20/08/2010

1

i r

1 1 2 2sin sinn n

Waves

Ensure Sound is on:

Light Doesn’t Just Bounce

It Also Refracts!

Reflected: Bounces (Mirrors!)

Refracted: Bends (Lenses!)

0

1 1 1

id d f

i r

1

2

n2

n1

i = r

n1 sin(1) = n2 sin(2)

cv

n

Speed of light in

mediumIndex of refraction

Speed of

light in

vacuum

1n v c so

Index of Refraction

300,000,000 m/second: it’s not

just a good idea, it’s the law!

always!

20/08/2010

2

n1

n2

When light travels from one medium to another the speed

changes v=c/n, but the frequency is constant. So the

light bends:

n1 sin(1)= n2 sin(2)

1

2

1) n1 > n2

2) n1 = n2

3) n1 < n2

Preflight

Compare n1 to n2.

Snell’s Law

1 < 2

sin1 < sin2

n1 > n2

n1

n2

Snell’s Law Practice

n1sin 1 n2 sin 2

no

rmal

2

A ray of light traveling through the air (n=1) is incident on water

(n=1.33). Part of the beam is reflected at an angle r = 60. The

other part of the beam is refracted. What is 2?

1 r

Usually, there is both reflection and refraction!

sin(60) = 1.33 sin(2)

2 = 40.6 degrees

1 =r =60

Total Internal Reflection

normal

2

1

n2

n1

Recall Snell’s Law: n1 sin(1)= n2 sin(2)

(n1 > n2 2 > 1 )

1 = sin-1(n2/n1) then 2 = 90

c

Light incident at a larger angle will

only have reflection (i = r)

ir

―critical angle‖

The light ray MUST

travel in the direction

from larger n to

smaller n for a

critical angle to be

formed.

Review Equations (1)

cn

v

1 1 2 2sin sinn n

c f

11 1 2

2 2 2 1

sin

sin

v n

v n

n is the refractive index

and v is the speed in the

medium

Snell’s Law

Wave equation

Alternative

versions of

Snell’s Law

21 2

1

sin ,c

nn n

n

Critical angle for

total internal

reflection

20/08/2010

3

Light travels from air into water. If it enters water at 600 to the surface, find the

angle at which it is refracted and the speed at which it moves in the water

(nwater=1.33)

Understanding

60n=1.00

n=1.33 1 1 2 2sin sinn n

30

12 1

2

1 12 1

2

sin sin

sin sin

n

n

n

n

1

2

1.00sin sin 30

1.33

22

cn

v

cv

n

8

8

3.0 10

1.33

2.3 10

m

sv

m

s

22

Determine the critical angle for an air-water boundary (nwater=1.33, nair=1.00)

Understanding

21 2

1

1 2

1

sin ,

sin

c

c

nn n

n

n

n

1 1.00sin

1.33

49

c

A wave moves from medium 1 to medium 2. given that θ2=300, 2=0.50 cm,

v1=3.0 cm/s, and the index of refraction for n2 = 1.3 and for n1 = 1.0, find v2, 1,

θ1, and f

Understanding

2 1

1 2

n

n

21 2

1

1.30.50

1.0

0.65

n

n

cm

cm

2 1

1 2

n v

n v

2 1

1 2

12 1

2

1.03.0

1.3

2.3

n v

n v

nv v

n

cm

s

cm

s

2

2

vf

2.3

0.50

4.6

cm

sfcm

Hz

12

1 2

sin

sin

n

n

22 1

1

1 21 2

1

1

sin sin

sin sin

1.3sin sin 30

1.0

41

n

n

n

n

n2

n1

d

d

d d

n2

n1

Apparent depth:

Apparent Depth

actual fish

apparent fish

d d

n2

n1

Apparent Depth

20/08/2010

4

Can the person standing on the edge of the pool be

prevented from seeing the light by total internal reflection ?

1) Yes 2) No

Understanding

“There are millions of light ’rays’ coming from the light. Some of the rays will be totally reflected back into the water,but most of them will not.”

As we pour more water into bucket, what

will happen to the number of people who

can see the ball?

1) Increase 2) Same 3) Decrease

Understanding Refraction

As we pour more water into bucket, what

will happen to the number of people who

can see the ball?

