3.3.2.3 Waves: Refraction at a plane surface

Question 1

How does optical density link with refractive index?

Answer 1

The more optically dense a material is, the higher refractive index it has.

Question 1

Define the phrase ‘optically dense’

Answer 1

The degree to which a material slows down the speed of light.
The more optically dense a medium, the more it slows down light

Question 2

What is absolute refractive index (n)?

Answer 2

The ration of the speed of light in a vacuum with the speed of light in a given material
- It is a property of the specific material ONLY.

Question 3

What is the formula for absolute refractive index?

Answer 3

n = c/c(s)
where c = speed of light in a vacuum

Question 4

What is the refractive index of light in air?

Answer 4

1
*** light in air is a TINY bit slower than in a vaccum

Question 5

What is the refractive index of light in glass?

Answer 5

1.50

Question 6

What is the refractive index of light in water?

Answer 6

1.33

Question 7

What is the refractive index of a boundary?

Answer 7

The relative refractive index between 2 materials, is the ratio of the speed of light in material 1, to the speed of light in material 2.

Question 8

What is the formula for the relative refractive index at a boundary?

Answer 8

1n2 = c1/c2
which gives
1n2 =n2/n1
where n1 = absolute refractive index of material 1
and n2 = absolute refractive index of material 2.

Question 9

What is the incident ray angle?

Answer 9

The angle that incoming light makes with the normal

Question 10

What is the refracted ray?

Answer 10

The refracted ray

Question 11

What is Snell’s law of refraction (equation)?

Answer 11

n₁sinθ₁=n₂sinθ₂
where θ₁ is the angle of incidence
and θ₂ is the angle of refraction

Question 12

Describe Snell’s law

Answer 12

As light moves across a boundary between two materials of different optical densities, its direction changes because its speed changes.
- it bends towards to normal if it goes from a less optically dense to more optically dense medium.
- it bends away from the normal if it goes from a more optically dense to less optically dense medium.

Question 13

What is a critical angle (θc)?

Answer 13

The value of the incident angle, for which the refracted angle = 90 degrees.
* This can happen for any boundary where the light is passing from a more optically dense medium, to a less dense one (n₁ > n₂)

Question 14

What is the equation for the critical angle?

Answer 14

sinθc =n₂/n₁, where (n₁ > n₂)

Question 15

What is total internal reflection?

Answer 15

At angles of incidence greater than the critical angle, refraction cannot happen, hence all light is reflected back into the material (TIR)

Question 16

What is an optical fibre?

Answer 16

A very thin flexible tube of glass or plastic fibre than can carry light signals over long distances or around corners.
** only need to know about step-index optic fibres

Question 17

What are step-index optic fibres?

Answer 17

Have a high refractive index (optically dense) core surrounded by cladding with a lower refractive index than the core to allow TIR.

Question 18

What is the purpose of the cladding?

Answer 18

Protects the fibre from scratches (so light doesn’t escape)
Has lower refractive index, so allows TIR to take place.

Question 19

Why is the fibre narrow?

Answer 19

So that the light going in always hits the boundary between core and cladding at an angle greater than the critical angle, so all light is totally internally reflected from boundary to boundary until it reaches the other end.

Question 20

Why are optic fibres better than copper cables with electricity?

Answer 20

• Light has a high frequency so signal can carry more information
• Light doesn’t heat up the fibre
• No electrical interference
• Fibre-optic cables are much cheaper to produce
• Signal can travel a long way, very quickly, with minimal signal loss.
• Light doesn’t heat up the fibre, so no energy lost as heat.

Question 21

What are the two types of signal degradation in optic fibres?

Answer 21

Absorption and Dispersion

Question 22

Describe signal degradation by apsorption

Answer 22

Absorption is where some of the signal’s energy is absorbed by the material the fibre is made from. Energy loss results in the amplitude of the signal being reduced.

Question 23

Describe signal degradation by dispersion (modal)

Answer 23

Caused by light rays entering the optical fibre at different angles, causing them to take different paths, with rays travelling straight through the fibre arriving quicker than those taking a longer reflected path.

Question 24

How can modal dispersion be reduced?

Answer 24

Using a single-mode fibre, in which light only follows a narrow path and reeuces multipath modal dispersion

Question 25

Describe signal degradation by dispersion (material)

Answer 25

Caused by different amounts of diffraction experienced by different wavelengths of light. Different wavelengths slow down by different amounts in a material. As white light consists of many wavelengths, this causes some parts of the signal to take a longer time to arrive than others.

Question 26

How can material dispersion be reduced?

Answer 26

Using monochromatic light

Question 27

How can degradation from both absorption and dispersion be reduced?

Answer 27

Using optical fibre repeaters to regenerate the signal every so often.

Question 28

How do refractive indexes differ for light of different wavelengths?

Answer 28

The refractive index for red is lower than for blue light, as red light travels faster than blue light, hence n = c/cs, has a bigger denominator, so the refractive index is lower

Question 29

What is the purpose of the core in an optic fibre?

Answer 29

It propagates (guides) the light inside the optic fibre by total internal reflection with low absorption.

Question 30

What are some problems posed by scratches?

Answer 30

Water could enter and increase refractive index, which may be higher than the core, so total internal reflection does not occur and signal leaves the fibre.

Scratch may also alter the angle at which the signal interacts with the core’s boundar, such that the angle is lowered below the critical angle and hence no tir and no signal