Optics: Questions Flashcards

1
Q

What is light?

A

Light is both a particle (photon) and an electromagnetic wave that has a given energy, wavelength, and frequency.

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2
Q

What are the wavelengths of visible light?

A

400 - 700 nm

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3
Q

What are the frequencies of visible light?

A

10¹⁴ - 10¹⁵ Hz

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4
Q

What is Maxwell’s equation for the (charged) source of an electric field in a vacuum?

A

∇ = del operator
E = electric field

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5
Q

What is Maxwell’s equation for the (charged) source of a magnetic field in a vacuum?

A

∇ = del operator
B = magnetic field

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6
Q

What is Maxwell’s equation for electromagnetic induction (Faraday’s Law)?

A

∇ = del operator
E = electric field
B = magnetic field

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7
Q

What is Maxwell’s equation for Ampere’s Law?

A

∇ = del operator
B = magnetic field
E = electric field
ε₀ = dielectric constant in a vacuum
µ₀ = permeability in a vacuum
c = speed of light

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8
Q

How can the wave equation for an electric field in a vacuum be derived using Maxwell’s equations?

A

1) Apply ∇ x to Faraday’s Law.
2) Use the identity ∇ x (∇ x A) = ∇ (∇ . A) - ∇² A for the LHS of this equation.
3) Knowing that ∇ . E = 0 and using Maxwell’s equation for Ampere’s Law, the wave equation can be found.

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9
Q

What is the wave equation for an electric field?

A

∇ = del operator
E = electric field
c = speed of light

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10
Q

How can the wave equation for a magnetic field in a vacuum be derived using Maxwell’s equations?

A

1) Apply ∇ x to Ampere’s Law.
2) Use the identity ∇ x (∇ x A) = ∇ (∇ . A) - ∇² A for the LHS of this equation.
3) Knowing that ∇ . B = 0 and using Maxwell’s equation for Faraday’s Law, the wave equation can be found.

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11
Q

What is the wave equation for a magnetic field?

A

∇ = del operator
B = magnetic field
c = speed of light

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12
Q

What is the speed of light in a vacuum?

A

c₀ = speed of light in a vacuum

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13
Q

What is the equation for the wavelength in a vacuum?

A

λ₀ = wavelength in a vacuum
c₀ = speed of light in a vacuum
ν = frequency

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14
Q

What is the equation for the wavelength in a medium?

A

λ’ = wavelength in a medium
c’ = speed of light in a medium
ν = frequency

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15
Q

What is the solution to the wave equation in one dimension?

A

E₀ = amplitude (could also be B₀ for a magnetic wave)
k = wavenumber = 2π/λ
ω = angular frequency = 2πν

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16
Q

What is the solution to the wave equation in three dimensions?

A

E₀ = amplitude
k = wave vector
ω = angular frequency = 2πν

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17
Q

What is the solution of the wave equation for a plane wave?

A

E₀ = amplitude
r = (x, y, z)
k = wave vector
ω = angular frequency = 2πν
c.c = complex conjugate

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18
Q

What is the solution of the wave equation for a spherical wave?

A

E₀ = amplitude
r = |r| = √(x² + y² + z²)
k = wave vector
ω = angular frequency = 2πν
c.c = complex conjugate

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19
Q

Plane waves have planes of _______ phase. The wave moves ____________ to these planes.

A

Constant
Orthogonally

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20
Q

Define the Poynting vector

A

A vector that describes the flow of energy. It is orthogonal to the electric and magnetic fields in an EM wave and points in the direction of travel of the wave.

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21
Q

What is the equation for the Poynting vector?

A

S = Poynting vector
E = electric field
B = magnetic field
µ₀ = permeability in a vacuum

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22
Q

Define the index of refraction

A

The ratio between the speed of light in a vacuum and the speed of light in a medium.

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23
Q

What is the equation for the index of refraction?

A

c = speed of light in a vacuum
c’ = speed of light in a medium
εᵣ = ε/ε₀ = dielectric constant in the medium
µᵣ = µ/µ₀ = permeability of the medium

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24
Q

What is Huygen’s Principle?