1) Increase 2) Same 3) Decrease

Understanding Refraction Fiber Optics

Telecommunications

Arthroscopy

Laser surgery

Total Internal Reflection only

works if noutside < ninside

At each contact w/ the glass air interface, if the light hits at

greater than the critical angle, it undergoes total internal

reflection and stays in the fiber.

ninside

noutside

20/08/2010

5

Fiber Optics

At each contact w/ the glass air interface, if the light hits at

greater than the critical angle, it undergoes total internal

reflection and stays in the fiber.

We can be certain that

ncladding < ninside

ninside

Add ―cladding‖ so

outside material doesn’t

matter!

ncladding

noutside

Polarization

Transverse waves have a polarization

(Direction of oscillation of E field for light)

Types of Polarization

Linear (Direction of E is constant)

Circular (Direction of E rotates with time)

Unpolarized (Direction of E changes randomly)

x

y

Linear PolarizersLinear Polarizers absorb all electric fields perpendicular to their

transmission axis.

Dochroic (Polaroid H)

materials have the property

of selectively absorbing

one of two orthogonal

components of ordinary

light. The crystal actual

align opposite to what you

would intuitively expect

waves to be blocked. I.e.

vertical alignment block

vertical electric fields, by

absorbing their energy and

oscillating the electrons in a

vertical manner. Horizontal

waves will pass through.

Reflected Light is Polarized

20/08/2010

6

Unpolarized Light on

Linear Polarizer

Most light comes from electrons accelerating in

random directions and is unpolarized.

Averaging over all directions: Stransmitted= ½ Sincident

Always true for unpolarized light!

2S E

Intensity is

proportional

to the cross

product of E

and B

Linearly Polarized Light on Linear

Polarizer (Law of Malus)

Etranmitted = Eincident cos()

Stransmitted = Sincident cos2()

TA

is the angle between the

incoming light’s polarization, and

the transmission axis

Transmission axisIncident E

ETransmitted =Eincidentcos()

Recall S is intensity

and2S E

Intensity

2

2

0

2 2

0

2

0

2

0

0

cos

cos

1

2

1

2

1

2

trans

S E

S E

E

E

E

S

The average of 2cos

2 22

0 0

2

0

1 cos 21 1cos

2 0 2 2

1 1 1sin 2

2 2 4

1

2

1

2

d d

2 2

2 2

2

2

cos sin 1

cos sin cos 2

2cos 1 cos 2

1 cos 2cos

2

Question

Unpolarized light (like the light from the sun) passes through a polarizing

sunglass (a linear polarizer). The intensity of the light when it emerges is

1. zero

2. ½ what it was before

3. ¼ what it was before

4. ⅓ what it was before

5. need more information

Now, horizontally polarized light passes through the same glasses (which are

vertically polarized). The intensity of the light when it emerges is

1. zero

2. ½ what it was before

3. ¼ what it was before

4. ⅓ what it was before

5. Need more information

20/08/2010

7

Answer for 1

Unpolarized light (like the light from the sun)

passes through a polarizing sunglass (a linear

polarizer). The intensity of the light when it

emerges is

1. zero

2. 1/2 what it was before

3. 1/4 what it was before

4. 1/3 what it was before

5. need more information 0

1

2transS S

Answer for 2

Now, horizontally polarized light passes through the

same glasses (which are vertically polarized). The

intensity of the light when it emerges is

• zero

• 1/2 what it was before

• 1/4 what it was before

• 1/3 what it was before

• need more information

Since all light hitting the

sunglasses was

horizontally polarized

and only vertical rays

can get thru the

glasses, the glasses

block everything

Law of Malus – 2 Polarizers

Cool Link

unpolarized

light

E1

45

I = I0

TA

TA

0

TA

E0

I3

B1

unpolarized

light

E1

45

I = I0

TA

TA

0

TA

E0

I3

B1

1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0

S = S0

S1

S2

2) Light transmitted through first polarizer is vertically polarized. Angle between it and

second polarizer is =90º. S2 = S1 cos2(90º) = 0

unpolarized

light

E1

45

I = I0

TA

TA

0

TA

E0

I3

B1

unpolarized

light

E1

45

I = I0

TA

TA

0

TA

E0

I3

B1

Law of Malus – 3 Polarizers

2) Light transmitted through first polarizer is vertically polarized. Angle between it and

second polarizer is θ=45º

3) Light transmitted through second polarizer is polarized 45º from vertical. Angle

between it and third polarizer is θ=45º

I2= I1cos2(45)

I3 = I2 cos2 (45º)

= ½ I0 cos4 (45º)

I1= ½ I0

. I2 = I1 cos2 (45º) = ½ I0 cos2 (45º)

Angle is 45 degrees with

respect to last TA

I1=½I01) Light will be vertically polarized with:

Remember you can use I

or S for intensity.

