F. OPTICS

23. Physical optics

Objectives 23.1 23.2 Interference 23.3 Two-slit interference pattern 23.4 Interference in a thin film 23.5 Diffraction at single slit 23.6 Diffraction gratings 23.7 Polarisation 23.8 Optical waveguides

Outcomes a) b)

and diffraction phenomena c) explain the concept of coherence d) explain the concept of optical path difference, and

solve related problems e) state the conditions for constructive and destructive

interferences f) -slit interference pattern g)

interference pattern

Outcomes h) explain the phenomenon of thin film

interference for normal incident light, and solve related problems

i) explain the diffraction pattern for a single slit j)

minimum in the diffraction pattern for a single slit

k)power of an aperture

Outcomes l) explain the diffraction pattern for a

diffraction grating m) use the formula d for a diffraction

grating n) describe the use of a diffraction grating to

form the spectrum of white light, and to determine the wavelength of monochromatic light

Outcomes o) state that polarisation is a property of transverse

waves p) explain the polarisation of light obtained by

reflection or using a polariser q) understand polarisation planes

law tan B = n r) I = I0cos2 s) explain the basic principles of fibre optics and

waveguides t) state the applications of fibre optics and

waveguides

Objectives (a) understand and use

principle to explain interference and diffraction phenomena

emitted at the opening and they will combine when expanding on the other side of the opening creating the diffraction pattern.

Every point on a wave front is a source of secondary wavelets. i.e. particles in a medium excited by electric field (E) re-radiate in all directions i.e. in vacuum, E, B fields associated with wave act as sources of additional fields

Given wave-front at t Allow wavelets to

evolve for time t

r = c t

New wavefront Construct the wave front tangent to the wavelets

Plane wave propagation New wave front is still a plane as long as dimensions of wave front are >> If not, edge effects become important Note: no such thing as a perfect plane wave, or collimated beam

23.2 Interference Objectives (b) understand the concept of coherence (c) understand the concept of optical path

difference (d) know the conditions for constructive

interference and destructive interference

Coherence

If the phase of a light wave is well defined at all times (oscillates in a simple pattern with time and varies in a smooth wave in space at any instant), then the light is said to be coherent.

If, on the other hand, the phase of a light wave

varies randomly from point to point, or from moment to moment (on scales coarser than the wavelength or period of the light) then the light is said to be incoherent.

Coherence For example, a laser produces highly coherent

light. In a laser, all of the atoms radiate in phase.

An incandescent or fluorescent light bulb produces incoherent light. All of the atoms in the phosphor of the bulb radiate with random phase. Each atom oscillates for about 1ns, and produces a coherent wave about 1 million wavelengths long. But after several ns, the next atom radiates with random phase.

Interference

Recall interference of sound waves. Light waves also display constructive and destructive interference.

For incoherent light, the interference is hard to

washed outrapid phase jumps of the light.

Soap films are one example where we can see

interference effects even with incoherent light.

Interference of light waves was first demonstrated by Thomas Young in 1801.

When two small apertures are illuminated with coherent light, an interference pattern of light and dark regions is observed on a distant screen:

Light

Path Difference

We can understand the interference pattern as resulting because light from the two apertures will, in general, travel a different distance before reaching a point on the screen. The difference in distance is known as the path difference.

Light

P

Constructive and Destructive Interference

Two waves (top and middle) arrive at the same point in space. The total wave amplitude is the sum of the two waves. The waves can add constructively or destructively

Constructive and Destructive Interference 23.3 Two-slit interference pattern

Objectives (e) -slit interference pattern (f) derive and use the formula y = D/a for

Two Slit Diffraction Two Slit Interference An incoherent light source illuminates the first slit. This creates a nearly-uniform but coherent illumination of the second screen (from side-to-side on the screen, the light wave has the same oscillating phase). The two waves from the two slits S1 and S2 create a pattern of alternating light and dark fringes on the third screen.

