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Interference, Diffraction Polarization

PHY232 Spring 2007 Jon Pumplin http//www.pa.msu

.edu/pumplin/PHY232 (Ppt courtesy of Remco

Zegers)

light as waves

- so far, light has been treated as if it travels

in straight lines - ray diagrams
- refraction, reflection
- To describe many optical phenomena, we have to

treat light as waves. - Just like waves in water, or sound
- waves, light waves can interact
- and form interference patterns.
- Remember c f ?

interference

constructive interference

destructive interference

at any point in time one can construct the total

amplitude by adding the individual components

demo interference

Interference III

constructive interference waves in phase

Interference in spherical waves

maximum of wave

minimum of wave

(No Transcript)

light as waves

it works the same for light waves, sound waves,

and small water waves

double slit experiment

- the light from the two sources is

incoherent (fixed phase with respect to each

other - in this case, there is
- no phase shift between
- the two sources
- the two sources of light must have identical wave

lengths

Youngs interference experiment

there is a path difference depending on its size

the waves coming from S1 or S2 are in or out of

phase

Youngs interference experiment

If the difference in distance between the screen

and each of the two slits is such that the waves

are in phase, constructive interference occurs

bright spot difference in distance must be a

integer multiple of the wavelength d sin?

m?, m0,1,2,3 m 0 zeroth order, m1 first

order, etc. if the difference in distance is off

by half a wavelength (or one and a half etc.),

destructive interference occurs (d sin?

m1/2?, m0,1,2,3)

path difference

demo

distance between bright spots

tan?y/L

L

if ? is small, then sin ? ? ? ? tan ? so d sin?

m?, m0,1,2,3 converts to dy/L m?

difference between maximum m and maximum

m1 ym1-ym (m1)?L/d-m?L/d ?L/d ymm?L/d

demo

question

- two light sources are put at a distance d from a

screen. Each source produces light of the same

wavelength, but the sources are out of phase by

half a wavelength. On the screen exactly midway

between the two sources will occur - a) constructive interference
- b) destructive interference

1/2?

question

- two narrow slits are illuminated by a laser with

a wavelength of 600 nm. the distance between the

two slits is 1 cm. a) At what angle from the beam

axis does the 3rd order maximum occur? b) If a

screen is put 5 meter away from the slits, what

is the distance between the 0th order and 3rd

order maximum?

- use d sin? m? with m3
- ?sin-1(m?/d)sin-1(3x600x10-9/0.01)0.01030
- b) Ym m?L/d
- m0 y0 0
- m3 y3 3x600x10-9x5/0.01 9x10-4 m 0.9

mm

other ways of causing interference

- remember

equivalent to

n1gtn2

n1ltn2

1

2

2

1

phase changes at boundaries

If a light ray travels from medium 1 to medium 2

with n1ltn2, the phase of the light ray will

change by 1/2?. This will not happen if n1gtn2.

n1gtn2

1

2

1

2

n1ltn2

1/2? phase change

no phase change

In a medium with index of refraction n, the

wavelength changes (relative to vacuum) to ?/n

thin film interference

n1

The two reflected rays can interfere. To analyze

this system, 4 steps are needed

n1.5

n1

- Is there phase inversion at the top surface?
- Is there phase inversion at the bottom surface
- What are the conditions for constructive/destructi

ve interference? - what should the thickness d be for 3) to happen?

thin film analysis

- top surface?
- bottom surface?
- conditions?
- d?

1

2

n1

n1.5

n1

- top surface n1ltn2 so phase inversion 1/2?
- bottom surface n1gtn2 so no phase inversion
- conditions
- constructive ray 1 and 2 must be in phase
- destructive ray 1 and 2 must be out of phase by

1/2? - if phase inversion would not take place at any of

the surfaces - constructive
- 2dm? (difference in path lengthinteger

number of wavelengths) - due to phase inversion at top surface

2d(m1/2)? - since the ray travels through film

2d(m1/2)?film (m1/2)?/nfilm - destructive 2dm?film m?/nfilm

Note

The interference is different for light of

different wavelengths

question

- Phase inversion will occur at
- top surface
- bottom surface
- top and bottom surface
- neither surface

na1

nb1.5

nc2

- constructive interference will occur if
- 2d(m1/2)?/nb
- 2dm?/nb
- 2d(m1/2)?/nc
- 2dm?/nc

note if destructive 2d(m1/2)?/nb this is used

e.g. on sunglasses to reduce reflections

another case

1

2

The air gap in between the plates has varying

thickness. Ray 1 is not inverted (n1gtn2) Ray 2 is

inverted (n1ltn2) where the two glasses touch no

path length difference dark fringe. if

2t(m1/2)? constructive interference if 2tm?

destructive interference.

question

Given h1x10-5 m 30 bright fringes are seen, with

a dark fringe at the left and the right. What is

the wavelength of the light?

2tm? destructive interference. m goes from 0

(left) to 30 (right). ?2t/m2h/m2x1x10-5/306.67

x10-7 m667 nm

newtons rings

demo

spacing not equal

quiz (extra credit)

- Two beams of coherent light travel different

paths arriving - at point P. If constructive interference occurs

at point P, - the two beams must
- travel paths that differ by a whole number of
- wavelengths
- b)travel paths that differ by an odd number of

half - wavelengths

question

- why is it not possible to produce an interference

pattern - in a double-slit experiment if the separation of

the slits - is less than the wavelength of the light used?
- the very narrow slits required would generate

different - wavelength, thereby washing out the

interference pattern - the two slits would not emit coherent light
- the fringes would be too close together
- in no direction could a path difference as large

as one wavelength be obtained

diffraction

In Youngs experiment, two slits were used to

produce an interference pattern. However,

interference effects can already occur with a

single slit.

