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Optics News: Kepler Mission successfully launched Saturday!

•Solar orbit•Watching 105 stars for planetary transits.

•Expects to find 50 Earth like planets in a few years.

•95Megapixel CCD•Why cant we directly image distant planets?

• Find out this week!

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Planet Mercury passing in front of the sun.

• Brief dimming of the sun by about 1 part in 105

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Kepler

Week 3

• Interference and Diffraction of Light• Young’s Double Slit Experiment• Thin Film Interference• Michelson Interferometer• Single Slit Diffraction• Rayleigh’s Criterion• Diffraction Gratings

Assignment 2

• Chap 33 # 14, 22, 27, 34,50, 59• Chap 34 #6, 88, 92

My Error?–How is the energy distributed for waves in a string?–Standing waves: I was right: energy alternates between kinetic and potential energy.–Travelling wave: text book argues that max KE coincides with Max PE because position of max velocity coincides with position of max stretching. Figure 16-12–Localisation of energy is often difficult in physics

Interference of Waves• Waves sum to

– a maximum (constructive)– or minimum (destructive)

– Use– Combined wave is

– New phase, new amplitude (zero for

y1(x,t) ym sin(kx t)

y2(x,t) ym sin(kx t )

)(2

1cos)(

2

1sin2sinsin BABABA

)2

1sin(]

2

1cos2[),( tkxytxy m

n

From week 1

Phasor Method for InterferenceFrom week 1, phasor: a vector, length equals wave amplitude, direction: relative phase angle (compared with some standard wave)For multiple waves, resultant is vector sum of their phasorsThis case: (2/3) radians

behind

Vector sum of the two phasors

Near destructive interference: =0.95

Interference from Thin Films

Fig. 35-15

• Light reflects off both faces of film.

• Light reflecting at b has to travel a bit further.

• There is a phase difference: outgoing waves may experience constructive or destructive interference.

Interference Thought Experiment•Light is quantised: photons : energy E=hf.• Since c=f E =hc/. h= Planck’s constant = 6 x 10-

34

•Green LED light (=500nm) power 1mW. Just visible from 1km distance.

•Energy per photon = 6 x 10-34 x 3 x 108/500 x 10-9 = 3.6 x 10-19 J

•Photons per second from LED = power/energy per photon = 10-3/3.6 x 10-19 ~ 3 x 1015 per second

•Photons per second entering your eye = number per second emitted x area of pupil/area of 1km sphere ~ 1000 photons/second. Or 1 photon/millisecond.

•But photons travel 300km in 1 millisecond.•So in 1km path no photons at all most of the time. •Similarly in a 10-6 meter thckness soap bubble film photons only present a tiny fraction of the time.

DEMO bubbles

How can there be interference when only 1

photon is present at a time?• Seems like interference of waves • Not interference of photons • Interference of possibilities

• We will continue to treat light as waves but remember that the reality is the strange and absurd concept of the interference of something intangible: we call it a wavefunction but we don’t know what it is.

Reflection Phase Shifts

Fig. 35-16

n1 n2

n1 > n2

n1 n2

n1 < n2

Reflection Reflection Phase ShiftOff lower index 0Off higher index 0.5 wavelength

(35-15)Fig. 35-15

Rope: high refractive index, low velocity

Wall or thin string: low refractive index, high velocity

No phase change on reflection

Important to understand phase change at reflection

Film Thickness Much Less Than

If L is much less than l, for example L < 0.1, then phase difference due to the path difference 2L can be neglected.

Phase difference between r1 and r2 will always be ½ wavelength destructive interference film will appear dark when viewed from illuminated side.

ie: zero reflection as film thickness goes to zero

r2

r1

(35-17)

Note choice of reference phase is arbitrary

Length of resultant phasor is independent of choice of reference phasor direction

Frequency, Wavelength in different media

•Frequency of waves remains constant (except from Doppler effect)

•Refractive index n = c/v•When waves travel slow wavelength

must reduce.

•Hence phase changes c/v times greater for a given distance.

nn n

v v

c c n

Equations for Thin-Film Interference

Fig. 35-17

Three effects can contribute to the phase difference between r1 and r2.

1. Differences in reflection conditions.

2. Difference in path length traveled.

3. Differences in the media in which the waves travel. One must use the wavelength in each medium (/ n) to calculate the phase.

2

odd number odd number2 wavelength = (in-phase waves)

2 2 nL

½ wavelength phase difference to difference in reflection of r1 and r2

2

0

22 integer wavelength = integer (out-of-phase waves)nL

22

n n

12

2

2 for 0,1,2, (maxima-- bright film in air)L m mn

2

2 for 0,1,2, (minima-- dark film in air)L m mn

(35-16)

Why Does Light Refract

• Life Saver on the Beach• Which trajectory?

sand

Seadrowning

Lifesaver

Last week

Law of Refraction

Index of Refraction:c

nv

Fig. 35-3

1 2 1 1

1 2 2 2

vt

v v v

11

22

sin (for triangle )

sin (for triangle )

hcehc

hcghc

1 1 1

2 2 2

sin

sin

v

v

1 21 2

and c c

n nv v

1 1 2

2 2 1

sin

sin

c n n

c n n

Law of Refraction: 1 1 2 2sin sinn n (35-3)

Why does angle of incidence equal angle of

reflection?–Consider trajectories where angle of incidence was not equal to angle of reflection.