Interference of waves from double slit

Each slit in the previous slide acts as a source of an outgoing wave. Notice that the two waves are coherent The amplitude of the light wave reaching the screen is the coherent sum of the wave coming from the two slits.

Why did Young (1800s) use single slit before the double slit? 1. The first slit forces the wave to be

coherent all the time 2. From moment to moment (after

many oscillations of wave) the wave is still incoherent, but at each moment in time, the wave has the same phase at the two slits.

3. He was too cheap to buy a 19th century laser.

Two Slit Diffraction If the two slits are separated by a difference d

and the screen is far away then the path difference at point P is l dsin

Light P

dsin

L

Two Slit Diffraction f we put a lens of Focal Length f=L, then the expression l dsin is exact.

If l = 2 etc, then the waves will arrive in phase and there will be a bright spot on the screen.

Light P

dsin

L

Fringes

Consider apertures made of tall, narrow slits. If at point P the path difference yields a phase difference of 180 degrees between the two beams a dark fringe will appear. If the two waves are in phase, a bright fringe will appear.

Interference Conditions For constructive interference,

the path difference must be zero or an integral multiple of the wavelength:

For destructive interference,

the path difference must be an odd multiple of half wavelengths:

m is called the order number

2...) 1, 0,( , = sin mmd

2...) 1, 0,()-( = sin = 21 mmd

Double Slit interference

If we know distance D, position y of mth bright fringe

dmDym

Could be used to measure the wavelength of light!

Dyd m

m

dd sin

Double Slit interference

dmDym

dDy

dDy

If m = 1,

Example If the distance between two slits is 0.050 m and the

distance to a screen is 2.50 m, find the spacing between the first- and second-order bright fringes for yellow light of 600 nm wavelength.

2...) 1, 0,( , = sin mmd

mmyymmmy

radmnm

dmLy

m

mm

030.0030.0)1012)(5.2(

10121012)05.0/()600(1sin

/sintan

12

61

61

61

mmyymyym

mnmyy

dDmym

03.05010605.0

50.2600)12(

12

712

12

Two Slit Diffraction: When green light ( = 505 nm) passes through a pair of double slits, the interference pattern shown in (a) is observed. When light of a different color passes through the same pair of slits, the pattern shown in (b) is observed.

Two Slit Diffraction: (a) Is the wavelength of the second color greater than or

less than 505 nm? Explain.

(b) Find the wavelength of the second color. (Assume that the angles involved are small enough to set sin = tan = .)

2...) 1, 0,( , = sin mmd

Solution green light ( = 505 nm) 4.5 orders of green light = 5 orders of mystery light 4.5 (505 nm) = (5)

< 505 nm, = (4.5/5)(505 nm) = 454 nm

2...) 1, 0,( , = sin mmd

Thin film - continued example of air wedge:

reflection from upper plate, na>nb, no phase shift reflection from lower plate, na<nb, phase shift of

23.4 Thin film

Description of phase shift during reflecting

Assume light travelling in medium with index of na, and hits interface to the medium with index of nb

if na>nb, no phase shift if na<nb, half circle phase shift

Interference conditions for thin films Non-normal incident ray

AB = BC = d/cos

n = sin /sin

Example 1: Soap film interference Maximum reflection or transmission happens to be a certain (colour) at different position (angle)

Example 1: Soap film interference The interference colours from a soap film can be related to the thickness of the film by using the interference condition and noting that there is a 180 degree phase change upon reflection from the film surface, but no phase change for the reflection from the back surface. The colour seen depends also upon the angle of view

Example 2: Oil film interference Example 2: Oil film interference

The interference colours from an oil film on water can be related to the thickness of the film by using the interference condition and noting that there is a 180 degree phase change upon reflection from the film surface, but no phase change for the reflection from the back surface. This presumes that the index of refraction of the oil is greater than that of the water. The colour seen depends also upon the angle of view

Applications:

