Section 3 - Waves: 3.2 - Refraction, Diffraction and Interference

Question 1

Define coherence.

Answer 1

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

Question 2

Why is a laser useful in showing interference and diffraction?

Answer 2

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

Question 3

What was Young’s double-slit experiment?

Answer 3

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

Question 4

Describe the interference pattern created using white light.

Answer 4

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

Question 5

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

Answer 5

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 completely out of phase destructive interference occurs making a dark fringe.

Question 6

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

Answer 6

False, the slit is not so close to the wavelength in size so less diffraction occurs - the central maximum becomes narrower and more intense.

Question 7

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

Answer 7

Single Slit

Question 8

What is the approximate refractive index of air?

Answer 8

1

Question 9

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

Answer 9

Towards the normal.

Question 10

When does total internal reflection occur?

Answer 10

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

Question 11

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

Answer 11

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

Question 12

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

Answer 12

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

Question 13

What is pulse broadening?

Answer 13

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

Question 14

How does modal dispersion cause pulse broadening?

Answer 14

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 15

What is material dispersion?

Answer 15

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 16

How can modal dispersion be reduced?

Answer 16

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

Question 17

How can material dispersion be reduced?

Answer 17

Use monochromatic light.

Question 18

How can both absorption and dispersion be reduced?

Answer 18

Use a optical fibre repeater to regenerate the signal now and then.

Question 19

State the advantages of optical fibres over traditional copper wires (5)

Answer 19

● Signal can carry more information as light has a high frequency.
● No energy lost as heat.
● No electrical interference.
● Cheaper.
● Very fast.

Question 20

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

Answer 20

It goes along the boundary ie. the angle of refraction is 90°.

Question 21

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

Answer 21

sinC = n2 / n1 | n2 = air = 1 | n1 = water = 1.33
C = sin^-1 (1 / 1.33)
C = 48.8°

Question 22

Using snell’s 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 RI = 1.23.

Answer 22

n1sini = n2sinr
1.23sin32 = 1.53sinr
sinr = 1.23sin32 / 1.53
sinr = 0.426
r = 25.2°

Question 23

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

Answer 23

Glass.

Question 24

State 2 applications of diffraction gratings.

Answer 24

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

Question 25

Derive the formula dsinθ = nλ

Answer 25

1. For the first order maximum, the path difference between two adjacent rays of light is 1λ (as shown), the angle between the normal to the grating and the light ray is θ.
2. A right angled triangle is formed, with side lengths d and λ. the upper angle is θ (the lower angle is 90-θ°) .
3. for the first maximum sinθ =λ/d, (sin θ = Opp/Hyp) rearrange to dsin =λ,
4. other maxima occur when the path difference between the two rays of light is nλ, where n is an integer, replace λ with nλ to get: d sin =nλ

Question 26

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

Answer 26

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

Question 27

What is diffraction?

Answer 27

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

Question 28

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

Answer 28

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

Question 29

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

Answer 29

● Wear laser safety goggles
● Don’t shine the laser at reflective surfaces
● Display a warning sign
● Never shine the laser at a person

Question 30

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

Answer 30

w = λD / s
w - fringe spacing
λ - wavelength of light used
D - distance from screen to slits
s - slit separation

Question 31

The maxima formed by shining a laser through 2 slits are 0.04m apart, the slits are 0.2mm apart and the distance from the slits to the screen is 15m, what is the wavelength of the light?

Answer 31

λ = ws / D
= (0.04 x 0.2 x10^-3) / 15
=5.3 x 10^-7 m

Question 32

What is path difference?

Answer 32

The difference in distance travelled by 2 waves.

Question 33

How could you investigate stationary sound waves?

Answer 33

Place a speaker at one end of a closed glass tube, lay powder across the bottom of the tube, it will be shaken from the antinodes and settle at the nodes. The distance between each node is half a wavelength.

Question 34

What is the frequency of the first harmonic of a string length 2m, mass 0.03kg with a mass of 2kg hanging off it?

Answer 34

T = tension = 2 x 9.81 = 19.62N
μ = mass / unit length = 0.03 / 2 = 0.015 kg/m
f = (1⁄4) x (sq.rt: 19.62/0.015)
F =9.0 Hz

Question 35

What is the speed of a wave with frequency 10GHz and wavelength 6cm?

Answer 35

c = fλ
c = (10x10^9) x (6x10^-2)
c = 6x10^8 m/s

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 formed by sound waves and all EM waves too.