OpticsOptics

ISAT 241

Analytical Methods III

Fall 2003

David J. Lawrence

Light as a WaveLight as a Wave Wavemotion (propagation) of a

disturbance. Waves transport energy. Mechanical waves

sound, earthquakes, and ocean waves require a “medium” that can be disturbed.

Light and radio waves are examples of “Electromagnetic waves”. Disturbance is variations in electric (E) and

magnetic (B) fields. Since E and B can exist in a vacuum,

electromagnetic waves do not require a medium.

For a “Harmonic Wave” (sinusoidal wave)

Wave speed = frequency x wavelength

v fFor light and other “Electromagnetic Waves”, in a vacuum:

v = c = 2.9979 108 m/sIn air:

vair = 2.9970 108 m/s

In glass:vglass = 2.0 108 m/s~

1 nanosecond = 1 ns = 1 x 10-9 s= one billionth of a second

In 1 ns, light travels a distance of:

d = (3 108 m/s) (1 10-9 s) = 0.3 m = 1 ft~

Speed and Wavelength of Light in Several MediaSpeed and Wavelength of Light in Several Media

(for light of frequency = 5.09 x 1014 Hz)

‘Yellow’

v = f Medium Speed of Light Wavelength of Light

vacuum* 2.9979 x 108 c 589 air 2.9970 x 108 589 water 2.25 x 108 442glass (crown) 1.97 x 108 387 diamond 1.24 x 108 243

(m/s) (nm)

* Light travels at its maximum speed in a vacuum

c f For the visible portion of the spectrum, the wavelength in vacuum (or in air) ranges from:

nm to nm

Å to Å

Color f (Hz)(nm)

red 630-760 4.5 x 1014

orange 590-630 4.9 x 1014

yellow 560-590 5.2 x 1014

green 500-560 5.7 x 1014

blue 450-500 6.3 x 1014

violet 380-450 7.1 x 1014

These are average values!!

Light as a Wave

Rays

Wavefronts

A “plane wave” propagating to the right.

The Ray Approximation

WavefrontsA plane wave propagating to the right.

Rays

“Rays” are straight lines perpendicular to the wavefronts. They point in the direction of motion of the wave.

Serway & Jewett, Principles of Physics

Figure 25.1

Serway & Jewett, Principles of Physics

Figure 25.2

Serway & Jewett, Principles of Physics

Figure 25.3

Specular Reflection = Reflection from a Smooth Surface

Law of Reflection

“Incident Ray”

“Reflected Ray”“normal”

The incident ray, the reflected ray, and the normal, all lie in the same plane

Angle of reflection = Angle of incidence

Serway & Jewett, Principles of Physics

Figure 25.5

Consider MRDs marching from afootball field into a swamp . . .

Swamp

v1

Field

v2

The Index of Refraction, n , of a medium is defined as the ratio:

speed of light in a vacuum cn speed of light in the medium =

v

(m/s)

vacuum 2.9979 x 108 589 1 air 2.9970 x 108 589 1.000293 water 2.25 x 108 442 1.333glass (crown) 1.97 x 108 387 1.52 diamond 1.24 x 108 243 2.419

Medium Speed of Light Wavelengthof Light (nm)

Index of Refraction, n

n > 1n is dimensionless

These values apply for light of frequency = 5.09 x 1014 Hz

(‘Yellow’)

What happens when light travels from one medium into another?

• The frequency does not change• The speed and wavelength do change

E.g. , when light goes from air into glass

rays

Air = #1

Glass = #2

wavefronts1

18

1

589

3 10

1

nm

m / sv

n

2

28

2

387

2 10

152

nm

m / sv

n .

vc

n

n

22

20

2

speed in vacuum

wavelengthin vacuum

vc

n

n

11

10

1

Serway & Jewett, Principles of Physics

Figure 25.11

Field

Swamp

What if columns of MRD’s enter theswamp at an angle?

v1

v2

Refraction of Light

“Reflected Ray”

1

AirGlass

“Incident Ray”

“Refracted Ray”

“normal”

n1

n2

2

1

All rays and the normal lie in the same plane.

The “Refracted Ray” is bent towards the normal.

