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

qi qr

q1

q2

n2

n1

qi = qr

n1 sin(q1) = n2 sin(q2)

cvn

Speed of light in medium

Index 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!

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(q1)= n2 sin(q2)

q1

q2

1) n1 > n2

2) n1 = n2

3) n1 < n2

Preflight

Compare n1 to n2.

Snell’s Law

q1 < q2sinq1 < sinq2

n1 > n2

n1

n2

Snell’s Law Practice

n1sin1n2 sin2

norm

al

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 qr = 60. The other part of the beam is refracted. What is q2?

1 r

Usually, there is both reflection and refraction!

sin(60) = 1.33 sin(q2)

q2 = 40.6 degrees

q1 =qr =60

Total Internal Reflection

normal

q2

q1

n2

n1

Recall Snell’s Law: n1 sin(q1)= n2 sin(q2)

(n1 > n2 q2 > q1 )

q1 = sin-1(n2/n1) then q2 = 90

qc

Light incident at a larger angle will only have reflection (qi = qr)

qiqr

“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)c

nv

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

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

60 n=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

12

1.00sin sin 30

1.33

22

cn

vc

vn

8

8

3.0 10

1.33

2.3 10

msv

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

1.00.65

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

0.504.6

cmsfcmHz

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

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

Fiber OpticsAt 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

PolarizationTransverse 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)

xz

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

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 EIntensity is proportional to the cross product of E and B

Linearly Polarized Light on Linear Polarizer (Law of Malus)

Etranmitted = Eincident cos(q)

Stransmitted = Sincident cos2(q)

TA

is the angle between the incoming light’s polarization, and the transmission axis

Transmission axisIncident E

ETransmitted

Eabsorbed

=Eincidentcos(q)

Recall S is intensityand

2S E

Intensity

2

2

0

2 20

20

20

0

cos

cos

1

2

1

21

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

21

2

d d

2 2

2 2

2

2

cos sin 1

cos sin cos 2

2cos 1 cos 2

1 cos 2cos

2

QuestionUnpolarized 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. zero2. ½ what it was before3. ¼ what it was before4. ⅓ what it was before5. Need more information

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

I = I0

TATA

0

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

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

I = I0

TATA

0

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

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.

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 qB = n2 sin (90-qB) n1 sin qB = n2 cos (qB)

horiz. and vert. polarized

qB qB

90º-qB

90º

horiz. polarized

only! n1

n2

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 cheaper

When glare is around qB, 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

tan glass

air

n

n

1.5tan

1.0

56

56

Air n=1.0

Glass n=1.5

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 AngleWhen 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

Wow look at the

variation in index of

refraction!

Which is red?

Which is blue?

Rainbow Power point

(1)

(1)

(2)

(2)

LIKE SO! In second rainbow pattern is reversed

Interference Requirements

Need two (or more) wavesMust have same frequencyMust 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 l:

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=l/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

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 l shift

The experiment is modified so that one of the waves has its phase shifted by ½ l. 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, l,2l, …, 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

d

Path length difference =

Young’s Double Slit Quantitative

d sin( )q

d

Constructive interference

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

sin(θ) tan(θ) = y/L

sind m

1sin

2d m

Need l < d

y

L

m Ly

d

12

m Ly

d

(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=0Central

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

Equations1 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

1sin , 0,1,2,

2d 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 mL

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

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 q = m l => sin q = m l / d

If l > d then l / d > 1

so sin q > 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 = 51 2 1.2 10m mP S P S m

1 2

1 2

5

7

1.2 10

6.00 1020

m m

m m

P S P S m

P S P Sm

m

m

We don’t know the angle, so we use

UnderstandingLight 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

16

sin

1 4.50 10sin

3.0 10

8.6

m

d

m

m

Since we are asked for the angle

1sin

2d m

1

7

16

12sin

10 4.50 10

2sin3.0 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

210

mm mm

mm

mm

1 2m mP S P S m sind m

1

1

sin

2 10sin

40

30

m

d

mm

mm

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

6.6 100.15

my L

d

mm

mm

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 sinq

sind m Constructive interference for all paths when

d

Path length difference 1-3= 2d sinqd

1

23

Path length difference 1-4= 3d sinq

=l

=2l=3l

4

d

UnderstandingL

d

1

23

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

23

2

3

23

2

dsin

Three slit interference

I0

9I0

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

nsi

ty

l 2l0

2 slits (N=2)

dsin

inte

nsi

ty

l 2l0

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

q

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

2.0 100.34

red

m my

mm

7

6

1.1 5.10 10

2.0 100.28

green

m my

mm

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

dslitsm

m

Now for the spacing, y

Ly

d

7

5

4.0 4.8 10

1.0 100.192

red

m my

mm

For max we will set the angle θ to 900

5

7

sin

sin 90

1.0 10

4.8 1020.8

20

d m

dm

d

m

m

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!

