Rarefaction, 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 wavelength so diffraction and interreference 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 interreference pattern

Question 4

Describe the interreference pattern created using white light

Answer 4

A bright light 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 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 7

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

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 is greater than the critical angle

Question 10

Purpose of cladding

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 received signal?

Answer 11

Part of the signals 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, his 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 model 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 an 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 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 critical angle?

Answer 19

It goes along the boundary

Question 20

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

Answer 20

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

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

Snells law of refraction

Answer 22

n1sini = n2sinr

Question 23

Formula used to determine the refractive index of a material

Answer 23

n = c/v
n = refractive index
c = speed of light in vacuum
v = speed of light in material

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

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

Answer 25

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

Question 26

What is diffraction?

Answer 26

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

Question 27

Double slit experiment providing evidence for the wave nature of light

Answer 27

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

Question 28

4 safety precautions that must be followed when using a laser

Answer 28

*Wear laser safety goggles
*Dont shine the laser at reflective surfaces
*Display a warning sign
*Never shine the laser at a person

Question 29

Youngs double slit experiment formula

Answer 29

w = lambda x D / s
w = fringe spacing
lambda = wavelength of light used
D = distance from screen to slits
s = slit separation

Question 30

What is path difference

Answer 30

The difference in distance travelled by 2 waves

Question 31

What is a phase?

Answer 31

The position of a certain point on a wave cycle