Law of Refraction - Snell’s Law

n2 > n1

n1 sin = n2 sin

Serway & Jewett, Principles of Physics

Figure 25.8

Refraction of Light

Air

“Incident Ray” “Reflected Ray”

“Refracted Ray”

Glass n

n1

2

“normal”

1

2

1

All rays and the normal lie in the same plane.

The “Refracted Ray” is bent away from the normal.

Law of Refraction - Snell’s Law

n1 sin = n2 sin n2 < n1

Dispersion

The speed of light in any material depends on the frequency (or wavelength, or color) of the light.

• n depends on wavelength.• Light of different wavelengths will be refracted or bent at different angles when it enters a material.

Therefore:

The material is said to exhibit “Dispersion”

“Normal”

“White light”

Total Internal Reflection

airn2 = 1.00

n1 = 1.50glass

0 %

c

the “criticalangle”

100 %

Snell’s Law : n1 sin 1 = n2 sin 2

For n1 > n2 , when c , 2 , and Snell’s Law gives

n1 sin c= n2 sin n2

Total Internal Reflection

airn2 = 1.00

n1 = 1.50glass

0 %

c

the “criticalangle”

100 %

angle critical thesin

for sin

1

21

211

2

n

n

nnn

n

c

c

Total Internal ReflectionTotal Internal Reflection (cont.) (cont.)Total internal reflection occurs only when light attemptsto go from a medium of higher index of refraction to amedium of lower index of refraction.

For this example, when light attempts to go from glasswith n1 = 1.50 into air with n2 = 1,

c = sin-1( ) = sin-1( )

c = 41.8°

n2

n1

11.50

For 1 > c , the beam is entirely reflected at the boundary.

Fiber Optics

end view

n1

n2

Cladding

CladdingCore

c

n1

n2

typically < 8 °

Light Ray

Fiber Optics (cont.)

Light must fall inside this angleto be guided in the fiber core.

2 × Acceptance Angle 2 accept

Acceptance

Angle , acceptair

nair = 1

Core

Cladding

n2

n2

n1c

accept = sin-1 [ n12 - n2

2 ]

Numerical Aperture = NA = sin accept = n12 - n2

2

Not all of the light that gets into one end of an opticalfiber gets out the other end.

Pin

Pout

Total Fiber Transmission Loss (dB) =Total Fiber Attenuation (dB) =

= -10 log10[ ]light power exiting the fiberlight power coupled into the fibera

= -10 log10[ ]Pout

Pina = 10 log10 [ ] Pin

Pout

What if the length of the fiber is l ?

Fiber Loss per Unit Length (dB/km)

= Fiber Attenuation per Unit Length (dB/km)

fiber length in km

10l= log10[ ]Pout

Pin

10l

= log10 [ ] Pin

Pout

al

Total Fiber Loss (dB) = Fiber Loss per Unit Length (dB/km) × l (km)

• Accounts for transmission loss in the fiber,

• NOT the losses encountered in getting the light into or out of the fiber (coupling losses).

Coupling Loss (in decibels)

Not all of the source light output gets into the fiber.

Coupling Loss (dB)= acoupling

[ ]power coupled into the fiberpower produced by the source= 10 log

10

[ ]power produced by the source

power coupled into the fiber= 10 log

10

Interference of Light WavesInterference of Light Waves “Superposition” In order for us to be able to observe sustained

interference between two or more light waves -(1) The sources of the waves must be

“monochromatic”, which means that they must have a single wavelength (or a single frequency, or a single “color”).

(2) The sources of the waves must be “coherent”, which means that the phase difference between the waveforms must remain constant, e.g.,

Huygens’ PrincipleHuygens’ Principle

All points on a given wave front are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate outward with speeds characteristic of waves in that medium. After some time has elapsed, the new position of the wave front is the surface tangent to the wavelets.

