Waves

Question 1

Define amplitude

Answer 1

A wave’s maximum displacement from its equilibrium position.

Question 2

Define frequency

Answer 2

The number of waves that pass a point in a unit time period. It is the
inverse of the time period.

Question 3

Define wavelength

Answer 3

The distance between two identical positions on two adjacent
waves. It is commonly measured from peak to peak or trough to trough

Question 4

Define wave speed

Answer 4

The distance travelled by a given point on a wave in a given interval of time

Question 5

Define phase difference

Answer 5

The difference in phase between two points on a wave. It is
usually expressed in radians.

Question 6

Nature of longitudinal waves

Answer 6

:A wave with oscillations that are parallel to the direction of
energy propagation

Question 7

Nature of transverse waves

Answer 7

A wave with oscillations that are perpendicular to the direction
of energy propagation

Question 8

What is polarisation

Answer 8

Where a wave passes through a filter such that it can only oscillate in one plane. Polarisation provides evidence for the nature of transverse waves because
polarisation can only occur if a wave’s oscillations are perpendicular to its direction of travel (as
they are in transverse waves)

Question 9

Applications of polarisation

Answer 9

Polaroid materials reduce glare by blocking partially
polarised light reflected from water and tarmac, as they only allow oscillations in the plane of the
filter, making it easier to see.
Another application of polarisation is TV and radio signals, which are usually plane-polarised by
the orientation of the rods on the transmitting aerial, so the receiving aerial must be aligned in the
same plane of polarisation to receive the signal at full strength.

Question 10

What is a stationary wave?

Answer 10

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.

Question 11

What are nodes and antinodes?

Answer 11

Nodes are regions where there is no displacement. Anti-nodes are points where the displacement of the standing wave is at its maximum

Question 12

How do stationary waves form?

Answer 12

Stationary waves are formed from the superposition of two waves of the same frequency and amplitude travelling in opposite directions. In a string, the stationary wave is formed from the superposition of a waves and its reflection.

Question 13

What is path difference?

Answer 13

Path difference is the difference in distance that two waves must travel from their sources to a given point. It is usually expressed in terms of wavelength.

Question 14

Define coherence

Answer 14

Waves are coherent if they have the same wavelength and
frequency, as well as there being a fixed phase difference between them

Question 15

What is diffraction?

Answer 15

The spreading of waves as they pass through a gap of a similar
magnitude to their wavelength.

Question 16

What is monochromatic light?

Answer 16

Light of one wavelength

Question 17

Describe the interference pattern produced when white light is passed through single/double slits.

Answer 17

Single slit- The central maximum would be white
All maxima would be composed of a spectrum
The shortest wavelength (violet / blue) would appear nearest to the central maximum
The longest wavelength (red) would appear furthest from the central maximum
The fringe spacing would be smaller and the maxima would be wider
Double slit- A central bright white fringe will be produced. All other bright fringes would consist of a spectra of light, with blue light on the side closest to the centre, and red on the far side of each fringe.

Question 18

How can you vary the width of the central maximum from single slit diffraction?

Answer 18

In order to vary the width of the central maximum, you can vary slit width and wavelength:
● 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 19

Derive the diffraction grating formula

Answer 19

1. Considering the first order maximum, where the path difference between two adjacent
   rays of light is one wavelength (as shown in the diagram below), name the angle
   between the normal to the grating and the ray of light θ.
2. As you can see a right angle triangle is formed, with side lengths d and λ. And by using the
   fact that a right angle is 90°, and angles in a triangle add up to 180°, you can see the upper
   angle in the triangle is θ (because the lower angle is 90-θ°).
3. By using trigonometry we can see that for the first maximum sin θ = , (as sin θ = d
   λ
   Opp/Hyp) which rearranges to , (for the first order). dsin θ = λ
4. We know that the other maxima occur when the path difference between the two rays of
   light is nλ, where n is an integer, therefore we can generalise the equation by replacing λ
   with nλ to get: d sinθ λ = n .

Question 20

Applications of diffraction gratings

Answer 20

There are several applications of diffraction gratings:

● You split up light from stars using a diffraction grating to get line absorption spectra
which can be used to show which elements are present in the star.
● X-ray crystallography, which is where x-rays are directed at a thin crystal sheet which actsas a diffraction grating to form a diffraction pattern, this is because the wavelength of x-rays is similar in size to the gaps between the atoms. This diffraction pattern can be used to measure the atomic spacing in certain materials.

Question 21

What is the refractive index of air?

Answer 21

Approximately 1

Question 22

What is the purpose of cladding in optical fibres?

Answer 22

The cladding has lower optical density than the core, allowing TIR to occur. Cladding also protect the core from damage, and prevents signal degradation from light escaping the core.

Question 23

What is material dispersion?

Answer 23

Material dispersion is a consequence of the signal containing several different wavelengths, each of which travels at slightly different speeds in the core. This results in pulse broadening. Solution is to use monochromatic light.

Question 24

What is modal dispersion?

Answer 24

Pulse broadening 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. Solution is to use a very narrow core, making the possible difference in path lengths smaller

Question 25

What is absorption?

Answer 25

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

Question 26

What is pulse broadening?

Answer 26

Where the received signal is broader
than the original transmitted signal. Broadened signals can overlap causing loss of
information.

Question 27

Single slit diffraction formula

Answer 27

W=λ/a X 2d

Question 28

Explain, with reference to the processes within an atom, the difference between an emission spectrum and an absorption spectrum.

Answer 28

Photon is an energy carrier. In an absorption spectra, the atom becomes excited by absorbing the photon. In an emission spectrum, the atom is de-excited by emitting a photon