3.2 Refraction, Diffraction And Interference

Question 2

Define coherence

Answer 2

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

Question 3

Why is a laser useful in showing interference and diffraction?

Answer 3

It produces monochromatic (same wavelength/colour) light so diffraction and interference patterns are more defined

Question 4

What was Young’s double-slit experiment?

Answer 4

A single light source is directed towards two slits, which each act as a coheret light source The light interferes constructively and destructively to create an interference pattern

Question 5

Describe the interference pattern created using white light

Answer 5

A bright white central maximum flanked by alternating spectral fringes of decreasing intensity with violet closest to the zero order and red furthest

Question 6

Why does an interference pattern form when light is passed through a single slit?

Answer 6

The light diffracts as it passes through the slit Where the waves are in phase constructive interference occurs making bright fringes Where the waves are out of phase destructive interference occurs making a dark fringe

Question 7

Increasing the slit width increases the width of the central diffraction maximum. True or false?

Answer 7

False, the slit is not so close to the wavelength in size so less diffraction occurs The central maximum would become narrower and more intense

Question 8

Is the following a double slit pattern, single slit pattern or a diffraction grating pattern?

Answer 8

Single slit

Question 9

What is the approximate refractive index of air?

Answer 9

1

Question 10

When light enters a more optically dense medium does it bend towards or away from the normal?

Answer 10

Towards the normal

Question 11

When does total internal reflection occur?

Answer 11

When light is at a boundary to a less optically dense medium and the angle of incidence is greater than the critical angle

Question 12

What is the purpose of the cladding in a step index optical fibre?

Answer 12

Protects core from scratches which would allow light to escape and degrade the signalAllows TIR as it has a lower refractive index than the core

Question 13

How does signal degradation by absorption in an optical fibre affect the recieved signal?

Answer 13

Part of the signal’s energy is absorbed by the fibre so its amplitude is reduced

Question 14

What is pulse broadening?

Answer 14

When the received signal is wider than the original. This can cause overlap of signals leading to memory loss

Question 15

How does modal dispersion cause pulse broadening?

Answer 15

Light rays enter the fibre at different angles so they take different paths along it, some may travel down the middle while others are reflected repeatedly, so the rays take different times to travel along the fibre, causing pulse broadening

Question 16

What is material dispersion?

Answer 16

When light with different wavelengths is used some wavelengths slow down more than others in the fibre so they arrive at different times causing pulse broadening

Question 17

How can modal dispersion be reduced?

Answer 17

Use a single mode fibre (very narrow) so the possible difference in path lengths is smaller

Question 18

How can material dispersion be reduced?

Answer 18

Use monochromatic light

Question 19

How can both absorption and dispersion be reduced?

Answer 19

Use an optical fibre repeater to regenerate the signal now and then

Question 20

State the advantages of optical fibres over traditional copper wires

Answer 20

Signal can carry more information as light has a high frequencyNo energy lost as heatNo electrical interferenceCheaperVery fast

Question 21

What path does a light ray take when the angle of incidence is equal to the critical angle?

Answer 21

It goes along the boundry

Question 22

What formula can be used to find the critical angle for 2 materials whose refractive indicies are known?

Answer 22

sinC = n2 / n1 where n1 > n2 C = critical angle n1 = refractive index of material 1 n2 = refractive index of material 2

Question 23

What is the critical angle of a water to air boundry if water has a refractive index of 1.33?

Answer 23

sinC = n2/n1 n2 = air and n1 = water C = sin-1(1/1.33) C = 48.8

Question 24

Using snells law of refraction, find the angle of refraction in a material with RI = 1.53 when the angle of incidence is 32 from a material with a RI = 1.23

Answer 24

n1sin(i) = n2sin(r) 1.23sin32 = 1.53sinr sinr = 1.23sin32/1.53 sinr = 0.426 r = 25.2

Question 25

Glass has a refractive index of 1.5, water has a refractive index of 1.33, which is more optically dense?

Answer 25

Glass

Question 26

What formula is used to determine the refractive index of a material?

Answer 26

n = c/(c_s) n = refractive index c = speed of light in a vaccum (c_s) = speed of light in material

Question 27

State 2 applications of diffraction gratings

Answer 27

Splitting up light from stars to make line absorption spectra - used to identify elements present in the starX-ray crystallography, a crystal sheet acts as the diffraction grating the X-rays pass through, used to find the spacing between atoms

Question 28

Derive the formula (d \sin \theta = n \lambda)

Answer 28

For the first order maximum, the path difference between two adjacent rays of light is (\lambda), the angle between the normal to the grating and the light ray is (\theta)A right angled triangle is formed, with side lengths (d) and (\lambda). The upper angle is (\theta).For the first maximum (\sin \theta = \frac{\lambda}{d}), rearrange to (d \sin \theta = \lambda)Other maxima occur when the path difference between the two rays of light is (n \lambda), where (n) is an integer, replace (\lambda) with (n \lambda) to get: (d \sin \theta = n \lambda)

Question 29

When light passing through a diffraction grating is changed from blue to red, do the orders get closer together?

Answer 29

The wavelength of the light has increased so it will diffract more, the orders will become further apart

Question 30

What is diffraction?

Answer 30

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

Question 31

How did Young’s double-slit experiment provide evidence for the wave nature of light?

Answer 31

Diffraction and intergerence are wave properties, hence the interference pattern of light shows light has wave properties

Question 32

What are 4 safety precautions that must be followed when using a laser?

Answer 32

Wear laser safety gogglesDont shine the laser at reflective surfacesDisplay a warning signNever shine the laser at a person

Question 33

What formula is associated with Young’s double-slit experiment?

Answer 33

(w = \frac{\lambda D}{s}) w - fringe spacing (\lambda) - wavelength of light used D - distance from screen to slits s - slit separation

Question 34

What is path difference?

Answer 34

The difference in distance travelled by 2 waves

Question 35

How could you investigate stationary sound waves?

Answer 35

Place a speaker at one end of a closed glass tubeLay powder across the bottom of the tubePowder will be shaken from the antinodes and will settle at the nodesThe distance between each node is half a wavelength

Question 36

What is ‘phase’?

Answer 36

The position of a certain point on a wave cycle Units are radians, degrees or fractions of a cycle

Question 37

True or False: Only light can produce interference patterns

Answer 37

False: interference patterns can be produced by sound waves and all EM waves too