–Now remember phasors–Consider the sum of all trajctories

–At edges phase angle increases steadily: phasors create circles that add to zero

–Near centre phase shifts approach zero:phasors add to give a finite resultant

Path length

Trajectory number

1 n… m… N ∞

Paint Black Stripes on the Mirror– Consider the same sum of all

trajectories.– Now black out mirror for bits that

make up phase shift of the phasors– Now the phasor circles are only half

circles.– At edges where the phase angle

increases steadily we have cut out the phasors components for half the circles so there is now a large resultant.

– Now trajectories for which angle of incidence is not equal to angle of reflection are allowed.

– This is a diffraction grating.

Path length

Trajectory number

1 n… m… N ∞

1 n… m… N… ∞ trajectory number

Wavelength and Index of Refraction

Fig. 35-4

nn n

v v

c c n

nn

v c n cf f

n The frequency of light in a medium is the same

as it is in vacuum.

Since wavelengths in n1 and n2 are different, the two beams may no longer be in phase.

11 1

1 1

Number of wavelengths in : n

L L Lnn N

n

22 2

2 2

Number of wavelengths in : n

L L Lnn N

n

2 22 1 2 1 2 1Assuming :

Ln Ln Ln n N N n n

(35-4)

Diffraction pattern from a single narrow slit.

Diffraction and the Wave Theory of Light

Centralmaximum

Side or secondarymaxima

Light

Fresnel Bright Spot.

Brightspot

Light

These patterns cannot be explained using geometrical optics (Ch. 34)!

(36-2)

Diffraction

Fig. 35-7

For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting.

(35-6)

DEMO ripple tank

Young’s Double Slit Experiment

Fig. 35-8

(35-7)

The phase difference between two waves can change if the waves travel paths of different lengths.

Calculating Fringes

Fig. 35-10

What appears at each point on the screen is determined by the path length difference L of the rays reaching that point.

Path Length Difference: sinL d

(35-8)

When the path length difference between rays r1 and r2 is /2, the two rays will be out of phase when they reach P1 on the screen, resulting in destructive interference at P1. The path length difference is the distance from the starting point of r2 at the center of the slit to point b.

For D>>a, the path length difference between rays r1 and r2 is (a/2) sin .

Fig. 36-4

Diffraction by a Single Slit: Locating the Minima

(36-3)

Repeat previous analysis for pairs of rays, each separated by a vertical distance of a/2 at the slit.

Setting path length difference to /2 for each pair of rays, we obtain the first dark fringes at:

Fig. 36-5

Diffraction by a Single Slit: Locating the Minima, cont'd

(first minimum)sin sin2 2

aa

For second minimum, divide slit into 4 zones of equal widths a/4 (separation between pairs of rays). Destructive interference occurs when the path length difference for each pair is /2.

(second minimum)sin sin 24 2

aa

Dividing the slit into increasingly larger even numbers of zones, we can find higher order minima:

(minima-dark fringes)sin , for 1,2,3a m m

(36-4)

Fig. 36-6

To obtain the locations of the minima, the slit was equally divided into N zones, each with width x. Each zone acts as a source of Huygens wavelets. Now these zones can be superimposed at the screen to obtain the intensity as a function of , the angle to the central axis.

To find the net electric field E (intensity E2) at point P on the screen, we

need the phase relationships among the wavelets arriving from different zones:

Intensity in Single-Slit Diffraction, Qualitatively

phase path length2

difference difference

2sinx

N=18

= 0

small

1st min.

1st sidemax.

(36-5)

Here we will show that the intensity at the screen due to a single slit is:

Fig. 36-7

Intensity in Single-Slit Diffraction, Quantitatively

2

sin (36-5)mI I

1where sin (36-6)

2

a

, for 1,2,3m m In Eq. 36-5, minima occur when:

sin , for 1,2,3

or sin , for 1,2,3

(minima-dark fringes)

am m

a m m

If we put this into Eq. 36-6 we find:

(36-6)

If we divide the slit into infinitesimally wide zones x, the arc of the phasors approaches the arc of a circle. The length of the arc is Em. is the difference in phase between the infinitesimal vectors at the left and right ends of the arc. is also the angle between the 2 radii marked R.

Proof of Eqs. 36-5 and 36-6

Fig. 36-8

12sin .

2

E

R The dashed line bisecting f forms two triangles, where:

.mE

R In radian measure:

121

2

sin .mEE

Solving the previous 2 equations for E one obtains:

22

2

sin m

m m

I EI I

I E

The intensity at the screen is therefore:

2sina

is related to the path length difference across the entire slit:

(36-7)

Diffraction by a Circular ApertureDistant point source, e,g., star

lens

Image is not a point, as expected from geometrical optics! Diffraction is responsible for this image pattern.

d

Light

a

Light

a

sin 1.22 (1st min.- circ. aperture)d

sin 1.22 (1st min.- single slit)a

(36-8)

Rayleigh’s Criterion: Two point sources are barely resolvable if their angular separation R results in the central maximum of the diffraction pattern of one source’s image centered on the first minimum of the diffraction pattern of the other source’s image.

Rayleigh’s Criterion

Fig. 36-10

R small1

R sin 1.22 1.22 (Rayleigh's criterion)d d

(36-9)

– Southern Cross and Pointers

– Brightest is nearest star -Cen

– Increase aperture d and it resolves into a binary

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