A

Each point on a wavefront is a source for a secondary, spherical wave. Cumulatively, the waves produced from these sources will form an envelope and produce a new wavefront.

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25
Q

What is Snell’s Law (the refraction law)?

A

When a ray of light travelling through an area with refractive index, n₁, hits an area with a different refractive index, n₂, the ray is refracted (or in some cases the ray splits into a reflected and refracted part).

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26
Q

What is the equation for Snell’s Law?

A

n₁ = refractive index of medium 1
n₂ = refractive index of medium 2
θ₁ = angle of incidence
θ₂ = angle of refraction

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27
Q

Give 3 examples of the applications of refraction in optical components

A
  • Lenses
  • Prisms
  • Optical fibres
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28
Q

One implication of Snell’s Law is that the angle of _________ is greater than the angle of _________ because n₁ > n₂.

A

Refraction
Incidence

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29
Q

Does the angle of incidence or the angle of refraction reach 90º first?

A

The angle of incidence

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30
Q

Define total internal reflection

A

When the angle of refraction is larger than the critical angle, the light will be fully reflected into the medium.

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31
Q

Give 4 examples of the applications of total internal reflection in optical components

A
  • Optical fibres
  • Beam splitters
  • Prisms
  • Polarisation optics
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32
Q

What is the equation for the critical angle?

A
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33
Q

Prisms use _______ ___ to separate light into its coloured components (spectroscopy).

A

Snell’s law

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34
Q

What are dispersive media?

A

Strongly absorptive materials whose index of refraction are dependent on wavelength. These materials are used for prisms.

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35
Q

Optical fibres combine materials with different _____ __ ________ in the core (centre) and the cladding (environment) so that, under the right angle of incidence, all of the light is ________ and stays in the core.

A

Index of refraction
Reflected

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36
Q

What are optical fibres used for?

A
  • Telecom applications
  • Lasers
  • Data transport
  • Endoscopes
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37
Q

Measurements of the critical angle can be used to calculate the ________ ______ of a material using the ____ _________.

A

Refractive index
Abbe refractometer

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38
Q

What is Fermat’s Principle (principle of least time)?

A

Light always takes the fastest way e.g. the way which minimises the total time for the distance. This is a variation all principle.

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39
Q

What is the equation for Fermat’s Principle?

A

t = time
d = distance
nd = L = optical path length (n = local refractive index)
c’ = speed in a medium
c = speed in a vacuum

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40
Q

What is the equation used to calculate the optical path length?

A

The minimal value of this equation is found.

L(r) = optical path length
n(r) = refractive index
A = initial point
B = final point

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41
Q

In a homogeneous medium (one with completely uniform composition), the chosen path length will be _ ______ ____. This means that ∂L = _.

A

A straight line
0

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42
Q

Is Fermat’s Principle reversible? Why?

A

Yes

For two arbitrarily chosen points, the path taken will be the same even no matter which point the light goes to/from.

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43
Q

What path will light take if there multiple paths that are all the same (minimum) optical path length?

A

The light will take all of the paths.

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44
Q

How can Snell’s Law be derived using Fermat’s Principle?

A

1) Write the optical path length equation in terms of the refractive index and length of each beam in each medium (using Pythagoras’ theorem for the length).
2) Differentiate this equation and set it equal to 0 to find the minimal length.
3) Use trigonometry to write this new equation in terms of angles. This is Snell’s Law.

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45
Q

What is a real image?

A

An image formed of converging rays. Light passes through these images so they can be projected onto a screen. Real images are always inverted.

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46
Q

For a lens, real images form ______ the lens. For a spherical mirror, real images form __ ___ ____ ____ __ the mirror.

A

Behind
On the same side as

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47
Q

What is a virtual image?

A

An image formed of rays that diverge from the optical system. The extension of these rays intersect to form a virtual image so light does not pass through it. This means that the image cannot be projected into a screen. Virtual images are always upright.

