# chapter23 wave optics

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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

d

mDym

Could be used to measure the wavelength of light!

D

yd m

m

dd sin

Double Slit interference

d

mDym

d

Dy

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

mmyy

mmmy

rad

mnm

dm

Ly

m

mm

030.0

030.0)1012)(5.2(

1012

1012)05.0/()600(1sin

/sin

tan

12

61

61

61

mmyy

myy

m

mnmyy

d

Dmym

03.0

50106

05.0

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, nanb, no phase shift if na

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 + ) 2ndCos = (m + ) ,

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

minimum; 2nd = 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 wa