Waves

Question 1

What are em waves?

Answer 1

Consist of electric wave and magnetic wave that are in phase and are at right angles to each other and to the direction in which they’re travelling.

Question 2

Electromagnetic waves types?

Answer 2

Radio waves, microwaves, IR light, UV radiation, X rays, gamma radiation

Question 3

EM spectrum - wave type:
Radio wave
Microwave
IR
Visible
UV
X rays
Gamma rays

Answer 3

λ range:
>0.1m
0.1m - 1mm
1mm - 700nm
700nm - 400nm
400nm - 1nm
10nm - 0.001nm
<1nm

Question 4

What’s polarisation?

Answer 4

Restriction of vibration to a single plane. Only transverse waves, bc longitudinal waves vibrate in direction of travel.

Question 5

Eq for phase difference in radians?

Answer 5

2π d/λ , where d is distance apart of two points along wave eg d = 3/4λ .

Question 6

Waves in ripple tank called?

Answer 6

Wavefronts (lines of constant phase)

Question 7

Ripple tank used to observe?

Answer 7

reflection, refraction, diffraction

Question 8

Ripple tank has sloping sides to?

Answer 8

to prevent wavefronts from reflecting off the sides of the rank (bc otherwise difficult to see waves).

Question 9

What is refraction?

Answer 9

When wave passes across a boundary at which wave speed changes,

Question 10

What is refraction? (What stays constant?)

Answer 10

When wave passes across a boundary at which wave speed changes, λ also changes. If wavefronts approach at an angle to boundary, they change speed and change direction (bc they change speed).

Question 11

Principle of superposition?

Answer 11

When two waves meet, the total displacement at a point is equal to the sum of the individual displacements at that point.

Question 12

Crest meets trough of same amplitude? Similar amplitude?

Answer 12

Resultant displacement is zero, the two waves cancel each other out. If not same amplitude, resultant is called a minimum.

Question 13

What are coherent waves? What happens when coherent sources of waves overlap?

Answer 13

Produce interference pattern where they overlap bc they vibrate at same frequency with constant phase difference (therefore cancellations and reinforcements occur at fixed positions).

Question 14

Microwave transmitter and receiver can be used to demonstrate?

Answer 14

Reflection, refraction, diffraction, interference, stationary waves, and polarisation of microwaves.

Question 15

Is energy transferred along standing wave?

Answer 15

No!

Question 16

Phase difference in stationary wave?

(+ wisdom from tom)

Answer 16

Phase difference between two vibrating particles is 0 if separated by even number of nodes (m), and 180 degrees / π radians if separated by odd number of nodes (m). Therefore phase difference = mπ

Basically when in phase both particles are at their max displacement (not necessarily the max amplitude of the wave) and moving in same direction, so in phase. When 180 out of phase, particles in max displacement travelling in opposite directions.

Question 17

Sound resonance?

Answer 17

A phenomenon in which an acoustic system amplifies sound waves whose frequency matches one of its own natural frequencies of vibration.

Question 18

When does sound resonance occur?

Answer 18

Sound resonates at certain frequencies in air-filled tube/pipe. In tube/pipe that’s closed at one end, resonance frequency occurs when we have antinode at open end and node at other end.

Question 19

First harmonic eq f(1)

Answer 19

λ(1) = 2L therefore f(1) = c/2L

Question 20

Second harmonic f(2)

Answer 20

f(2) = c/λ(2) = c/L = 2f(1)

Question 21

Third harmonic f(3)

Answer 21

f(3) = c/λ(3) = 3c/2L = 3f(1)

Question 22

Number of loops on stationary wave?

Answer 22

= number of nodes - 1

Question 23

Stationary wave key condition:

Answer 23

Time taken for a wave to travel along string and back should be equal to the time taken for a whole number of cycles of the vibrator.

t = 2L/c (time to travel whole distance and back)
t = m/f (time to pass whole number of vibrations, where m is whole number and f is frequency of vibrator).
Therefore 2L/c = m/f
f = mc/2L = mf & λ= c/f = 2L/m.

In other words, stationary waves form at frequencies f, 2f, 3f etc and L = mλ/2!!!

Question 24

What is pitch?

Answer 24

Pitch corresponds to frequency.
Increasing T or decreasing L will increase pitch.

f = 1/2L sqrt T/μ

Question 25

Tuning and tuning fork:

Answer 25

The sound from a vibrating string inc all the harmonic frequencies, whereas a tuning fork vibrates only at a single frequency. The wire is tuned when its first harmonic is the same as the tuning fork frequency.

Question 26

Oscilloscope values

Answer 26

Y-sensitivity/ y -gain and
time base (value per division)

Y-shift and x-shift is just moving the trace

Question 27

Refraction at boundary between two transparent substances, light ray bends:

Answer 27

towards the normal when it passes into a more dense substance

away from the normal when it passes into a less dense substance

Question 28

Imagine glass cube. Light enters at angle i(1) and refracts at angle r(1), it then travels in straight line through glass and leaves on other side at angle i(2) and refracts into air at angle r(2). Explain relationship.

Answer 28

n = sin 1(1) / sin r(1)
when leaving:
sin i(2) / sin r(2) = 1/n

Because two sides of block are parallel, i(2) = r(1) and r(2) = i(1)

Question 29

Link refractive index with angle, speed, wavelength, frequency.