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48
Q

For a lens, virtual images form __ ___ ____ ____ __ the lens. For a spherical mirror, virtual images form ______ the mirror.

A

On the same side as
Behind

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49
Q

The position of a real image from the optical system, s’, is ____ _____ (_) 0. This means that the distance to the image is ________.

A

Greater than (>)
Positive

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50
Q

The position of a virtual image from the optical system, s’, is ____ _____ (_) 0. This means that the distance to the image is ________.

A

Less than (<)
Negative

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51
Q

What is the optical axis?

A

The line passing through the centre of the optical system. This axis is perpendicular on the reflecting/refracting surfaces of the system. This means that if an incident ray travels along the optical axis then it will be reflected/refracted by 0º.

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52
Q

What is paraxial approximation?

A

When only ray close to the optical axis are considered so that only small angles are dealt with. This means that the small angle approximation tan θ ~ sin θ ~ θ can be used.

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53
Q

What is the small angle approximation used in paraxial approximation?

A

tan θ ~ sin θ ~ θ

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54
Q

Where are parallel rays focused on a parabolic mirror?

A

A parabolic mirror has a curve of y² = 2px and all incoming parallel rays are collected in one focus, F, where F = p/2.

p = radius of mirror

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55
Q

How are spherical mirrors different to parabolic mirrors?

A

Spherical mirrors don’t focus all of the incoming rays into one focus, unlike the parabolic mirror, only the ones close to the optical axis. Spherical mirrors are also easier to make than parabolic mirrors.

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56
Q

Where are spherical mirrors good approximations for parabolic mirrors?

A

When rays are close to the optical axis because paraxial approximation can be used (tanθ ~ sinθ ~ θ).

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57
Q

What is a concave mirror?

A

A mirror whose reflecting surface is curved inwards towards the source of light.

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58
Q

What is a convex mirror?

A

A mirror whose reflecting surface is curved outwards away from the source of light.

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59
Q

What are the sign conventions for the radius and focal point of a concave mirror?

A

r > 0
f > 0

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60
Q

What are the sign conventions for the radius and focal point of a convex mirror?

A

r < 0
f < 0

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61
Q

What are the general sign conventions for mirrors?

A

All distances along the natural beam path of a ray are counted positive. This means that, as rays are reflected, a focal point is counted as positive if it is on the same side of the mirror as the incident light.

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62
Q

What is the imaging equation?

A

s = distance from object to optical element
s’ = distance from image to optical element
f = r/2 = focal length

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63
Q

How can the imaging equation be derived using the properties of a spherical mirror and paraxial approximation?

A

1) Calculate the angles between a given point on the mirror, y, and the object (β), the radius of the mirror (γ), and the image (δ).
2) Use trigonometric rules to relate all of these angles/ratios to another angle, α.
3) Divide the ratio equation by y to form the imaging equation.

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64
Q

What is the equation for magnification?

A

M = magnification
h’ = image height
h = object height
s’ = distance from image to optical element
s = distance from object to optical element

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65
Q

Negative magnification means that the image is ________ and _____ (s, s’ >0).

A

Inverted
Real

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66
Q

Positive magnification means that image is ______ and _____ (s’ < 0).

A

Virtual
Upright

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67
Q

What are 3 rays should be considered when ray tracing for a mirror? What path do they take?

A
  • Parallel rays: come out of infinity and are turned into rays that pass through the focal point.
  • Central rays: pass through the centre of the mirror and are reflected back into the same path.
  • Focal rays: pass through the focal point and are turned into parallel rays.
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68
Q

Convex mirrors can only form ______ images. For a convex mirror, f _ 0 and s’ _ 0.

A

Virtual
<
<

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69
Q

Concave mirrors can form _____ __ _____ images as given by the imaging equation.

A

Real or virtual

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70
Q

What is the general sign convention for a transparent spherical surface?

A

All distances along the natural beam path are counted positive, meaning that distances behind the optical element are counted positive. This means that a convex element has a negative curvature, r, and a concave element has a positive curvature, r.

