Refraction, Diffractions And Inteference

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 are more defined.

Question 3

What was Young’s double-slit experiment?

Answer 3

A single light source is directed towards two slits, 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 secreting intensity with 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 of 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

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

Question 10

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

Answer 10

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

Question 11

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

Answer 11

Parts of the signal’s energy is absorbed but 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 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 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 critical angle?

Answer 19

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

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

Using snell’s law of retraction, 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

N1sin(i = n2sin(r
1.23sin(32 = 1.53sin(r
Sin(r = 1.23sin(32 / 1.53
Sin(r = 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

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

Answer 24

n = c / v

n = refractive index
c = speed of light un vacuum, 3x10^8 m/s
v = speed of light in material

Question 25

State 2 applications of diffraction gratings.

Answer 25

• 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 26

When 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 diffraction more, the orders will become further apart