Optics

Question 1

What is refraction?

Answer 1

Refraction is the change of direction that occurs when light crosses a boundary between two TRANSPARENT substances AT AN ANGLE

Question 2

What happens when light travels from a more optically dense medium to a less optically dense medium?

(enters at an angle)

Answer 2

When light travels from a more optically dense medium to a less optically dense medium, it bends away from the normal.

Question 3

What happens when light travels from a less optically dense medium to a more optically dense medium?

(enters at an angle)

Answer 3

When light travels from a less optically dense medium to a more optically dense medium, it bends towards the normal.

Question 4

What happens if the incident ray travels along the normal?

Answer 4

No refraction takes place as there is no change in direction of the ray

Question 5

What does refraction depend on?

Answer 5

It depends on the refractive index of the materials that the ray is leaving and entering

Question 6

What is the refractive index of a material?

Answer 6

The ratio of sin(i) to sin(r), where i is the angle of incidence in air, and r is the angle of refraction in the material.

Question 7

How do you calculate the refractive index of a material?

Answer 7

Moving from air to material:

Refractive index of the material, n = sin(i) / sin(r)
^^This is SNELL’S LAW
where:
i = angle of incidence in a vacuum or air
r = angle of refraction in the material.

Moving from material to air:

Refractive index of the material, sin (i) /sin (r) = 1/n

where:
i = angle of incidence in the material
r = angle of refraction in a vacuum or air

Question 8

The refractive index of a material can also be defined in terms of speed of light in air and the material.

State the equation.

Answer 8

Refractive index of the material, n = c/cₛ

where:
c = 3 x 10⁸ms⁻¹ (speed of light in vacuum)
cₛ = speed of light in substance

*light refers to the whole EM spectrum

Question 9

What happens to the frequency of a wave as it moves between two mediums of different optical density?

Answer 9

Frequency remains the same in both mediums, regardless of their optical densities. This means the change in speed between mediums is directly proportional to the wavelength of the wave. If speed increases, the wavelength also increases.

Question 10

What is the equation to calculate the refractive index of a substance using the wavelength of the wave?

Answer 10

Refractive index of the material, n = λ/λₛ

where:
λ = wavelength of light in vacuum
λₛ = wavelength of light in the material

Question 11

What is the refractive index of air and vacuum?

Answer 11

The refractive index of air and vacuum is 1.

Question 12

If neither substances are air. what happens to Snell’s law?

Answer 12

n = sin(i)/sin(r) becomes n₁sin(i) = n₂sin(r)

where: 
n₁ = refractive index of material 1
n₂ = refractive index of material 2
i = angle of incidence in material 1
r = angle of refraction in material 2

If one of the substances were air, i.e. the incidence ray is in air, then the second equation would change into the first one, as n₁ would become the refractive index of air which is 1.

Question 13

What happens if white light is directed at a prism at an angle?

Answer 13

It will split up into the colours of the visible light spectrum.
Shorter wavelength light e.g. violet/blue will bend towards the normal more than longer wavelength light, like red.
Therefore shorter wavelength light, will travel slower in glass than longer wavelength light.

Question 14

When light is travelling from a less optically dense substance to a more optically dense susbtance what always happens to the light FOR ALL ANGLES OF INCIDENCE?

Answer 14

When light is travelling from a less optically dense susbtance to a more optically dense substance, for all angles of incidence, light is always:

• refracted in the denser susbtance
• while the rest is reflected off the surface

Question 15

When light is travelling from a more optically dense substance to a less optically dense susbtance what happens to the light?

Answer 15

When light is travelling from a more optically dense susbtance to a less optically dense substance, the strength of the rays refracted and (internally) reflected depends on the angle of incidence.

Question 16

When light is travelling from a more dense substance to a less dense susbtance, what happens to the light as the angle of incidence increases?

Answer 16

As the angle of incidence increases, the refracted ray gets weaker (i.e. less waves get refracted) and the internally reflected rays get stronger (i.e. more waves get reflected).

Question 17

What happens at the critical angle?

Answer 17

The critical angle is the angle of incidence for which refraction occurs along the boundary.