20/08/2010

8

0

TA

TA

S1

S2

S0

60

TATA

S1

S2

S0

60

ACT: Law of Malus

A B

1) S2A > S2

B 2) S2A = S2

B 3) S2A < S2

B

S1= S0cos2(60)

S2= S1cos2(30) = S0 cos2(60) cos2(30)

S1= S0cos2(60)

S2= S1cos2(60) = S0 cos4(60)

E0E0

Brewster’s Angle

Light

The incident light can have the electric field of the

light waves lying in the same plane as the that is is

traveling. Light with this polarization is said to be

p-polarized, because it is parallel to the plane.

Light with the perpendicular polarization is said to

be s-polarized, from the German senkrecht—

perpendicular

Brewster’s angle

…when angle between reflected

beam and refracted beam is

exactly 90 degrees, reflected

beam is 100% horizontally

polarized !

Reflected light is partially polarized (more horizontal than

vertical). But…

2

1

tan B

n

n

n1 sin B = n2 sin (90-B)

n1 sin B = n2 cos (B)

horiz. and vert.

polarized

B B

90º-B

90º

horiz.

polarized

only! n1

n2

20/08/2010

9

Polarized so that the oscillation is in

same direction as to be flat with

reflecting surface. (polarized in the

plane of reflecting surface)

Brewster’s Angle

Since then

1 1 2 2

1 2

2 1

1 2

1 1

21

1

sin sin

sin

sin

sin

cos

tan

n n

n

n

n

n

n

n

1 2 90

2 1sin cos

Polarizing sunglasses are often considered to be better

than tinted glasses because they…

Preflight

block more light

block more glare

are safer for your eyes

are cheaperWhen glare is around

B, it’s mostly horiz.

polarized!

Polarizing sunglasses (when worn by someone standing

up) work by absorbing light polarized in which direction?

horizontal

vertical

Brewster’s Angle from air off glass

tanglass

air

n

n

1.5

tan1.0

56

56

Air n=1.0

Glass n=1.5

20/08/2010

10

How Sunglasses Work

Any light ray that comes from a reflection will

be slightly horizontally polarized, the glasses

will only allow vertically (or vertical

component of the unpolarized light) to pass

through to the eye. Thus reducing the glare.

ACT: Brewster’s Angle

When a polarizer is placed between the light source and

the surface with transmission axis aligned as shown, the

intensity of the reflected light:

(1) Increases (2) Unchanged (3) Decreases

T.A.

Dispersion

prism

White light

Blue light gets

deflected more

nblue > nred

The index of refraction n depends on color!

In glass: nblue = 1.53 nred = 1.52

Skier sees blue coming up

from the bottom (1), and

red coming down from the

top (2) of the rainbow.

Understanding

Which is red?

Which is blue?

Rainbow Power point

(1)

(1)

(2)

(2)

20/08/2010

11

LIKE SO! In second rainbow

pattern is reversedInterference Requirements

Need two (or more) waves

Must have same frequency

Must be coherent (i.e. waves must have definite

phase relation)

Interference for Sound …

For example, a pair of speakers, driven in phase,

producing a tone of a single f and :

l1 l2

But this won’t work for light rays--can’t get coherent pair of

sources

hmmm… I’m just

far enough away that l2-l1=/2, and I

hear no sound at

all!

Interference for Light …

Can’t produce coherent light from

separate sources. (f 1014 Hz)

Single source

Two different paths

Interference

possible here

• Need two waves from single source

taking two different paths

– Two slits

– Reflection (thin films)

– Diffraction

20/08/2010

12

ACT: Young’s Double Slit

Screen a distance

L from slits

Single source of

monochromatic

light

d

2 slits-

separated

by d

1) Constructive

2) Destructive

3) Depends on L

L

Light waves from a single source travel through 2 slits

before meeting on a screen. The interference will be:

The rays start in

phase, and travel the

same distance, so

they will arrive in

phase.