Anti-reflection coating Reducing reflection

Reflective coating Increasing the reflection

Applications: For destructive interference

Path deference = (m + ½) 2ndCos = (m + ½) ,

Take 0 , for nearly normal light 2nd = (m + ½) If m = 0 , then thickness of the film is

minimum; 2nd = ½ d = /4n

d

Applications: For constructive interference

Path deference = m 2ndCos = m ,

Take 0 , for nearly normal light 2nd = m If m = 1 , then thickness of the film is

minimum; 2nd = d = /2n

d

Anti-reflection coating

1/4

Anti-reflection coating A single layer anti-reflection coating can be made non-reflective only at one wavelength, usually at the middle of the visible. Multiple layers are more effective over the entire visible spectrum.

Anti-reflection coating One coating layer is only for a certain wavelength, normally chosen in the central yellow-green portion of the spectrum ~550nm (most sensitive to the eyes) Overall reflection can be reduced from 4~5% to 1%, the reflectivity can be reduced further by multiple-layer coatings

Anti-reflection coating widely used for

highly corrected photographic lenses usually using many pieces of glasses devices of solar cell or phototdetectors use this coating to increase light amount transmitted LED device to reduce Fresnel lossses

Reflective coating If the quarter-wavelength layer has greater index than the glass

half-cycle phase shift happens at the air-film interface no phase shift at the interface of film-glass half-cycle phase shift during travelling in film the interference is constructive

Reflective coating The reflectivity is increased, for a film with index of 2.5 increases reflectivity of 38% 100% reflectivity can be achieved by multiple-coating layers widely used in modern optoelectronics to construct micro-cavity

23.5 Diffraction at single slit Objectives (i) know the diffraction pattern for a single slit (j) /a for

the first minimum in the diffraction pattern for a single slit

Diffraction The bending of light around objects into what would otherwise be a

known as diffraction. Diffraction occurs when light passes through very small apertures or near sharp edges.

Single Slit Diffraction

We have seen how we can get an interference pattern when there are two slits. The interference pattern with a single slit can

size is approximately (neither too small nor too large)

Light

Single Slit Diffraction To understand single slit diffraction, we must consider each point along the slit (of width a

) to be a point source of light. There will be a path difference between light leaving the top of the slit and the light leaving the middle. This path difference will yield an interference pattern.

Light

P

(a/2) sin L

a

Single Slit Diffraction

Sin = BC/AB = 2BC/a BC = (a/2)sin

Light P

(a/2) sin a a

B

A

C

If a = , ray AP and BP are completely out of phase, destructive int.

a/2 A

B C

P

Single Slit Diffraction Path difference of rays to P from top and bottom edge of slit, L = asin destructive if

L = m

ma

ma

m

sin where angleat obtain isminimun a Thus,

eDestructiv 2...) 1,( = sin

Light

P

(a/2) sin a

Single Slit Diffraction Notice that central

maximum is twice as wide as secondary maxima

Sin = m / W, Destructive

Dark Fringes on screen y = L tan L (m /W) Maxima occur for y= 0 and, y L (m 1/2)( /W)

m=1

m= 1

L

Diffraction from a pinhole

Dark fringes occur at zeros of Bessel function, (2-D geometry).

First dark fringe Sin = 1.22 (l/D) D = diameter of pinhole

Single-Slit Diffraction from a large aperture (telescope, microscope, camera).