Question 18

Define critical angle?

Answer 18

The critical angle for an interface between two transparent substances is the angle of incidence in the denser substance for which the angle of refraction for the less dense substance is 90°

Question 19

What happens if the angle of incidence is increased beyond the critical angle?

Answer 19

The light ray undergoes total internal reflection.

Question 20

What are the two conditions that must be satisfied for total internal reflection to take place?

Answer 20

1) The incident substance must have a larger refractive index than the other substance.
2) The anfle of incidence must be greater than the critical angle.

*Larger refractive index = more optically dense substance

Question 21

Equation and derivation calculate the critical angle?

Answer 21

sin(i꜀) = n₂/n₁

Since, at the critical angle, angle of refraction is equal to 90°, sin(r) in n₁sin(i) = n₂sin(r) becomes sin(90°) = 1. Therefore the equation becomes n₁sin(i) = n₂ x 1, which rearranged gives sin(i) = n₂/n₁ i.e. sin(i꜀) = n₂/n₁

Question 22

Applications of total internal reflection?

Answer 22

Optical fibres in medical endoscopes and communications.

Question 23

What is an optical fibre?

Answer 23

A very thin flexible glass/plastic tube that carries light signals over long distances and around corners/curved paths, using total internal reflection.

Question 24

How do optical fibers allow total internal reflection for communication?

Answer 24

Optical fibres have a high refractive index but are surrounded by cladding of lower refractive index, which allow total internal reflection (Condition 1 satisfied).
Optical fibres are very narrow so that the light always hits the boundary between the fibre/core and cladding at an angle bigger than the critical angle (Condition 2 satisfied).

Question 25

What else does cladding do apart from allowing total internal reflection?

Answer 25

• Provides strength and support for the fibers
• Protects fibre from scratches = lead to light loss (i.e. loss of energy) from fibre which = reduces amplitude of pulses.
• Increases the critical angle. Benefit of this is that only light rays travelling close to the axis of the fibre pass through to the other end as only they will have an angle of incidence greater than the critical angle. Another benefit is that there are fewer total internal reflections + distance travelled is smaller than in the case of multiple reflections (due to smaller critical angle) and so there is less energy loss and the time of transmission is shorter.
  Cladding also prevents light from moving between fibers which in turn prevent them from reaching the wrogn destination.

http://www.schoolphysics.co.uk/age16-19/Optics/Refraction/text/Fibre_optics/index.html

Question 26

Why are optical fibres transparent?

Answer 26

Optical fibres need to be transparent to minimise absorption of light by the fibre which reduces the amplitude of pulses.

Question 27

Optical fibres are used in communication to send signals. However signal degradation can occur. What is a signal and what does signal degradation lead to?

Answer 27

Signals are pulses of light that carry information. Therefore, signal degradation leads to the loss of information.

Question 28

How does signal degradation occur?

Answer 28

1) Absorption of light by optical fibre

2) Dispersion of light in the optical fibre

Question 29

What does absorption of light by the optical fibre mean and what does it result in?

Answer 29

As a signal travels, some of its energy is lost through absorption by the material of the fibre.
The further the signal travels, the greater the loss of energy by absorption.
This causes the amplitude of the signal to decrease.

Question 30

What does dispersion of light in the optical fibre mean?

Answer 30

The spreading of light pulses as they travel in the fibre.

Question 31

What are the two types of dispersion that can occur?

Answer 31

Modal Dispersion

Material Dispersion

Question 32

What is modal dispersion?

Answer 32

Modal dispersion is dispersion caused by light rays entering at different angles and thus take different paths. i.e. light travelling along the axis of fibre travel faster/a shorter distance, than light that repeatedly undergoes total internal reflection. This means the pulses could merge together if they become longer than they should be. This is known as pulse broadening.

Question 33

How to prevent modal dispersion?

Answer 33

It occurs in a wide core/fibre so to prevent it, a very narrow core/fibre must be used.

Question 34

How to prevent loss of amplitude due to absorption?

Answer 34

Use an optical fibre repeater to boost the signal (and thus its amplitude) at periodic positions along the fibre.