Understanding

Screen a distance

L from slits

Single source of

monochromatic

light

d

2 slits-

separated

by d

1) Constructive

2) Destructive

3) Depends on L

The rays start out of

phase, and travel the

same distance, so

they will arrive out of

phase.

L

½ shift

The experiment is modified so that one of the waves has its

phase shifted by ½ . Now, the interference will be:

Young’s Double Slit Concept

Screen a distance

L from slits

Single source of

monochromatic

light

d

2 slits-

separated

by d

L

At points where the

difference in path

length is 0, ,2, …,

the screen is bright.

(constructive)

At points where the

difference in path

length is

the screen is dark.

(destructive)

3 5, ,

2 2 2

Young’s Double Slit Key IdeaL

Two rays travel almost exactly the same distance.(screen must be very far away: L >> d)

Bottom ray travels a little further.

Key for interference is this small extra distance.

d

20/08/2010

13

d

Path length difference =

Young’s Double Slit Quantitative

d sin(

d

Constructive interference

Destructive interference(Where m = 0, 1, 2, …)

sin(θ) tan(θ) = y/L

sind m

1

sin2

d m

Need < d

y

L

m Ly

d

1

2m L

yd

(Where m = 0, 1, 2, …)

L is much larger than d so yellow lines are parallel

If You only know the Distances between

Sources and Point (or Large Distances)

Pm

1 2 , 0,1,2m mP S P S m m

1 2

1, 0,1,2

2m mP S P S m m

Constructive Interference

Destructive interference

m=0

Central

Maximum m=1 (first order maximum) m=0 (first order minimum)

Equations

1 2 , 0,1,2m mP S P S m m Constructive interference when angle not given

1 2

1, 0,1,2

2m mP S P S m m

Destructive interference when angle not given

sin , 0,1,2,d m m Constructive interference, angle given

1

sin , 0,1,2,2

d m m

Destructive interference, angle given

, 0,1,2,y

d m mL

Constructive interference, distance from

screen given

1, 0,1,2,

2

yd m m

L

Destructive interference, distance from

screen given

d

L

Understanding

y

When this Young’s double slit experiment is placed under

water. The separation y between minima and maxima

1) increases 2) same 3) decreases

…wavelength is shorter under water.

1 1

2 2m L L

m Ly

d d d

1 2

2 1

n

n

20/08/2010

14

UnderstandingIn the Young double slit experiment, is it possible to see interference

maxima when the distance between slits is smaller than the wavelength

of light?

1) Yes 2) No

Need: d sin = m => sin = m / d

If d then / d > 1

so sin > 1

Not possible!

Understanding

If one source to the screen is 1.2x10-5 m farther than the other source

and red light with wavelength 600 nm is used, determine the order

number of the bright spot.

Path length difference =5

1 2 1.2 10m mP S P S m

1 2

1 2

5

7

1.2 10

6.00 10

20

m m

m m

P S P S m

P S P Sm

m

m

We don’t know the

angle, so we use

Understanding

Light of wavelength 450 nm shines through openings 3.0 um apart. At what

angle will the first-order maximum occur? What is the angle of the first order

minimum?

sind m

1

7

1

6

sin

1 4.50 10sin

3.0 10

8.6

m

d

m

m

Since we are asked for the angle

1

sin2

d m

1

7

1

6

1

2sin

10 4.50 10

2sin

3.0 10

4.3

m

d

m

m

First order max starts at m=1,

first order min starts at m=0

Understanding

A point P is on the second-order maximum, 65 mm from one source and 45 mm

from another source. The sources are 40 mm apart. Find the wavelength of the

wave and the angle the point makes in the double slit.

65 45 2

20

2

10

mm mm

mm

mm

1 2m mP S P S m sind m

1

1

sin

2 10sin

40

30

m

d

mm

mm

20/08/2010

15

Understanding

For light of wavelength 650 nm, what is the spacing of the bright bands on a screen

1.5 m away if the distance between slits is 6.6x10-6 m?

Since we want the spacing on the

bands and are given the distance

the screen is from the slits

y m

L d

7

6

1 6.50 101.5

6.6 10

0.15

my L

d

mm

m

m

Therefore the distance between the principle bright band and the first order

band is 0.15m (this is actually the distance between any two bright bands

d

Path length difference 1-2

Multiple Slits(Diffraction Grating – N slits with spacing d)

L

= d sin

sind m

Constructive interference for all paths when

d

Path length difference 1-3 = 2d sind

1

2

3

Path length difference 1-4 = 3d sin

2

3

4

d

UnderstandingL

d

1

2

3

All 3 rays are interfering constructively at the point shown. If the intensity

from ray 1 is I0 , what is the combined intensity of all 3 rays?

