3: Waves Flashcards

Question

Absorption in Fibre Optics

Answer 1

Some of the pulse's energy is absorbed by the material, which reduces the pulse's amplitude, meaning the receiver may not be able to pick up the signal

Answer 2

There are two types of dispersion – modal and material. They both result in pulse broadening – where the pulse spreads out as it travels through the core. Broadened pulses can overlap leading to information loss

Answer 3

- Caused by the light rays entering the optical fibre at different angles. This means they take different paths with varying distances down the fibre so there is a wide range of times at which the rays arrive at the receiver resulting in pulse broadening - Can be reduced by using cladding with a higher refractive index, which lets rays with small angles of incidence (on the boundary between the core and cladding) escape the core - Can be reduced by narrowing the core so there's less variation in the angles of incidences of the rays

Answer 4

- Given that light consists of different wavelengths, different rays experience different amounts of refraction so they slow down by different amounts. Thus, there's a range of times where they reach the receiver resulting in pulse broadening - Using monochromatic (all one wavelength) light can stop material dispersion

Answer 5

Regenerate the signal at regular intervals, which reduces signal degradation

Answer 6

When two or more waves cross, the resultant displacement equals the vector sum of the individual displacements

Answer 7

- The difference, in wavelengths, between the distances of the paths of two waves to a point - If it is an integer (0, λ, 2λ) the waves superpose to produce a maximum - If it is an integer plus 1/2 (λ/2, 3λ/2), the waves cancel each other out

Answer 8

Where waves have the same wavelength and frequency and a fixed phase difference

Answer 9

The superposition of two progressive waves, with the same frequency, wavelength and amplitude, travelling in opposite directions

Answer 10

Points on the stationary wave where the progressive waves always superpose to give 0 amplitude / displacement as they have a phase difference of π radians

Answer 11

Points on the stationary wave where the progressive waves superpose to produce an oscillation between maximum positive and negative displacements (progressive waves are always in phase)

Answer 12

- A stationary wave is only formed at a resonant frequency (when an exact number of half wavelengths fits on the string) - First harmonic: • Lowest possible resonant frequency • Wave has one "loop" with a node at each end and an antinode in the middle • 1/2 wavelength fits on string - Second harmonic is twice the frequency of the first harmonic

Answer 13

f = (1 / 2l) √(T / μ) f is frequency in Hz l is length in m T is tension in N μ is mass per unit length in kg m⁻¹

Answer 14

The use of two coherent sources or the use of a single source with double slits to produce an interference pattern

Answer 15

w = λ D / s w is fringe spacing in m λ is wavelength in m D is distance between slits and screen in m s is distance between slits in m

Answer 16

d sin θ = n λ d is distance between slits in m θ is angle to the normal made by the maximum n is order of maximum λ is wavelength of light source in m

Answer 17

- https://th.bing.com/th/id/OIP.FPwz0Tf_bHcdHpVxKEoe8QAAAA?pid=ImgDet&rs=1 - The angle between the nth order maximum and incoming light is θ - d is slit spacing - n λ is the path difference at the nth order maximum - sin θ = n λ / d → d sin θ = n λ

Answer 18

Diffraction gratings contain equally spaced slits. The interference pattern produced using monochromatic light is really sharp as there are lots of rays reinforcing the pattern

Answer 19

- Astronomers and chemists analyse the spectra of stars and materials respectively, produced by a diffraction grating, to see what elements are present - Crystals can be used as diffraction gratings for X-rays (which have a similar wavelength to the crystals atom separation) and the interference pattern gives information about the crystal structure

Answer 20

Investigation into the variation of the frequency of stationary waves on a string with length, tension and mass per unit length of the string

Answer 21

https://docs.google.com/document/d/1dnOyhtbTycL6i5XJ8f8RwxPVPK0cZC-GfugFbyWu4tU/edit?usp=sharing 1. Set the length of the string to a known distance 2. Measure the mass of the hook 3. Adjust the frequency on the vibrator until the wave on the string is the first harmonic. Read the frequency 4. Repeat step 4, incrementing the mass on the hook

Answer 22

Investigation of interference effects to include the Young’s slit experiment and interference by a diffraction grating

Answer 23

1. https://docs.google.com/document/d/1MzsvzAsm11bKLadcWOdFqeUHiMMljszp2DKmvPQgcuA/edit?usp=sharing 2. Read the slit spacing of the double slit 3. Set a known distance between the double slit and screen 4. The fringe width can be measured by measuring across a large number of visible fringes 5. Repeat step 4, changing the distance between the slits and screen

Answer 24

https://docs.google.com/document/d/1JB0MDui0cTgAxu9-LY8g2XVcII4Pp4FKdq-gX_BFUWI/edit?usp=sharing 1. Record the slit spacing of the diffraction grating 2. Calculate θ₁, θ₂ and θ₃ by measuring h₁, h₂ and h₃ and using the distance between the screen and slits and note the maximum order for each angle

Answer 25

- High-intensity central maximum with low-intensity subsidiary maxima on either side - Width of central maximum is double the width of other maxima

Answer 26

- High-intensity white central maximum with low-intensity spectra for subsidiary maxima on either side - Width of central maximum is double the width of the subsidiary maxima

Answer 27

Increasing wavelength increases the width of the central maximum

Answer 28

Decreasing slit width increases the width of the central maximum

3: Waves Flashcards

(52 cards)