Question 35

What is material dispersion?

Answer 35

Material dispersion is dispersion caused by the use of white light rather than monochromatic light. The different wavlengths of light travel at different speeds in the fibrso reach the end of the fibre at different points. For example, red light travels faster than violet light in glass and this diffference in speed causes the pulses of white light to become longer, as the component of violet light falls behind the faster red component of each pulse. This also results in pulse broadening.

Question 36

Draw the effect on amplutide of both absorption and dispersion (modal + material).

Answer 36

ANSWER ON CGP BOOK PG.105

Question 37

What is monochromatic light?

Answer 37

Monochromatic light is light of a single wavelength.

Question 38

How does a medical endoscope work?

Answer 38

Contains two fibre bundles - illumination fibre optic and viewing fibre optic. First, endoscope is inserted into a body cavity, which is then illuminated using light sent through illumination fibre optic.
Lens system at each end of the viewing fibre optic is then used to form an image of the body cavity.
The light produced from the illumination fibre optic lights up the inside of the patient and after being scattered by the patient’s insides, returns via the viewing fibre optic to form an image on the lens outside of the body.

Question 39

What must the viewing fibre optic/second bundle be and why?

Answer 39

The second bundle must be narrow and coherently arranged so that the fibre ends at each end are in the same relative position to prevent the image produced being distorted.

Question 40

We looked at diffraction in the waves topic. Light shone through a narrow slit and will siffract sometimes to produce a diffraction pattern. How can we produce a clear diffraction pattern?

Answer 40

We can produce a clear diffraction patterns by using a light source that produce waves which are monochromatic and coherent.
For example laser light.

Question 41

Monochromatic light is light of a single wavelength (As discussed). Can you remember what coherent waves are?

Answer 41

Waves are perfectly coherent if their frequency and waveform are identical and their phase difference is constant.

Question 42

Describe and draw the diffraction pattern formed when monochromatic and coherent light is shone through a narrow slit.

Answer 42

If the monochroamtic and coherent light is shone through a slit of a size similar to the wavelength of the light:

1) The diffraction pattern formed has a bright central fringe known as the central maximum with alternating bright and dark fringes on either side.
2) The central fringe is twice as wide as each of the outer fringes.
3) The peak intensity of each fringe decreases with distance from the centre.
4) Each of the outer fringes is the same width.

Question 43

What is the diffraction pattern formed due to?

Answer 43

The diffraction pattern formed when light passes through the slit is due interference.

Question 44

The diffraction pattern formed is due to interference - what does this mean?

Answer 44

• The bright fringes of a diffraction pattern are due to constructive interference where waves from across the width of the slit arrive at the screen in phase.
• The dark fringes of a dffraction pattern are due to total destructive interference, where waves from across the width of the slit arrive at the screen in antiphase.

Question 45

What is the diffraction pattern formed when white light is passed through a slit?

Answer 45

White light consists of many colours and thus many different wavelengths. Different wavelengths of light are diffracted by different amounts so you get a spectrum of colours in the diffraction pattern. The diffraction pattern for white light has a​ central white maximum​ with alternating bright fringes which are ​spectra​. The central maximum is twice as large as the other maxima. For the inner fringes, ​violet/blue is closest to the central maximum and red is the furthest away​.
The outer fringes of the pattern merge into an indistinct background of white light becoming fainter with increasing distance from the centre. This is because where the fringes merge, different colours reinforce and therefore overlap. The peak intensity of each fringe decreases with distance from the centre.

*Why is violet on the inner side and red on the outer side?

Question 46

Why is the central maximum/fringe of the white light white?

Answer 46

This is because every colour contributes to the centre of the pattern.

Question 47

If violet/blue light are found on the inner side/side closest to the central maximum of eahc fringe and red light is found on the other end, what does this tell us?

Answer 47

Unsure. What does it tell us about wavelengths and amount of diffraction.

Question 48

Which part is the brightest part of the single slit diffraction pattern for a monochromatic, coherent light source? What does this tell you about the intensity here?