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

1

1

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

2

2

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 sinw

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

1

1

wsin( )

4

2

2

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 (q=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 = l w sin q @ w q. The wavelet pair (1, 2) has a path length difference Dr12 = l/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

w2

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 SummaryCondition 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 >

my mL w

If you do not

know the angle

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

15

sin

sin

sin

1 5.50 10sin

2.0 10

1.6

m

m

m

m w

m

wm

w

m

m

The full width is twice this, 3.20

mwym

L

7

5

1 5.50 10 0.30

2.0 100.00825

m

m Ly

w

m m

mm

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 (l = 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

31 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-SlitDiffraction 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

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?

l1

l2

(a) The slit width w is the same for both; l1>l2.(b) The slit width w is the same for both; l1<l2.(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

1.5430

nm

nm

q (degrees)

q (degrees)

q (degrees)

Combined Diffractionand 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

22slit 1slit

sin4 cos ;

sin ;

sin .

I I

a ay

Ld 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

q

Hole with diameter D

light

Diffraction from Circular Aperture

1st diffraction minimum

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 PowerTo see two objects distinctly, need qobjects » qmin

qmin

qobjects

Improve resolution by increasing qobjects or decreasing qmin

qobjects is angle between objects and aperture:

tan qobjects d/y

sin qmin qmin = 1.22 /D

qmin is minimum angular separation that aperture can resolve:

D

dy

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 qmin = 1.22 l / D

Want qobjects > qmin

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(q) = m l(m=1,2,3…) Destructive d sin(q) = (m + 1/2) l 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(q) = m l (m=1,2,3…)

oppo

site

!

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.

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 l/2.)

Incident wave Reflected wave

Thin Film Summary (1)

n1 (thin film)

n2

n = 1.0 (air)

t

1 2

Ray 1: d1 = 0 or ½

Determine ,d number of extra wavelengths for each ray.

If |(d2 – d1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructive

If |(d2 – d1)| = 0, 1, 2, 3 …. (m) constructive

Note: this is wavelength in

film! (lfilm= lair/nfilm)+ 2 t/ lfilm

Reflection Distance

Ray 2: d2 = 0 or ½

This is important!

1 2

2 1

n

n

1.0filmair

film

n

Thin Film Summary (2)

n1 (thin film)

n2

tvacuumfilm

filmn

1 2n n

1 2n n

2 filmt m

12

2 filmt 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

d1 = ½

d2 = 0 + 2t / lglass = 2t nglass/ l0= 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 = d2 – d1 = ½ wavelength

Reflection at air-film interface only

ACT: Thin Film

nglass =1.5

nplastic=1.8

n=1 (air)

t

1 2

d1 = ½

d2 = ½ + 2t / lglass = ½ + 2t nglass/ l0= ½ + 1

Blue light l = 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 = d2 – d1 = 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 = l

nwater=1.3

ngas=1.20

nair=1.0

noil=1.45

d1,gas = ½

The oil looks: • bright• dark

d2,gas = ½ + 2 d1,oil = ½ d2,oil = 2

| d2,gas – d1,gas | = 2 | d2,oil – d1,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 2Ray 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

1.40340

airfilm

filmn

m

nm

2

20

2 10200

3406

film

t

nm

nm

d1 = ½

Phase shift = |d2 – d1|= 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 2Ray 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.

d1 = ½ d2 = ½ + 2t / lfilm

0

0

2 1

2

4

550

4 1.25

110

film

film

tn

tn

nm

nm

We need d1 and d2 to be out by a minimum ½ if we require destructive interference

airfilm

filmn

Therefore minimum thickness is 110 nm

02 1

2 1

2filmtn

LightColor

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 wavelengthcalled 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 colorcombination—workswith light sources

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

dissecting its constituent colorsA prism is one way to do thisA diffraction grating also does the job

The spectrum represents the wavelength-by-wavelength content of lightcan represent this in a color graphic like

that aboveor 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.