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71
Q

What is the imaging equation for image formation at a spherical surface?

A

n₁, n₂ = refractive index
s = distance from object to optical element
s’ = distance from image to optical element

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72
Q

How can the imaging equation for image formation at a spherical surface be derived using paraxial approximation and Snell’s Law?

A

1) Calculate the angles between a given point on the mirror, y, and the object (φ₁), the radius of the mirror (γ), and the image(φ₂).
2) Use trigonometric rules to relate all of these angles/ratios to another angle, α, on the same side as the image and an angle, β, on the same side as the image.
3) Substitute these angles into Snell’s Law and use paraxial approximation.
3) Write this equation in terms of ratios and divide by y to form the imaging equation.

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73
Q

What is the magnification equation for a spherical refractive surface?

A

M = magnification
h’ = image height
h = object height
n₁, n₂ = refractive index
s’ = distance from image to optical element
s = distance from object to optical element

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74
Q

Why is there a minus sign in the magnification equation for a spherical refractive surface?

A

It is convention because the a image is inverted.

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75
Q

How can the focal point on either side of a spherical refractive lens be found?

A

1) Set s (or s’) to infinity.
2) Substitute all known values into the imaging equation for image formation at a spherical surface.
3) rearrange to find s’ (or s), which is equal to the focal length.

76
Q

What is the focal length of a thin lens that is curved on both sides?

A

f = total focal length
f₁ = focal length (side 1)
f₂ = focal length (side 2)

77
Q

How can the focal length of each side of a thin lens be found?

A

1) Set s₁’ (or s₂) to infinity.
2) Substitute all known values into the imaging equation for image formation at a spherical surface.
3) rearrange to find s₁ (or s₂’), which is equal to the focal length.

78
Q

What is the imaging equation for a thin lens in a medium?

A

n₁ = refractive index of the medium
n₂ = refractive index of the lens
r₁ = radius of lens (side 1)
r₂ = radius of lens (side 2)
s = distance from object to optical element
s’ = distance from image to optical element
f = focal length

79
Q

What is Lensmaker’s equation?

A

n₁ = refractive index of the medium
n₂ = refractive index of the lens
r₁ = radius of lens (side 1)
r₂ = radius of lens (side 2)
f = focal length

80
Q

What are 3 rays should be considered when ray tracing for a thin lens? What path do they take?

A
  • Parallel rays: come out of infinity and are turned into rays that pass through the focal point.
  • Central rays: pass through the centre of the lens and will not be refracted so will carry on travelling in a straight line.
  • Focal rays: pass through the focal point and are turned into parallel rays.
81
Q

What is the magnification equation for a thin lens?

A

m = magnification
m₁ = magnification of lens (side 1)
m₂ = magnification of lens (side 2)
n₁ = refractive index of the medium
n₂ = refractive index of the lens
s = distance from object to optical element
s’ = distance from image to optical element

82
Q

What are aberrations?

A

When a perfect image cannot be formed because the approximations made in calculations have been violated.

83
Q

Define monochromatic aberration

A

A type of aberration that occurs when the paraxial approximation is violated (because rays are not concentrated around the centre of the imaging system).

84
Q

Define astigmatism

A

A type of aberration where the foci for the horizontal and vertical are in different positions.

85
Q

Define chromatic aberration

A

The focus of a lens depends on the refractive index of the materials used (shown through Lensmaker’s equation). When dispersive materials are used, the index of refraction depends of the wavelength incident on the lens, hence, different wavelengths have different focal lengths and aberrations occur.

86
Q

How can the effects of chromatic aberrations be minimised?

A

By combining different types of lenses or by using lenses where the index of refraction varies with the radius.

87
Q

What is the method of image-object propagation?

A

The method of forming an image for a set of lenses. The object forms an image through the first lens which acts as the image for the second lens and so on until a final image has been obtained.

88
Q

What are the 3 main examples of optical instruments?