Answer 48

The central maximum is the brightest part of the single slit diffraction pattern. This tells us the intensity of the light is greatest here (specifically at the centre of the central maximum).

Question 49

What is intensity?

Answer 49

Intensity is the power per unit area.

*Power is rate of energy transfer

Question 50

For monochromatic light, all photons have the same _______, therefore an increase in intensity at the central maximum means?

Answer 50

An increase in the number of photons hitting the area each second (in comparison to other fringes).

Question 51

What does the width of the central maximum for a single slit diffraction pattern depend on - tell me how you can increase it?

Answer 51

• The greater the wavelength, the greater the diffraction that occurs and thus the greater the width of the central maximum.
• The narrower the gap, the greater the diffraction that occurs and thus the greater width of the central maximum.

What about slit-screen distance???

Question 52

If the central maximum becomes wider, what happens to the intensity of the central maximum?

Answer 52

The intensity of the central maximum decreases.

Question 53

What is the equation to calculate the width of the central maximum for a single slit (or double slit) diffraction pattern?

Answer 53

W = λ2D/a

W = width of central maximum (m_
λ = wavelength of light (m)
a = width of single slit (m)
D = slit-screen distance

*Do we need to know this equation?

Question 54

What is two source interference?

Answer 54

Two source interference is when two sources interfere to produce a pattern (NOT A DIFFRACTION PATTERN - this is a seperate sub-topic).

Question 55

To get clear interference patterns from two-source interference, the two waves must be?

Answer 55

Monochromatic and coherent

Question 56

Whether you get constructive or destructive interference from two-source interference at a point depends on what?

Answer 56

It depends on the path difference of the two waves that are interfering.

Question 57

What is path difference?

Answer 57

Path difference is the amount by which the path travelled by one of the waves that are interfereing is longer than the path travelled by the other.

Question 58

For two source interference, when do you get constructive interference and what is this called?

Answer 58

When the path difference between the two waves = nλ (where n = 0, 1, 2…), so when the waves are in phase.
This produces a maxima.

Question 59

For two source interference, when do you get destructive interference and what is this called?

Answer 59

When the path difference between the two waves = (n+1/2)λ (where n = 0, 1, 2…), so when the waves are in phase.
This produces a minima.

Question 60

To see two source interference with light, you can either:

Answer 60

You can use two coherent, monochromatic light sources or you can use one coherent, monochromatic light source and pass it through two slits.

*The first method is just two-source interference, the second method is double slit diffraction method. Both produce the same diffraction pattern, but one of them involves diffraction of waves after passing through a gap. All the same things apply as two source interference still.

Question 61

If using two slits to look at two source interference, what do you need to remember?

Answer 61

The slits need to be of similar size to the wavelength of the monochromatic source, so that they diffract.

Question 62

Describe and draw the diffraction pattern formed when monochromatic, coherent light is passed through a double slit?
This is the same pattern that is formed for two-source interference.

Answer 62

Alternating bright and dark fringes. The fringes are evenly spaced, equal widths and parallel to the double slits. The peak intensity of each fringe decreases with distance from the centre (more gradual than single slit).

Question 63

What is the diffraction pattern formed when white light is passed through a double slit?

Answer 63

The diffraction pattern for white light has a​ central white maximum​ with alternating bright fringes which are ​spectra​. The central maximum has the same width as the subsidiary maximas. For the inner fringes, ​violet/blue is closest to the central maximum and red is the furthest away​. The outer fringes of the pattern merge into an indistinct background of white light becoming fainter with increasing distance from the centre. This is because where the fringes merge, different colours reinforce and therefore overlap. The peak intensity of each fringe decreases with distance from the centre (more gradual than single slit).

Question 64

How does the diffraction pattern formed by a single slit different to a double slit for white light?

Answer 64

For single slit, the central maximum is much brighter (much greater intensity) than the subsidiary maximas.
For single slit, the central maxima has double the width of the subsidiary maximas.
For singel slit, the peak intensity of each fringe from the centre decreases much more rapidly, than the double slit.

Question 65

When using lasers what safety precautions must be taken?