1) I0 2) 3 I0 3) 9 I0

When rays 1 and 2 are interfering destructively, is the intensity

from the three rays a minimum?

1) Yes 2) No

Each slit contributes amplitude

Eo at screen. Etot = 3 Eo. But I a

E2. Itot = (3E0)2 = 9 E0

2 = 9 I0this one is still there!

these add to zero

3

2

3

2

3

2

3

2 dsin

Three slit interference

I0

9I0

20/08/2010

16

For many slits, maxima are still at sin md

Region between maxima gets suppressed more and

more as no. of slits increases – bright fringes become

narrower and brighter.

10 slits (N=10)

dsin

inte

nsity

20

2 slits (N=2)

dsin

inte

nsity

20

Multiple Slit Interference (Diffraction Grating)

Peak location

depends on

wavelength!

N slits with spacing d

Constructive Interference Maxima are at:

sin md

* screen

VERY far

away

Diffraction Grating

Same as for Young’s

Double Slit !

UnderstandingCompare the spacing produced by a diffraction grating with 5500 lines

in 1.1 cm for red light (620 nm) and green light (510 nm). The distance

to the screen is 1.1 m

First we need to

determine d the distance

between the grates

2

6

1.1 10

5500

2.0 10

md

lines

m

Now for the spacing, y

Ly

d

7

6

1.1 6.20 10

2.0 10

0.34

red

m my

m

m

7

6

1.1 5.10 10

2.0 10

0.28

green

m my

m

m

Note: the larger

spacing for red than

for green

UnderstandingA diffraction grating has 1000 slits/cm. When light of wavelength 480

nm is shone through the grating, what is the separation of the bright

bands 4.0 m away? How many orders of this light are seen?

First we need to

determine d the distance

between the grates

5

distance (m)

0.01

1000

1.0 10

dslits

m

m

Now for the spacing, y

Ly

d

7

5

4.0 4.8 10

1.0 10

0.192

red

m my

m

m

For max we will

set the angle θ

to 900

5

7

sin

sin 90

1.0 10

4.8 10

20.8

20

d m

dm

d

m

m

20/08/2010

17

Single Slit Interference?!Wall

Screen with opening

shadow

bright

This is not what is actually

seen!

Diffraction Rays

Huygens’ Wave Model

In this theory, every point on a wave front is considered a point source for tiny

secondary wavelets, spreading out in front of the wave as the same speed of

the wave itself.

•

•

Diffraction/ HuygensEvery point on a wave front acts as a source of tiny

wavelets that move forward.

We will see maxima and

minima on the wall.

Light waves originating at

different points within

opening travel different

distances to wall, and can

interfere!

20/08/2010

18

Huygens’ Principle

The Dutch scientist Christian

Huygens, a contemporary of

Newton, proposed Huygens’

Principle, a geometrical way

of understanding the behavior

of light waves.

Huygens Principle: Consider a wave front of light:

1. Each point on the wave front is a new source of

a spherical wavelet that spreads out spherically

at wave speed.

2. At some later time, the new wave front is the

surface that is tangent to all of the wavelets.

Christian Huygens

(1629 - 1695)

1st minima

Central

maximum

Single Slit Diffraction

As we have seen in the

demonstrations with light

and water waves, when light

goes through a narrow slit, it

spreads out to form a

diffraction pattern.

Now, we want to understand

this behavior in more detail.

W

sin2

w

2

W

Rays 2 and 2 also start W/2 apart and have the same path

length difference.

1st minimum at:

When rays 1 and 1

interfere destructively.

sin2 2

w

Under this condition, every ray originating in top half of slit interferes destructively

with the corresponding ray originating in bottom half.

Single Slit Diffraction

sin2 2

w sinW sin

w

20/08/2010

19

w

Rays 2 and 2 also start w/4 apart and have the same path

length difference.

2nd minimum at

Under this condition, every ray originating in top quarter of slit interferes destructively

with the corresponding ray originating in second quarter.