A
  • Magnifying glass
  • Microscopes
  • Telescopes (Keplerian and Galilean types)
89
Q

What is the near point for the human eye?

A

The distance that an object can be seen with magnification M = 1 when being viewed by the human eye; it is 25cm.

90
Q

What is the equation for the visual angle (or angular diameter / apparent size) of an object?

A

h = object height
ε₀ = visual angle

91
Q

If an object is placed so that s > 25cm, it appears to be _______. If it is placed so that s < 25cm, it appears to be ________.

A

Demagnified
Larger

92
Q

What is the angle resolution for a normal eye?

A

1/60 degree = 0.3 mrad ~ 1mm over 3m

93
Q

What is the shortest distance a young eye can focus on?

A

~ 10cm

94
Q

What is the equation for the (angular) magnification for an optical instrument?

A

MP = magnification for the instrument

95
Q

What is the equation for the (angular) magnification for an optical instrument when placed at the eyes focal point?

A

ε = h/f = visual angle with instrument
ε₀ = visual angle without instrument
h = object height
f = focal length
s₀ = near point

96
Q

What are the two lenses found in a microscope?

A
  • Objective lens
  • Eyepiece lens (ocular)
97
Q

How does a microscope work?

A

The object is positioned close to the focal point of the objective lens, and an intermediate image is created inside the tube of the microscope. This image is then magnified again by the eyepiece lens which creates a virtual image at infinity.

98
Q

Why does a microscope use a lens with a wide opening angle? What are the downsides of this?

A

To collect as much information as possible.

This means that paraxial approximation is often violated.

99
Q

What is the objective lens of a microscope?

A

A microscope lens with a short focal length that creates a real image. The object is positioned close to the focal point to create a large image.

100
Q

What is the eyepiece lens?

A

A microscope lens that is used to observe the image created by the objective lens; it acts as a magnifying glass and produces a virtual image at infinity.

101
Q

What is the equation for the numerical aperture of a magnifying glass?

A

NA = numerical aperture
n = refractive index
θₘₐₓ = the half-angle of the maximum light cone picked up by the objective lens.

102
Q

What is the numerical aperture in air?

A

1

103
Q

What is the equation for magnification by the objective lens?

A

M = magnification
s’ = distance from image to lens
s = distance from object to lens
t = tube length
f = focal length

104
Q

What is the equation for magnification by the eyepiece lens?

A

f = focal length
s₀ = near point

105
Q

What is the equation for the total magnification of a microscope?

A

t = tube length
f = focal length
s₀ = near point

106
Q

How does a telescope work?

A

A converging lens obtains an image of an object that is a large distance away and an additional eyepiece lens is used to increase the size of the image further.

107
Q

What is the objective lens of a telescope?

A

A lens with a long focal length that creates a real image approximately at the focal length. The object is far away so s ~ infinity.

108
Q

What is the equation for the total magnification of a telescope?

A

f = focal length
s₀ = near point

109
Q

What is a Kepler telescope?

A

A telescope that consists of two converging lenses positioned at a distance L = f₁ + f₂ from each other. The total magnification is negative so the image created is inverted.

110
Q

What is a Galileo telescope?

A

A telescope that consists of a converging objective lens and a diverging eyepiece lens. It creates an upright image and the overall length of the telescope is shorter than a Kepler telescope because f₂ is negative.

111
Q

What are ray transfer matrices?

A

A practical method of calculating the beam path and position of a ray at each point on the path.

112
Q

What is the vector used to describe the initial position of a ray?

A

u = ray vector
y = distance from the optical axis
α = angle with respect to the optical axis

113
Q

What is the transfer matrix equation that changes an initial ray vector into the final ray vector?

A

u = initial vector
u’ = final vector
y = initial distance from the optical axis
y’ = final distance from the optical axis
α = initial angle with respect to the optical axis
α’ = final angle with respect to the optical axis
A, B, C, D = elements of the transfer matrix

114
Q

What is the equation for the transfer matrix of a complex optical system?