(s = substance)

Answer 29

n(s) = sin i / sin r = c / c(s)
n(s) = c / c(s) = λ / λ (s) bc frequency doesn’t change when refracted therefore c(s) = λ(s) f

Question 30

equation for refractive index

Answer 30

n = speed of light in vacuum / speed of light in transparent substance

Question 31

What is white light?

Answer 31

White light is composed of a light with a continuous range of wavelength from red (650nm) to violet (350).

Question 32

Which light is refracted more?

Answer 32

Glass prisms refract light by diff amounts - shorter λ = greater refraction. This is bc diff λ have diff speed therefore diff refractive index.

Question 33

What is critical angle?

Answer 33

Light from glass to air refracts away from normal. if angle of incidence increases to critical angle, light ray refracts along boundary.

Question 34

Two conditions for TIR (total internal reflection)?

Answer 34

1) Incidence substance has larger refractive index than other substance.
2) Angle of incidence exceeds critical angle.

Question 35

Two optical fibre uses?

Answer 35

1) Medical endoscopes to see inside body (fibre bundles need to be coherent bundles - fibre ends at each end are in the same relative position).

2) In communications to carry light signals (allows pulses of light that enter at one end from a transmitter to reach a receiver at other end).

Question 36

Why are the fibres transparent?

Answer 36

To minimise absorption of light, which would otherwise reduce the amplitude of the pulses progressively the further they travel in the fibre.

Question 37

Core purpose?

Answer 37

Transmission medium for an em wave to progress (by TIR).

Question 38

Cladding has lower n than core because..?

Answer 38

To reduce light loss from core therefore reducing decrease in amplitude, by ensuring TIR.

Question 39

Cladding purpose? (2)

Answer 39

1) Keeps signal secure - otherwise if two fibres in contact, crossover of light will occur.
2) Protection of core from scratching which could lead to light leaving the core.

Question 40

Narrow core and monomodal fibre.

Answer 40

Prevent multi-path modal dispersion. Diff rays propagating at diff angles. Non axial rays take longer time to travel same distance along fibre as axial rays. If wide core, pulse will become longer = merge with next pulse.

Question 41

Use shorter repeaters.

Answer 41

So that pulse is reformed before significant pulse broadening has taken place.

Question 42

Why use monochromatic light?

Answer 42

Prevent material dispersion. When using white light, red travels faster bc smaller refractive index / bc λ determines c therefore pulse broadening and pulse merging.

Question 43

Light source + single slit + double slit interference pattern?

Answer 43

Alternating bright and dark fringes, evenly spaced.

Question 44

If white light ?

Answer 44

Central white fringe, fringes either side showing range of colours/spectrum, red furthest and blue/violet closest to centre.

Fringe spacing less/maxima wider/dark fringes smaller (or not present).

Question 45

Purpose of single slit?

Answer 45

Makes it coherent bc constant phase difference. Alternatively use laser rather than light blub + single slit.

Question 46

If single slit used is too wide what happens?

Answer 46

Each part of it produces a fringe pattern, which is displaced slightly from the pattern due to adjacent parts of single slit. As a result, dark fringes become narrower than bright, and contrast between dark and light is lost. Use laser instead.

Question 47

w equation and how to measure w

(w = centre of bright fringe to centre of next bright fringe)

Answer 47

w = λD/s
(assumption D&raquo_space; s)

Measure across several fringes from the centres of the dark fringes, and divide by the number of fringes measured across.

Question 48

Shorter λ effect on w?

Answer 48

Narrower w because D/s constant (ref. above eq).
Or say bc smaller λ means smaller path difference therefore smaller w.

Question 49

What’s diffraction?

Answer 49

The spreading of waves when they pass through a gap or by an edge, therefore increase w is increase diffraction.

Question 50

Single slit diffraction pattern and eq?

Answer 50

Much brighter central maxima is twice as wide (W) as subsequent maximas (w).

W = 2λD/a

(a = width of single slit)

Question 51

Max number of orders?

Answer 51

When θ = 90° therefore
n = dsin90 /

Question 52

Max number of orders?

Answer 52

When θ = 90° therefore
n = dsin90 /λ = d/λ

Question 53

Number of maximas?

Answer 53

= 2n + 1, where n is max number of orders.

Question 54

Three types of spectra using spectrometer?

Answer 54

Continuous spectra, line emission spectra, line absorption spectra

Question 55

Line emission spectra?

Answer 55

A glowing gas in a vapour lamp emits light of specific λ therefore narrow vertical lines of diff colours spectra. Wavelengths of lines characteristic of element that produced light. If several element, can identify by observing line spectrum.

Question 56

Continuous spectra?

Answer 56

Filament lamp. The hotter the light source, the shorter the λ of the brightest part of the spectra therefore by measuring λ of brightest part we can measure the temp.

Question 57

Line absorption spectra?

Answer 57

Filament lamp light through glowing gas. Continuous spectrum with narrow dark lines at certain λ’s. Elements in glowing gas absorb light of the same λ they can emit therefore transmitted light missing these λ’s. (Bc it’s then emitted but not necessarily in same direction).

Question 58

Application of spectrometer?

Answer 58

Analysis of light from stars, analyse composition of stars, chemical analysis, measuring red shift/rotation of stars.