A lens images parallel rays to a point at the focal distance f. All parallel rays experience the same phase change from a incident plane wave to the focus. A image formed by a lens of diameter a is fuzzed out by the single slit diffraction pattern, whose central maximum is of width: = 1.22 /a

Rayleigh Criterion Diffraction limited Resolution

Two objects can be resolved (barely) if the diffraction maximum of one object lies in the diffraction minimum of the second object. min = 1.22 [wavelength]/[diameter of lens or mirror]

23.6 Diffraction gratings Objectives (k) know the diffraction pattern for diffraction

gratings (l) use the formula d n for diffraction

gratings (m) describe the use of diffraction gratings to

form the spectrum of white light and measure the wavelength of monochromatic light

Diffraction Gratings

Each slit is a source of a wave Observe the outgoing wave at an angle , the contributions from all slits add up coherently if d sin = m ,

Diffraction Gratings If the incident wave uniformly and coherently excites N slits, then the contribution from all of the slits will exactly cancel if d sin = (m+1/N) , (m=0,

By virtue of using many slits, the diffraction grating reduces the width of each maximum by a factor 1/N.

Sharpening of Diffraction Pattern Diffraction pattern with N=5 Width of each principal maximum is

= /(Nd), d = spacing of grating N = number of slits illuminated by source.

Resolution of Diffraction Grating A grating can be used to measure the wavelength of a spectral line from an atomic or molecular transition. A grating has 5000 rulings/cm, Our light source makes a spot 5mm across on the grating. We observe the diffraction pattern in 3rd order.

Resolution of Diffraction Grating With what precision can we measure the wavelength of incident light?

d sin = (m+1/N) , N = (5000/cm)(0.5cm) = 2500

Consider two wavelengths 1 and 2 such that Sin = (m) 1/d = (m+1/N) 2/d 1 2 = measurement precision

Resolution of Diffraction Grating

1 = sin (d/m) 2 = sin [d/(m+1/N)] ( 1 2) / 1 = relative precision

mNNmN

Nmm

mdNmdmd

1/1

/1/1

1

sin)/(sin)]/1/([sin)/(

2

21

Precision improves with larger values of either N or m, But diffraction maxima get weaker and weaker as m increases

Diffraction Grating Resolution N = 2500, m = 3 Relative precision = 1/(2500 3) = 1.3e-4 Red light l = 800 nm

Absolute precision = (800 nm) 1.3e-4 = 0.1nm (One atomic diameter!!!!!)

23.7 Polarisation Objectives (n) understand that polarisation is a property of

transverse waves (o) understand the production of polarised

light by polaroid and by reflection (p) understand polarisation planes (q) use the formula I = I0cos2

EM waves Light is a transverse wave - like waves on a string, or ripples on the surface of water The associated electric and magnetic fields, E and B are at right angles to the direction the wave is traveling and to each other

E

B

Direction wave is traveling

electric

magnetic

Plane of polarisation The plane of polarisation is the plane containing the electric field E and the direction of the wave

E

B

Direction the wave is traveling

Production of radio waves Radio waves are produced by electric currents in an aerial The waves are polarised in the same direction as the aerial.

A horizontal aerial makes waves that are polarised in the horizontal plane

Production of radio waves Radio waves are produced by electric currents in an aerial The waves are polarised in the same direction as the aerial.

A vertical aerial makes waves that are polarised in the vertical plane

Light waves Light waves are produced by electric currents within atoms Usually atoms are oriented randomly and the produced light is unpolarised

Polarised vs unpolarised Observed along the direction the wave is traveling,

A polarised wave looks like this:

An unpolarised wave looks like this:

E has a definite direction and oscillates up and down

Direction of E changes randomly with time

Production of polarized light This is most easily done by placing a polarising filter in front of an unpolarised light source

Unpolarised incident light

Optical axis of polarising filter

E

E Transmitted polarised light

polariod

Polarising filters A polarising filter has conducting lines of molecules Electric fields along these lines generate electric currents and are absorbed Only the component perpendicular to the conducting lines is transmitted

Conducting lines of molecules

NB: the optical axis is perpendicular to the conducting lines E

E

Transmitted intensity A polarising filter absorbs half the intensity of unpolarised light

100%

50% E

E

Combining polarising filters A second polarising filter with the same optical axis as the first does not absorb additional light

E

E

100% 50%

50%

E

Combining polarising filters A second polarising filter with an optical axis perpendicular to the first absorbs all remaining light

