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 and 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 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

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

Answer 21

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

Question 22

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

Answer 22

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

Question 23

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 23

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

Question 24

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

Answer 24

Glass.

Question 25

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

Answer 25

n = c / v
n = refractive index
c = speed of light in vacuum, 3x 10^8 m/s
v = speed of light in material

Question 26

State 2 applications of diffraction
gratings

Answer 26

● 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 27

Derive the formula dsinθ = n�

Answer 27

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 =n�

Question 28

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

Answer 28

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

Question 29

What is diffraction?

Answer 29

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

Question 30

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

Answer 30

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

Question 31

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

Answer 31

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

Question 32

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

Answer 32

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

Question 33

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 33

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

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
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 36

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 36

T = tension = 2 x 9.81 = 19.62N
μ = mass / unit length = 0.03 / 2 = 0.015 kg/m
f = (¼) x (sq.rt: 19.62/0.015)
F =9.0 Hz

Question 37

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

Answer 37

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

Question 38

What is ‘phase’?

Answer 38

The position of a certain point on a wave
cycle, (units are radians, degrees or
fractions of a cycle).

Question 39

True or False:
‘Only light can produce interference
patterns’

Answer 39

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