A

M = matrix

115
Q

What happens to the transfer matrix when A = 0?

A

y’ = Bα₀

The final height is independent of the initial height. All incident rays coming at a fixed angle converge at the same point to form an image.

116
Q

What happens to the transfer matrix when B = 0?

A

y’ = Ay₀

All incident rays emerging from one point pass through the optical system and all converge at the same point, forming an image.

This shows that A must represent the magnification of the image.

117
Q

What happens to the transfer matrix when C = 0?

A

α’ = Dα₀

All incident rays coming at a given angle, α₀, will all leave the optical system at the same angle as they are independent of y₀. These rays form a parallel beam on both sides of the optical system.

This shows that D must represent the angular magnification.

118
Q

What happens to the transfer matrix when D = 0?

A

α’ = Cy₀

All rays emerging from a given point will leave the optical system at the same angle as y₀ is fixed, forming a parallel beam.

119
Q

Define geometrical optics

A

Optics involving light as straight rays that always add up; sharp shadows; perfect imaging; and streams of particles known as photons. This is Newtonian optics where light is thought of as a particle.

120
Q

Define wave optics

A

Optics involving the solutions of the wave equation; Huygens’ principle; and interference (patterns that can only be explained by the wave nature of light). This is Huygens’ optics where light is considered as a wave.

121
Q

What is meant by wave-particle duality?

A

Light behaves as both a wave (physical optics) and a particle (geometrical optics), following discussions by Huygens and Newton and experiments by Einstein and Young.

122
Q

Define interference

A

The creation of light patterns that cannot be explained with rays, only by the wave nature of light.

123
Q

What are the 5 key applications of interference?

A
  • Anti-reflection coatings
  • Sensitive measurement devices (like gyroscopes)
  • Distance measurements
  • Interferometers
124
Q

What is the equation for the intensity of light?

A

I = intensity
E₀ = amplitude
c = speed of light
ε₀ = permittivity in a vacuum

125
Q

What is the principle of superposition?

A

When two (or more) waves meet at one point in space, their solutions can be added to form a new solution with the total output depending on their relative phase (if they have the same frequency).

126
Q

What is the equation for the intensity of two superposed waves?

A

I = intensity
E(r) = wavefunction

127
Q

What is the equation for the intensity of two superposed waves with the same frequency but different phases (E₁ = √(I₁) exp(iφ₁) and E₂ = √(I₂) exp(iφ₂))?

A

E₁ = √(I₁) exp(iφ₁)
E₂ = √(I₂) exp(iφ₂)
E₁* = √(I₁) exp(-iφ₁)
E₂* = √(I₂) exp(-iφ₂)

128
Q

What is the interference equation for two waves with the same frequency but different phases?

A

I(r) = intensity
∆φ = phase difference

129
Q

What is the phase difference for fully constructive interference?

A

∆φ = 0

130
Q

What is the phase difference for fully destructive interference?

A

∆φ = π

131
Q

What is the intensity for fully constructive interference?

A

Maximum intensity: I(r) = 4I₀

132
Q

What is the intensity for fully destructive interference?

A

Minimum: I(r) = 0

133
Q

What is the average intensity over all phases for two interfering waves of the same phase?

A

I(r) = 2I₀

134
Q

Describe the double slit experiment

A

The double slit experiment is based on Huygens’ elementary waves. Two parallel slits are illuminated with a coherent plane wave and coherent circular waves emerge. These overlap at several points and form an interference pattern due to the constructive and destructive interference.

135
Q

For the double slit experiment, the observed intensity depends on the ______ _____ of the two waves.

A

Relative phase

136
Q

In the double slit experiment the relative phase difference depends on the difference in ______ ____ _______.

A

Optical path length

137
Q

What is the equation that relates phase difference to optical path length?

A

Phase difference = ∆φ
Path difference = ∆r
Wavelength = λ

138
Q

What is the path length difference for constructive interference?

A

∆r = path length difference
λ = wavelength

139
Q

What is the path length difference for destructive interference?

