Waves and optics- optics (5)

Question 1

Define coherence

Answer 1

Coherent waves have a fixed phase relationship with the same frequency and wavelength

Question 2

Why is a laser useful in showing interference and diffraction?

Answer 2

It produces monochromatic 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 centeral maximum flanked by alternating spectral fringes of dereasing 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 wave are in phase constructive interference occurs making bright fringes and where the waves are antiphase destructive interference occurs making a dark fringe

Question 6

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

Answer 6

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

Question 7

What is the approximate refractive index of air?

Answer 7

1

Question 8

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

Answer 8

Towards the normal

Question 9

When does total internal reflection occur?

Answer 9

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

Question 10

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

Answer 10

• 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 11

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

Answer 11

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

Question 12

What is pulse broadening?

Answer 12

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

Question 13

How does modal dispersion cause pulse broadening?

Answer 13

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 14

What is material dispersion?

Answer 14

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 15

How can modal dispersion be reduced?

Answer 15

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

Question 16

How can material dispersion be reduced?

Answer 16

Use monochromatic light

Question 17

How can both absorption and dispersion be reduced?

Answer 17

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

Question 18

State the advantages of optical fibres over traditional copper wires

Answer 18

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

Question 19

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

Answer 19

It goes along the boundary ie. the ngle of refraction is 90 degrees

Question 20

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

Answer 20

It goes along the boundary ie. the ngle of refraction is 90 degrees

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
C= sin^-1 (1 / 1.33)
C= 48.8 degrees

Question 23

Using snell’s law of refraction, find the angle of reflection in a material with RI = 1.53 when the angle of incidence is 32 degrees from a material with RI = 1.23

Answer 23

n1sin(i) = n2sin(r)
1.23sin(32) = 1.53sin(r)
sin(r) = 1.23sin(32) / 1.53
sin(r) = 0.426
r = 25.2 degrees

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 dsinx = nλ

Question 28

When light passing through a diffraction grating is changed from blue to red, do the others 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 four safety precautions that must be followed when using a laser?

Answer 31

• Wear laser safety goggles
• Don’t shine the laser at refractive 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 the the screen is 15m, what is the wavelength of the light?

Answer 33

λ = ws / D
=(0.04 x 0.2 x 10^-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 power 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

F= 9.0Hz

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 = 6x10^8m/s

Question 38

What is ‘phase’?

Answer 38

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