E

E

100%

0%

50%

The intensity I0 the light is proposional to its amplitude

I0 A02

When a polarized light of amplitude A0 insidents to a polaroid whose transmission axis is inclined a an angle of , only the light parallel to the transmission axis is allowed to pass through, where the component amplitude, A = A0 cos

A0

Optical transmission

axis

A = A0 cos ,

The intensity of the component wave I is related to the component amplitude, A, where

I A2 Then

I/Io = A2/A02

= A02cos2 /A0

2

I = I0cos2

A0

Optical transmission

axis

E 25%

100% I = I0

I = I0cos2 = I0cos2600

=0.25I0

600

E

E

100%

12.5%

50%

I = I0 I2 = I1cos2 = I1cos2600

= 0.25I1

= 0.25 0.5I0

= 0.125I0

600

I1 = 0.5I0

Polarisation by scattering in air Sunlight induces electric current oscillations in air molecules

Sunlight (unpolarised) Induced electric

currents in air molecules

observer

Polarised light

Scattered light

Polarisation by scattering in air If one observes the scattered light at right angles to the incident light direction, it will be polarised because oscillations along the observation direction cannot produce transverse scattered light

Sunlight (unpolarised) Induced electric

currents in air molecules

observer

Polarised light

Scattered light

Polarisation by reflection Light reflected off non-metallic surfaces (e.g. water, glass) is partly polarised with E parallel to the surface

Brewster angle The amount of polarisation depends on the incidence angle At the Brewster angle of incidence the reflection is fully polarised At the Brewster angle of incidence, the reflected and refracted light are at right angles The Brewster angle is calculated as tan B=n2/n1

B

normal

n2

n2 B

1.3 52±

1.4 54±

1.5 56±

n1

for air, n1=1.0

Reflection off non-metallic surfaces

refle

ctio

n

0.2

0.4

0.6

0.8

1.0

0o 20o 40o 60o 80o

56.3o Normal incidence

component parallel

to surface

component perpendicular to surface

Brewster angle tan B=n2/n1

n=1.5

04.05.25.0 22

12

12nnnnR

0.04

Polarising sun-glasses Some sunglasses are made with a polarising filter The optical axis is vertical This reduces the reflection from horizontal surfaces (e.g. water on the road)

Randot and Titmus tests The left and right eye see slightly different images: close objects are displaced more between the left and right eye than far objects. This allows depth perception. One way to simulate this is overlaying two images with different polarisations and watching them through polarizing glasses (3D movies)

Randot and Titmus tests If the optical axis differs by 90o between the two filters, the left eye sees a different image than the right eye This is used to test stereoscopic vision in the Randot and Titmus tests Further study: 3D Movies http://wiki.answers.com/Q/Why_do_Real_D_3D_glasses_work_in_the_cinema_but_they_wont_work_with_any_other_3D_images_at_home_Are_there_any_images_that_do_work

23.8 Optical waveguides explain the basic principles of fibre optics and waveguides state the applications of fibre optics and waveguides.

History of Fiber Optics

Total Internal reflection is the basic idea of fiber optic

John Tyndall demonstration in 1870

History of Fiber optics During 1930, other ideas were developed with this fiber optic such as transmitting images through a fiber. During the 1960s, Lasers were introduced as efficient light sources In 1970s All glass fibers experienced excessive optical loss, the loss of the light signal as it traveled the fiber limiting transmission distance.

History of Fiber optics This motivated the scientists to develop glass fibers that include a separating glass coating. The innermost region was used to transmit the light, while the glass coating prevented the light from leaking out of the core by reflecting the light within the boundaries of the core. Today, you can find fiber optics used in variety of applications such as medical environment to the broadcasting industry. It is used to transmit voice, television, images and data signals through small flexible threads of glass or plastic.