A

∆r = path length difference
λ = wavelength

140
Q

What is the path length difference for a given angle, θ, between two paths emerging from two slits?

A

∆r = path length difference
d = slit separation
θ = angle

141
Q

Define far field approximation?

A

When a screen that an interference pattern is being projected onto is much further away from the slits than the separation distance between the two slits, the curvature of the waves is neglected and they are approximated as plane waves.

142
Q

What is the equation used to find points of constructive interference on a screen?

A

y = L tanθ = position on screen (tan θ ~ sin θ ~ θ)
L = perpendicular distance from slit to screen

This equation uses ∆r = d sinθ

143
Q

What is the diffraction angle?

A

The angle, θ, for the first-order interference pattern (when m=1). It is given by the equation θ ~ sin θ = λ/d.

144
Q

What is the equation for phase difference in terms of wavenumber and path difference?

A

k = 2π/λ = wavenumber
∆r = path difference

145
Q

What happens when the double-slit experiment is carried out with polychromatic light (a light source that emits several wavelengths)

A

The waves with the same wavelength are coherent and form an interference pattern which all add up on the screen and form a complex interference pattern.

146
Q

Define coherence

A

Light is a stream of photons where each photon behaves like a wave. For an interference pattern to be observed, all photons must be the same.

147
Q

Define temporal coherence

A

For a clear interference pattern to be observed, all photons must have the same wavelength (colour). This is because there has to be a fixed phase relation between different photons.

A source is temporally coherent if two points on the wavefront are emitted at different times and have a fixed phase relation.

148
Q

Define spatial coherence (or longitudinal coherence)

A

For a clear interference pattern to be observed, all photos must propagate in the same direction. This is because there has to be a fixed phase relation between two emitting points.

A source is spatially coherent if two points along the wavefront at different positions but at the same time have a fixed phase relation.

149
Q

Fermat’s principle states that optical paths are reversible. This means that they must also be invariant under ____ reversal.

A

Time

150
Q

What are the coefficients r and t in terms of an incident field? How do they relate to a field with amplitude E₀?

A

r = reflection coefficient
t = transmission coefficient

rE₀ = amplitude of reflected field
tE₀ = amplitude of transmitted field

151
Q

What are the Stokes relations?

A

t = transmission coefficient
t’ = transmission coefficient entering from the other plane
r = reflection coefficient
r’ = reflection coefficient entering from the other plane

152
Q

What is a beam splitter?

A

A material that partly transmits and partly reflects a beam of light.

153
Q

Using Stokes relations describe how changing the entrance direction of a beam changes the direction light is reflected/transmitted in.

A
154
Q

What is an interferometer?

A

A highly sensitive device that splits a laser beam into two parts so that the phase difference of the two parts can be compared when they recombine later. This allows for distances to be measured.

155
Q

What are the applications of an interferometer?

A
  • Distance measurement
  • Position stabilisation
  • Gravity sensing
  • Gravitational wave detection
  • Vibrational sensing
  • Rotation sensing
156
Q

Describe how the Mach-Zehnder interferometer works to determine the phase shift due to an optical element

A

The beam is initially split into two spatially separated (but otherwise equivalent) paths. One of these paths typically has an optical element within it, the other acts as a ‘phase reference’ for the phase-shifted beam.

157
Q

What are the wave equations for each path in the Mach-Zehnder interferometer?

A
158
Q

Assuming that an equal amount of light is reflected and transmitted, what is the equation for the intensity on the positive and negative ports of the Mach-Zendler interferometer?

A
159
Q

How can the total intensity from the beams in a Mach-Zehnder interferometer be calculated?

A

I₀ = I₊ + I₋

160
Q

Describe how the Michelson interferometer works

A

The Michelson interferometer is an interferometer where the two arms are collapsed so only one beam splitter is required. This means that the interference pattern is only displayed in one place rather than two. A lens is generally used to transform the plane wave pattern into a spherical wave pattern (increasing the size of the interference pattern). One of the arms can be moved to observe changes in the interference pattern with changing path difference.