Single Slit Diffraction

w

4

wsin( )

4

When rays 1 and 1

will interfere destructively.

wsin( )

4 2

wsin( )

4 2

sin( ) 2w

2sin( )

w

Analyzing Single Slit Diffraction

For an open slit of width w, subdivide the opening into segments and imagine a

Hyugen wavelet originating from the center of each segment. The wavelets

going forward (=0) all travel the same distance to the screen and interfere

constructively to produce the central maximum.

Now consider the wavelets going at an angle such that w sin @ w . The

wavelet pair (1, 2) has a path length difference Dr12 = /2, and therefore will

cancel. The same is true of wavelet pairs (3,4), (5,6), etc. Moreover, if the

aperture is divided into m sub-parts, this procedure can be applied to each sub-

part. This procedure locates all of the dark fringes.

thsin ; 1, 2, 3, (angle of the m dark fringe)m mm mw

@

w w

w

2

Conditions for Diffraction Minima

th

sin ; 1, 2, 3,

(angle of the m dark fringe)

m mm mw

@

w w

w w w w w w

w 2w w

Condition for quarters of slit to destructively interfere

sin mw

(m=1, 2, 3, …)

Single Slit Diffraction Summary

Condition for halves of slit to destructively interfere sin( )w

sin( ) 2w

Condition for sixths of slit to destructively interfere sin( ) 3w

THIS FORMULA LOCATES MINIMA!!

Narrower slit => broader pattern

All together…

Note: interference only occurs when the (w)idth >

mym

L w

If you do not

know the angle

20/08/2010

20

UnderstandingA slit is 2.0x10-5 m wide and is illuminated by light of wavelength 550 nm.

Determine the width of the central maximum, in degrees and in centimetres when

L=0.30m?

We will find the position of the first order minimum, which is

one-half the total width of the central maximum.

sin mm w

1

7

1

5

sin

sin

sin

1 5.50 10sin

2.0 10

1.6

m

m

m

m w

m

w

m

w

m

m

The full width is twice this, 3.20

mwym

L

7

5

1 5.50 10 0.30

2.0 10

0.00825

m

m Ly

w

m m

m

m

Therefore the width of Central

Maximum is 2(.825cm)=1.65 cm

UnderstandingLight is beamed through a single slit 1.0 mm wide. A diffraction pattern is

observed on a screen 2.0 m away. If the central band has a width of 2.5 mm,

determine the wavelength of the light.

To use the equations,

we need half of the

width

33

1

2.5 101.25 10

2

my m

mwym

L

1

3 3

7

1.0 10 1.25 10

2.0 1

6.25 10

wy

Lm

m m

m

m

Example: Diffraction of a laser

through a slit

Light from a helium-neon laser ( = 633 nm) passes through a narrow slit and

is seen on a screen 2.0 m behind the slit. The first minimum of the diffraction

pattern is observed to be located 1.2 cm from the central maximum.

How wide is the slit?

11

(0.012 m)0.0060 rad

(2.00 m)

y

L

74

3

1 1

(6.33 10 m)1.06 10 m 0.106 mm

sin (6.00 10 rad)w

@

1.2 cm

m

Lmw

y

7

4

2

2.0 1 6.33 101.06 10

1.2 10

m mm

m

or

Width of a Single-Slit

Diffraction Pattern

; 1,2,3, (positions of dark fringes)m

m Ly m

w

2(width of diffraction peak from min to min)

Lwidth

w

width

-y1 y1 y2 y30

20/08/2010

21

Understanding

Two single slit diffraction patterns are shown. The distance from

the slit to the screen is the same in both cases.

Which of the following could be true?

1

2

(a) The slit width w is the same for both; 12.

(b) The slit width w is the same for both; 12.

(c) The wavelength is the same for both; width w1<w2.

(d) The slit width and wavelength is the same for both; m1<m2.

(e) The slit width and wavelength is the same for both; m1>m2.

sin mw

(m=1, 2, 3, …) my

mL w

Understanding

What is the wavelength of a ray of light whose first side diffraction

maximum is at 15o, given that the width of the single is 2511 nm?

sin mw

sinw

m

We want the first maximum, therefore m is approximately 1.5

2511 sin 15

1.5

430

nm

nm

(degrees)

(degrees)

(degrees)

Combined Diffraction

and Interference (FYI only)

So far, we have treated

diffraction and interference

independently. However, in a

two-slit system both

phenomena should be present

together.