Page 1119 of the text

Huygens’ PrincipleHuygens’ Principle

Interference of WavesInterference of Waves

Huygen’s Principlep. 1119 of the text

Young’s Double-Slit ExperimentYoung’s Double-Slit Experiment

viewing screen

Sourced

two slits in metal plate

Light exiting the two slits interferes. In Phase Constructive Interference

Young’s Double-Slit ExperimentYoung’s Double-Slit Experiment

Sourced

two slits in metal plate

viewing screen

Light exiting the two slits interferes. Out of Phase Destructive Interference

Young’s Double-Slit ExperimentYoung’s Double-Slit Experiment Light exiting the two slits interferes. Constructive interference bright band

(fringe) Destructive interference dark fringe

Sourced

two slits in metal plate

viewing screen

Sourced

P

d sin path difference

r1

r2

Young’s Double-Slit ExperimentYoung’s Double-Slit Experiment In order to have a bright area at point P (called a

“bright fringe” or “constructive interference”), we must have

d sin = m (m = 0, ±1, ±2, ...)

Sourced

P

d sin path difference

r1

r2

In order to have a dark area at point P (called a “dark fringe” or “destructive interference”), we must have

d sin = (m + 1/2) (m = 0, ±1, ±2, …)

Young’s Double-Slit ExperimentYoung’s Double-Slit Experiment

The Diffraction GratingThe Diffraction Grating

The Diffraction GratingThe Diffraction Grating

A diffraction grating consists of a large number of equally spaced parallel slits, e.g., 5000 slits/cm.

Light diffracts (spreads out) as it leaves each slit.Light from each slit interferes with light from the other slits.

Each wavelength experiences constructive interference at a different angle .

blue

red

The Diffraction GratingThe Diffraction Grating

Just as with the double slit, for the diffraction grating we get constructive interference (“maxima” in the interference pattern) when

d sin = m(m = 0, 1, 2, 3, ...)

or sin =md

or = sin-1( )md

e.g., for m=1 and = blue= 475 nm

blue = sin-1 ( )475 nmd

and for m=1 and = red= 660 nm

red = sin-1 ( )660 nmd

red> blue

Interference in Thin FilmsInterference in Thin Films Interference effects are commonly

observed in “thin films”. Examples:

(1) thin layer of oil on water, (2) soap bubbles, (3) coatings on camera lenses and solar cells.

The varied colors observed when white light is incident on a thin film result from the interference of waves reflected from the two surfaces of the film.

Interference in Thin FilmsInterference in Thin FilmsRay 2 reflects off the bottom surface of the film, so it travels 2t farther than ray 1 .

1 2

If the thickness of the film is constant, then for some color (wavelength), rays 1 and 2 will add constructive interference.

For another color (wavelength), rays 1 and 2 will cancel destructive interference.

air n = 1

air n = 1

soap filmn 1.3

t = thickness

In reality, a soap film varies in thickness, so different parts have different colors.

Nonreflective Coatings for Solar CellsNonreflective Coatings for Solar Cells Solar cells

Convert sunlight directly into electricity. Frequently made of silicon.

t = thickness

Air n = 1SiO2

n 1.45

Sin = 3.5

1 2

Often coated with a thin, transparent film Such as silicon dioxide (SiO2).

Reduces the amount of reflected (and hence lost) sunlight.

Ray 2 travels 2t farther than Ray 1 .

Solar Cell Example (cont.)Solar Cell Example (cont.)

What must the film thickness, t, be so that we will get the least amount of green light (o = 550 nm) reflected from the surface?

The light wavelength inside the SiO2 is

SiO2 = = = 379 nm

o

nSiO2

550 nm1.45

(n = )o

n

Air

SiO2

Si

n = 1

n = 1.45

n = 3.5

0 1 20 = 550 nm

These two wavestend to cancel. The least reflectionof green light

SiO2 = 379 nm

t

Solar Cell Example (cont.)Solar Cell Example (cont.)

Ray 2 travels 2t farther than ray 1, and it does so in SiO2, where

SiO2 =

0

nSiO2

550 nm

1.45= = 379 nm for green light.

In order to make waves 1 and 2 cancel, the path difference, 2t ,

must equalSiO2

2, so we have

2t =SiO2

2

t = = = 94.8 nm SiO2

4

379 nm

4

For an SiO2 coating of this thickness, we get the minimum amount of green light reflected. However, other colors (wavelengths) are reflected.

Solar Cell Example (cont.)Solar Cell Example (cont.)