fringes are narrower with shorter wavelength of light.

Question 44

Why Youngs Double slit doesn’t work if slits are too far apart:

Answer 44

The light must overlap to interfere.

Question 45

Requirements for Young’s double slit to work:

Answer 45

each slit must be narrow to diffract enough.
They must be close enough together to interfere.

Question 46

number of slits per metre on diffraction grating:

Answer 46

N = 1/d (d is grating spacing)

Question 47

Equation for no. of slits on a grating:

Answer 47

n = d/lambda

Question 48

Emission spectra and absorption spectra:

Answer 48

Emission shows coloured lines that are from the wavelength of light from the element that emitted it.

Absorption spectra shows continuous spectrum with narrow dark lines at certain wavelengths. This shows the wavelengths a certain elements absorbs hence it is missing from the spectra. (The element may then emit the light but not necessarily in the same direction as the transmitted light.)