2

2

2slit 1slit

sin4 cos ;

sin ;

sin .

I I

a ay

L

d dy

L

a

a

a

d

a a

Notice that when d/a is an integer, diffraction minima will fall on top of ―missing‖

interference maxima.

Maxima and minima will be a series of bright and dark

rings on screen

Central maximum

sin 1.22D

First diffraction minimum is at

Hole with diameter D

light

Diffraction from Circular Aperture

1st diffraction minimum

20/08/2010

22

UnderstandingA laser is shone at a screen through a very small hole. If you

make the hole even smaller, the spot on the screen will get:

(1) Larger (2) Smaller

Which drawing correctly depicts the pattern of light on the

screen?

(1) (2) (3) (4)

sin 1.22D

sin

1.22D

I

Intensity from Circular Aperture

First diffraction minima

These objects are just resolved

Two objects are just resolved when the maximum of one is

at the minimum of the other.

Resolving Power

To see two objects distinctly, need objects » min

min

objects

Improve resolution by increasing objects or decreasing min

objects is angle between objects and aperture:

tan objects d/y

sin min min = 1.22 /D

min is minimum angular separation that aperture

can resolve:

D

dy

20/08/2010

23

Understanding

Astronaut Joe is standing on a distant planet with binary

suns. He wants to see them but knows it’s dangerous to look

straight at them. So he decides to build a pinhole camera by

poking a hole in a card. Light from both suns shines through

the hole onto a second card.

But when the camera is built, Astronaut Joe can only see one

spot on the second card! To see the two suns clearly, should

he make the pinhole larger or smaller?

Decrease min = 1.22 / D

Want objects > min

Increase D !

min minsin 1.22D

ACT: Resolving Power

How does the maximum resolving power of your eye

change when the brightness of the room is decreased.

1) Increases 2) Constant 3) Decreases

When the light is low, your pupil dilates (D can increase

by factor of 10!) But actual limitation is due to density of

rods and cones, so you don’t notice an much of an

effect!

Summary

Interference: Coherent waves

Full wavelength difference = Constructive

½ wavelength difference = Destructive

Multiple Slits

Constructive d sin() = m m=1,2,3…)

Destructive d sin() = (m + 1/2) 2 slit only

More slits = brighter max, darker mins

Huygens’ Principle: Each point on wave front acts as

coherent source and can interfere.

Single Slit:

Destructive: w sin() = m m=1,2,3…)

Thin Film Interference

n1 (thin film)

n2

n0=1.0 (air)

t

1 2

Get two waves by reflection off of two different interfaces.

Ray 2 travels approximately 2t further than ray 1.

20/08/2010

24

Reflection + Phase Shifts

n1

n2

Upon reflection from a boundary between two transparent

materials, the phase of the reflected light may change.

• If n1 > n2 - no phase change upon reflection.

• If n1 < n2 - phase change of 180º upon reflection.

(equivalent to the wave shifting by /2.)

Incident wave Reflected wave

Thin Film Summary (1)

n1 (thin film)

n2

n = 1.0 (air)

t

1 2

Ray 1: 1 = 0 or ½

Determine , number of extra wavelengths for each ray.

If |(2 – 1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructive

If |(2 – 1)| = 0, 1, 2, 3 …. (m) constructive

Note: this is

wavelength in film!

(film= o/n1)+ 2 t/ film

Reflection Distance

Ray 2: 2 = 0 or ½

This is important!

Thin Film Summary (2)

n1 (thin film)

n2

tvacuum

filmn

1 2n n

1 2n n

2 filmt m

12

2filmt m

Phase shift after reflection

No phase shift after reflection

No Phase shift

Phase shift

Thin Film Practice

nglass = 1.5

nwater= 1.3

n = 1.0 (air)

t

1 2

1 = ½

2 = 0 + 2t / glass = 2t nglass/ 0= 1

Blue light (o = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167 nm)

floating on top of water (nwater = 1.3).

Is the interference constructive or destructive or neither?

Phase shift = 2 – 1 = ½ wavelength

Reflection at air-film interface only

20/08/2010

25

ACT: Thin Film

nglass =1.5

nplastic=1.8

n=1 (air)

t

1 2

1 = ½

2 = ½ + 2t / glass = ½ + 2t nglass/ 0= ½ + 1

Blue light = 500 nm is

incident on a very thin film

(t = 167 nm) of glass on

top of plastic. The

interference is:

(1) constructive

(2) destructive

(3) neither

Phase shift = 2 – 1 = 1 wavelength

Reflection at both interfaces!