Source and transmitters

A basic fiber optic communications system consists of three basic elements:

Fiber media Light sources Light detector

Several applications of fiber optic

Configuration of a Fiber Optic Sensor System

A Light Sources

LED (Light emitting diode) ILD (injection laser diode)

Detectors Detector is the receiving end of a fiber optic link. There are two kinds of Detectors 1.PIN (Positive Intrinsic Negative) 2.APD (Avalanche photo diodes)

PIN APD

Fiber media There are three types of fiber optic cable commonly used Optical fibers are the actual media that guides the light

Single Mode Step-index Multimode fiber

Plastic optic fiber

How Does fiber optic transmit light The loss of fiber optic Material obsorption Material Scattering Waveguide scattering Fiber bending Fiber coupling loss

Idea of Modulation When sending information by an optical fiber, the information must be encoded or transformed somehow into information that capable of being transmitted through a fiber. The signal needs to be modulated. There are two types of modulation Analog and digital.

The advantages of fiber optic over wire cable

Thinner Higher carrying capacity Less signal degradation Light signal Low power Flexible Non-flammable Lightweight

Disadvantage of fiber optic over copper wire cable

Optical fiber is more expensive per meter than copper Optical fiber can not be join together as easily as copper cable. It requires training and expensive splicing and measurement equipment.

Optical fiber transmits light. But, what prevents the light from escaping from the fiber?

Total internal Reflection

Critical angle: sin c = n2/n1 (n1>n2)

Laws of Reflection & Refraction

Reflection law: angle of incidence=angle of reflection

2211 sinsin nn [2-18]

Optical Fiber communications, 3rd ed.,G.Keiser,McGrawHill, 2000

Total Internal Reflection & Critical Angle

1

n 2

n 1 > n 2

Incident light

Transmitted (refracted) light

Reflected light

k t

TIR

Evanescent wave

k i k r

( a ) ( b ) ( c )

Light wave travelling in a more dense medium strikes a less dense medium. Depending on the incidence angle with respect to which is determined by the ratio of the refractive indices, the wave may be transmitted (refracted) or reflected. (a) 1 < c (b) 1 = c (c) 1 > c and total internal reflection (TIR).

2

1 c

902

c1Critical angle

1

2sinnn

c

Phase shift due to TIR

The totally reflected wave experiences a phase shift however which is given by: Where (p,N) refer to the electric field components parallel or normal to the plane of incidence respectively.

2

1

1

122

1

122

sin1cos

2 tan;

sin1cos

2tan

nn

n

nnn

n pN

Optical waveguiding by TIR: Dielectric Slab Waveguide

Propagation mechanism in an ideal step-index optical waveguide. Optical Fiber communications, 3rd ed.,G.Keiser,McGrawHill, 2000

TIR supports that angle minimum ;sin1

2min n

n

22

211max0 sinsin nnnn c

Maximum entrance angle, 0max is found

end face.

max0

Launching optical rays to slab waveguide

Numerical aperture:

1

21

12

22

1max0 2sinNA

nnn

nnnn

Optical rays transmission through dielectric slab waveguide

ccnn2

;21

sincos

2sintan

1

22

2211

nnnmdn

For TE-case, when electric waves are normal to the plane of incidence must be satisfied with following relationship:

[2-25]

Optical Fiber communications, 3 rd ed.,G.Keiser,McGrawHill, 2000

O

Optical Fibers: Modal Theory (Guided or Propagating modes) & Ray Optics Theory

1n 2n

21 nn

Step Index Fiber

Optical Fiber communications, 3rd ed.,G.Keiser,McGrawHill, 2000

Ray Optics Theory (Step-Index Fiber)

Skew rays

Each particular guided mode in a fiber can be represented by a group of rays which Make the same angle with the axis of the fiber.

Optical Fiber communications, 3rd ed.,G.Keiser,McGrawHill, 2000

Different Structures of Optical Fiber

Optical Fiber communications, 3rd ed.,G.Keiser,McGrawHill, 2000