161
Q

What is the condition for destructive interference with a Michelson interferometer?

A

d = distance moved = (∆L/2)
∆L = change in path length

162
Q

What is the condition for constructive interference with a Michelson interferometer

A

d = distance moved = (∆L/2)
∆L = change in path length

163
Q

How can the wavelength of light be calculated using a Michelson interferometer?

A

n = number of fringes counted
d = distance moved

164
Q

What is a diffraction grating?

A

An expansion of double slits to many slits (N) in the form of a grating or a thin film. The distance between two consecutive slits and the path length difference between two slits are expressed in the same way as for double slits.

165
Q

What are diffraction gratings used for?

A
  • Interference of atoms and molecules
  • Analysis of crystal structures
  • Spectrometers
166
Q

What is the equation for the intensity distribution of a diffraction grating?

A

I = intensity
I₀ = Amplitude
k = wavenumber
N = number of slits
∆r = path length difference

167
Q

What is the path length difference equation for two reflected beams from a thin film?

A

∆D = path length difference
n₂ = index of refraction inside the film
d = width of thin film
θ₂ = angle of refraction

168
Q

What are the conditions for constructive and desctructive interference of a thin lens?

A

Top = destructive interference
Bottom = constructive interference

169
Q

What are anti-reflection coatings?

A

Thin coatings added to materials (such as glass for glasses) that leads to an additional phase shift so that destructive interference occurs and there is minimal reflection of light from the material.

170
Q

What is the condition for destructive interference for anti-reflection coatings?

A

∆D = path length difference

171
Q

What is the condition for constructive interference for anti-reflection coatings?

A

∆D = path length difference

172
Q

What is the Huygens-Fresnel Principle?

A

The idea that, in order to find the diffraction pattern from an opening at a given point on a screen, the waves arriving at that point must be summed up and the intensity must be found.

173
Q

What is the equation used to find the sum of all the spherical waves arriving at a point?

A

E₀ = amplitude
b = slit width
k = wavenumber
r = √((x’ - x)² + z₀²)

174
Q

What is far-field diffraction?

A

The intensity distribution given by the absolute square of the Fourier transform of the opening function.

175
Q

How is near-field diffraction different to far-field diffraction?

A

Far field diffraction, also known as Fraunhofer diffraction, is used when the distance between a diffracting object and a screen is very large. It approximates a square wave as a plane wave.

Near field diffraction cannot do this as the curvature of the wavefront is not negligible. It can be found by calculating a Fresnel transform.

176
Q

What is a Fourier transform?

A

A mathematical property that can convert from space to reciprocal space or from time to frequency. It is reversible.

177
Q

What shape is the Fourier transform of single slit diffraction?

A

A sinc function, because the slit is the same shape as a square wave.

178
Q

What shape is the Fourier transform of a cosine function?

A

Two vertical lines at ±frequency.

179
Q

What shape is the Fourier transform of double slit diffraction?

A

The square of the Fourier transform of a single slit multiplied by the Fourier transform of the intensity equation for double slits.

180
Q

What shape is the Fourier transform of a diffraction grating?

A

The Fourier transform for a single slit multiplied by the Fourier transform for the intensity of the diffraction pattern.

181
Q

What shape is the Fourier transform for a circular opening?

A

An airy pattern

182
Q

The definition for optical resolution is given by the ______ ______.

A

Rayleigh criterion

183
Q

Define the Rayleigh criterion

A

Two points can be resolved when the first minimum of the first Airy pattern coincides with the maximum of the second Airy pattern.

184
Q

What is the resolution angle of a telescope?

A

R = radius of the diffracting aperture
D = diameter of the diffracting aperture

185
Q

What is the equation for the numerical aperture of a microscope?

A

NA = numerical aperture
D = diameter of aperture
f = focal length

186
Q

What is the equation for the resolution of a microscope?

A

x = smallest resolvable distance (spatial resolution)
NA = numerical aperture
f = focal length
D = diameter of aperture