Understanding

The gas looks:

• bright

• dark

A thin film of gasoline (ngas=1.20) and a thin film of oil (noil=1.45) are floating on water (nwater=1.33). When the thickness of the two films is exactly one wavelength…

t =

nwater=1.3

ngas=1.20

nair=1.0

noil=1.45

1,gas = ½

The oil looks:

• bright

• dark

2,gas = ½ + 2 1,oil = ½ 2,oil = 2

| 2,gas – 1,gas | = 2 | 2,oil – 1,oil | = 3/2

constructive destructive

UnderstandingLight travels from air to a film with a refractive index of 1.40 to water (n=1.33). If

the film is 1020 nm thick and the wavelength of light is 476 nm in air, will

constructive or destructive interference occur?

nfilm=1.4

nwater=1.33

n=1.0 (air)

t

1 2

Ray 1: Since nair<nfilm, we have a phase shift of /2 when it reflects

of the film.

Ray 2: Since nfilm>nwater, we have

no phase shift. So we need only

determine the number of extra

wavelengths travelled by the ray.

74.76 10

1.40

340

airfilm

filmn

m

nm

2

20

2 10200

340

6

film

t

nm

nm

1 = ½

Phase shift = |2 – 1|= 5½ wavelength Destructive interference

UnderstandingA camera lens (n=1.50) is coated with a film of magnesium fluoride (n=1.25).

What is the minimum thickness of the film required to minimize reflected light of

wavelength 550 nm?

nfilm=1.25

nlens=1.50

n=1.0 (air)

t

1 2

Ray 1: Since nair<nfilm, we have a phase shift of /2 when it reflects

of the film.

Ray 2: Since nfilm<nlens, we have a phase shift of /2 when it

reflects of the lens.

1 = ½

2 = ½ + 2t / film

0

0

2 1

2

4

550

4 1.25

110

film

film

tn

tn

nm

nm

We need 1 and 2 to be

out by a minimum ½ if

we require destructive

interference

vacuumfilm

n

Therefore minimum thickness is 110 nm

0

2 1

2

filmtn

20/08/2010

26

Light

Color

Color Addition & Subtraction

Spectra

What do we see? Our eyes can’t detect intrinsic light from objects

(mostly infrared), unless they get ―red hot‖

The light we see is from the sun or from artificial light (bulbs, etc.)

When we see objects, we see reflected light

immediate bouncing of incident light (zero delay)

Very occasionally we see light that has been absorbed, then re-emitted at a different wavelength

called fluorescence, phosphorescence, luminescence

Colours Light is characterized by frequency, or more

commonly, by wavelength

Visible light spans from 400 nm to 700 nm

or 0.4 m to 0.7 m; 0.0004 mm to 0.0007 mm, etc.

White light White light is the combination of all wavelengths, with

equal representation

―red hot‖ poker has much more red than blue light

experiment: red, green, and blue light bulbs make

white

RGB monitor combines these colors to display white

blue light green light red lightwavelength

combined, white light

called additive color

combination—works

with light sources

20/08/2010

27

Introduction to Spectra We can make a spectrum out of light,

dissecting its constituent colors

A prism is one way to do this

A diffraction grating also does the job

The spectrum represents the wavelength-by-wavelength content of light

can represent this in a color graphic like that above

or can plot intensity vs. wavelength

Why do things look the way

they do?

Why are metals shiny? Recall that electromagnetic waves are generated from

accelerating charges (i.e., electrons)

Electrons are free to roam in conductors (metals)

An EM wave incident on metal readily vibrates electrons on

the surface, which subsequently generates EM radiation of

exactly the same frequency (wavelength)

This indiscriminate vibration leads to near perfect reflection,

and exact cancellation of the EM field in the interior of the

metal—only surface electrons participate

What about glass? Why is glass clear?

The clear piece of glass is transparent to

visible light because the available electrons in

the material which could absorb the visible

photons have no available energy levels

above them in the range of the quantum

energies of visible photons. The glass atoms

do have vibrational energy modes which can

absorb infrared photons, so the glass is not

transparent in the infrared. This leads to the

greenhouse effect.