Answer 65

• Never shine a laser towards a person
• Wear laser-safety goggles
• Don’t shine at reflective surfaces
• Have a warning sign on display
• Turn off when unecessary.

Question 66

What is Young’s double-slit formula?

Answer 66

w = λD/s

w = Fringe spacing (distance between two adjacent maxima or minima in m)
λ = wavelength (in m)
D = slit-screen distance (in m)
s = distance between slits (in m)

Question 67

Why must the two slits in double slit diffraction need to be close enough?

Answer 67

So that the diffracted waves overlap sufficiently on the screen.

Question 68

If you don’t have a coherent source of light, what can you do to carry out Young’s Double slit experiment?

Answer 68

If you do not have a coherent source of light for example a light bulb, you could place a single slit before the double slit to make the light have a fixed path difference, and a filter to make the light monochromatic.

Question 69

What is a diffraction grating?

Answer 69

A slide containing many ​equally spaced slits very close together.

Question 70

What happens when monochroamtic light is passed through a diffraction grating?

Answer 70

Same interference pattern as double slit, but the fringes are brighter, sharper and narrower.

Question 71

Why is the interference pattern sharper?

Answer 71

This is because the fringes are reinforced by many more rays of light.

Question 72

What is the zero order beam?

Answer 72

This is the central beam which is in the same direction as the incident beam. The ohter transmitted beams are numbered outwards from the zero order beam.

Question 73

What is the name of the beams on each side.

Answer 73

The first pair of beams on either side of the zero order beam is the first order beam. The next pair is second order beam and so on.

Question 74

What is the diffraction grating equation?

Answer 74

dsinθ = nλ

d = distance between slits (in m)
sinθ = angle to the normal made by a maximum (in degress or radians) - this is the angle of diffraction. 
n = order of maximum
λ = wavelength of wave (in m)

Question 75

You need to be able to derive the diffraction gradient equation. Derive it.

Answer 75

1) The angle between the first order maximum and the incoming light is defined as θ.
2) The triangle formed in the figure above, has the same angle θ (using basic trig), and the angle opposite it is the path difference (in this case, λ), and the hypotenuse of this triangle is the slit spacing.
3) So for the first maximum we can say that sinθ = λ/d –> dsinθ =λ
4) The first order maximm occurs when teh path differnce is 1λ, the second order maximum oxxurs when the path differece is 2λ, so the general equal becomes dsinθ = nλ, where n is the order.

Question 76

How can you calculate the number of slits per metre on the grating?

Answer 76

N = 1/d

N = The nuber of slits per meter on the grating
d = grating space /distance between slits
*1 refers to 1 metre.

Question 77

For a given order and wavelength, the smaller the value of d…

Answer 77

…the greater the angle of diffraction = the greater the value of sinθ and thus the greater the value of θ. In other words, the larger the no of slits per metre the greater the angle of diffraction.

Question 78

How can you calculate the maximum number of orders produce?

Answer 78

Substiute θ = 90degrees (sinθ = 1) in the grating equation and caluclate n using n = d/λ (Round down to nearest whole number)

Question 79

Applications of diffraction gratings includes splitting up light from stars and x-ray crystallography. How does splitting up light from stars work?

Answer 79

• You split up light from ​stars​ using a diffraction grating to get different orders in which each order become a spectrum. These spectrums can be analysed to find dark spectral lines that indicate which wavelengths of light has been absorbed. Different stars have different elements in their atmosphere, and thus produce different absorption spectra which can then be compared with the spectra from the diffraction gratings to identify which star is being observed.

Question 80

Applications of diffraction gratings includes splitting up light from stars and x-ray crystallography. How does x-ray crystallography work?

Answer 80

• X-ray crystallography​, which is where x-rays are directed at a thin crystal sheet which acts as a diffraction grating to form a diffraction pattern, this is ​​because the wavelength of x-rays is similar in size to the gaps between the atoms. This diffraction pattern can be used to measure the atomic spacing in certain materials.

Question 81

Why is diffraction gratings used to split up light from stars than a prism?

Answer 81

This is because it is much more accurate.