Wave Optics

Particle nature vs Wave nature of light

• Prior to 1800, most scientists thought that light behaved like a particle, including Isaac Newton…this was the basis for using rays of light.

• However, in 1801, Thomas Young demonstrated a phenomena of light that gave evidence that light was a wave.

Recall interference between 2 waves• Bright spots are C.I., dark are D.I.

There can also be interference between 2 sources of light

• Thomas Young performed double-slit experiment to prove the wave nature of light by passing light through 2 slits

• Only waves could do this...bending around obstacles

Particles or Rays of light would only yield 2 bright spots on screen…not a series of brights!

Young’s Double Slit Experiment• The narrow slits, S1 and

S2 act as sources of waves

• The waves emerging from the slits originate from the same wave front and therefore are always in phase

Resulting Interference Pattern• The light from the two slits form a

visible pattern on a screen• The pattern consists of a series of

bright and dark parallel bands called fringes

• Constructive interference occurs where a bright fringe appears

• Destructive interference results in a dark fringe

How interference pattern is created

• Constructive interference occurs at the center point

• The two waves travel the same distance, therefore, they arrive in phase or in step

Constructive Interference• In the picture at right, the

upper wave travels farther than the lower wave to reach the screen at Q.

• If the difference in path length is on the order of one wavelength multiples, then the waves arrive IN PHASE (C.I.)

• A bright fringe occurs here

Destructive Interference• If the picture at right, the

upper wave travels a longer path length, but this time the difference is on the order of one-half of a λ farther than the lower wave versus a whole λ.

• The trough of the bottom wave overlaps the crest of the upper wave which yields destructive interference (dark fringe)

General Rule for C. I.• If a difference in path length exists between 2 waves,

then the difference must be in integer (1, 2, 3..) multiple wavelengths of each other (in phase).

General Rule for D.I.

• Waves have to be a difference of ODD ½ integer (1/2, 3/2, 5/2, etc) wavelengths (out of phase)

Geometry of Interference

• The path difference, δ, is found from the tangent triangle• δ = r2 – r1 = d sin θ

– This assumes the paths are parallel– Not exactly parallel, but a very good approximation since L >> d

Interference Equations• For a bright fringe, produced by constructive

interference, the path difference must be either zero or some integral multiple of the wavelength

• δ = d sin θbright = m λ

– m = 0, ±1, ±2, … – m is called the order number

• When m = 0, it is the zeroth order maximum which is at center of screen

• When m = ±1, it is called the first order maximum (±1 because it’s on either side of 0)

Interference Equations• The positions of the fringes can be measured

vertically from the zeroth order maximum

• y = L tan θ L sin θ

• Assumptions– L>>d– d>>λ

• Approximation– θ is small and therefore the approximation tan θ

sin θ can be used

Interference Equations• When destructive interference occurs, a dark

fringe is observed• This needs a path difference of an odd half

wavelength• δ = d sin θdark = (m + ½) λ

– m = 0, ±1, ±2, …

Interference Equations• For bright fringes

• For dark fringes

• Importance of all of this…it allows us to measure Importance of all of this…it allows us to measure wavelength of light. Thus, it allows us to measure wavelength of light. Thus, it allows us to measure the size of very tiny objects!the size of very tiny objects!

0, 1, 2bright

Ly m m

d

10, 1, 2

2dark

Ly m m

d

Example1• A screen containing 2 slits 0.100mm apart is 1.2m

from a screen. Light (500nm) falls on the slits. How far apart will the bright fringes be? (6mm)

Hint: Looking for y’s

Example2 Monochromatic light falling on 2 slits

0.042mm apart produces a 5th order bright fringe at a 7.8o angle. What is the wavelength of light used? (1.14x10-6m)

Example3 Light of wavelength 680nm falls on 2 slits and

produces a pattern where the 4th-order DARK fringe is 48mm from the central fringe on a screen 1.5m away. What is the separation of the 2 slits? (0.074mm)

Questions:

b) What happens to the interference pattern for 2 slits if wavelength of incident light is increased?

a) What happens to interference pattern if the 2 slits are moved further apart?

Thin film interference• Color swirls seen on

soap bubbles & gasoline slicks.

• This occurs due to a very thin layer of film between 2 media.

• Different thickness yields different colors.

Reflection from thin filmsAs incident light wave strikes film, some energy of wave reflects from top layer of film. The remaining energy is transmitted into film whichthen reflects from bottom layer. These 2 waves then combine at the eye for either a combination of either C.I. or D.I.

Phase change due to REFLECTIONIf a light wave encounters a more dense medium, a phasechange will occur where the wave will flip its orientation or CHANGE PHASE 180o (or ½ λ)

In the case of a thin film, this CAN occur twice, once at top of film and once at bottom of film. It depends on the index values that surround the thin film.

This either leads to C.I. or D.I.

Phase change due to PATH DIFFERENCE (2t)A P.C. can also occur because of the difference in length of travel or path difference. Ray 2 travels farther than Ray 1.Comparing the distance in the film (2t) to the size of the wavelength of the light in film could yield either D.I. or C.I. when the 2 waves meet up at the eye.

If waves are meeting in C.I. we see a colorIf waves are meeting in D.I. we see dark

Thin film formula

2t = mλ / nfilm

Wavelength of light changes in film. λo refers to wavelength of light prior to entering thin film.

Must determine what ‘mode’ the 2 waves are in when they are meeting at the eye and then decide if it’s in the mode the problem wants, either C.I. or D.I.

If the mode is already satisfied, use formula above. If the mode is not satisfied, use (m + ½ ).

Example1 A thin layer of oil (n = 1.25) is floating on water (n =

1.33). How thick can the oil be such that it strongly reflects green light (λ = 525 nm)? (210nm)

How many phase changes occur in the problem?

HINTS:

What mode are the reflected and transmitted waves ‘in’ when they mix at the eye?

Do we have to do anything to the ‘mode’ based on what problem wants?

Example2Light (λ=550nm) movesfrom air to film of siliconoxide which sits on silicon. What minimum thickness of film must be present to get zero reflection? (94nm)

HINTS:

How many phase changes occur in the problem?

What mode are the reflected and transmitted waves ‘in’ when they mix at the eye?

Do we have to do anything to the ‘mode’ based on what problem wants?

Polarization• Nonpolarized light oscillates in all planes prior

to interaction with matter

We can selectively filter light into one plane of oscillation or polarize it

Vertically polarized light Horizontal polarized light

This happens by either using a polarizing filter or by reflection. A polarized filter only allows ONE plane of light to pass through

If 2 polarized filters are used successively, one vertical, one horizontal, then light can be removed all together.

Polarized filters• All LCD screens emit polarized light

Polarized by Reflection

• Light will plane polarize after reflecting off a surface. It will polarize in the same plane as the reflecting surface.

Reason for Polarized Sunglasses

• Your sunglasses are polarized vertically to block horizontal light which is the glare off of the road.