This is due to diffraction the capability of

light to be deflected by edges/small openings.

In fact, every point in the slit opening acts as

the source of a new wave front

(No Transcript)

interference pattern from a single slit

pick two points, 1 and 2, one in the top top half

of the slit, one in the bottom half of the

slit. Light from these two points

interferes destructively if ?x(a/2)sin??/2

so sin??/a we could also have divided up the

slit into 4 pieces ?x(a/4)sin??/2 so

sin?2?/a 6 pieces ?x(a/6)sin??/2 so

sin?3?/a Minima occur if sin? m?/a m1,2,3

In between the minima, are maxima sin?

(m1/2)?/a m1,2,3

AND sin?0

or ?0

slit width

a

a

if ?gta sin??/a gt 1 Not possible, so no patterns

if ?ltlta sin?m?/a is very small diffraction

hardly seen

?lta interference pattern is seen

the diffraction pattern

The intensity is not uniform II0sin2(?)/?2

??a(sin?)/ ?

a

a

a

a

a

a

question

light with a wavelength of 500 nm is used to

illuminate a slit of 5?m. At which angle is the

5th minimum in the diffraction pattern seen?

sin? m?/a ? sin-1(5x500x10-9/(5x10-6))300

diffraction from a single hair

instead of an slit, we can also use an

inverse image, for example a hair! demo

double slit interference revisited

The total response from a double slit system is a

combination of two single-source slits, combined

with a diffraction pattern from each of the slit

due to diffraction

minima asin?m?, m1,2,3 maxima asin?(m1/2)?,

m1,2,3 and ?0 a width of

individual slit

due to 2-slit interference

maxima dsin?m?, m0,1,2,3 minima

dsin?(m1/2)?, m0,1,2,3 d distance between

two slits

double-slit experiment

a

d

if ?gtd, each slit acts as a single source of

light and we get a more or less prefect

double-slit interference spectrum

if ?ltd the interference spectrum is folded with

the diffraction pattern.

question

A person has a double slit plate. He measures the

distance between the two slits to be d1 mm. Next

he wants to determine the width of each slit by

investigating the interference pattern. He finds

that the 7th order interference maximum lines up

with the first diffraction minimum and thus

vanishes. What is the width of the slits?

7th order interference maximum dsin?7? so

sin?7?/d 1st diffraction minimum asin?1? so

sin??/a sin? must be equal for both, so ?/a7?/d

and ad/71/7 mm

diffraction grating

consider a grating with many slits, each

separated by a distance d. Assume that for each

slit ?gtd. We saw that for 2 slits maxima appear

if d sin? m?, m0,1,2,3 This condition is not

changed for in the case of n slits.

d

Diffraction gratings can be made by scratching

lines on glass and are often used to analyze light

instead of giving d, one usually gives the number

of slits per unit distance e.g. 300

lines/mm d1/(300 lines/mm)0.0033 mm

separating colors

d sin? m?, m0,1,2,3 for maxima (same as for

double slit) so ? sin-1(m?/d) depends on ?,

the wavelength.

cds can act as a diffraction grating (DVDs work

even better because their tracks are more

closely spaced.)

question

- If the interference conditions are the same when

using a double slit or a diffraction grating with

thousands of slits, what is the advantage of

using the grating to analyze light? - a) the more slits, the larger the separation

between maxima. - b) the more slits, the narrower each of the

bright spots and thus easier to see - c) the more slits, the more light reaches each

maximum and the maxima are brighter - d) there is no advantage

question

An diffraction grating has 5000 lines per cm. The

angle between the central maximum and the fourth

order maximum is 47.20. What is the wavelength of

the light?

d sin? m?, m 0,1,2,3 d 1/5000 2x10-4 cm

2x10-6 m m 4, sin(47.2)0.734 so ? d sin?/m

2x10-6x0.734/4 3.67x10-7 m 367 nm

polarization

- We saw that light is really an electromagnetic

wave with electric and magnetic field vectors

oscillating perpendicular to each other. In

general, light is unpolarized, which means that

the E-field vector (and thus the B-field vector

as long as it is perpendicular to the E-field)

could point in any direction

E-vectors could point anywhere unpolarized

propagation into screen

polarized light

- light can be linearly polarized, which means that

the E-field only oscillates in one direction (and

the B-field perpendicular to that) - The intensity of light is proportional to the

square of amplitude of the E-field. IEmax2

How to polarize?

- absorption
- reflection
- scattering

polarization by absorption

- certain material (such as polaroid used for

sunglasses) only transmit light along a certain

transmission axis. - because only a fraction of the light is

transmitted after passing through a polarizer the

intensity is reduced. - If unpolarized light passes through a polarizer,

the intensity is reduced by a factor of 2

polarizers and intensity

polarization axis

direction of E-vector

For unpolarized light, on average, the

E-field has an angle of 450 with the

polarizer. II0cos2?I0cos2(45)I0/2

?

If E-field is parallel to polarization axis, all

light passes

If E-field makes an angle ? pol. axis only the

component parallel to the pol. axis passes

E0cos? So II0cos2?

question

- unpolarized light with intensity I0 passes

through a linear polarizer. It then passes

through a second polarizer (the second polarizer

is usually called the analyzer) whose

transmission axis makes and angle of 300 with the

transmission axis of the first polarized. What is

the intensity of the light after the second

polarizer, in terms of the intensity of the

initial light?

After passing through the first polarizer,

I1I0/2. After passing through the second

polarizer, I2I1cos2300.75I10.375I0

polarization by reflection

- If unpolarized light is reflected, than the

reflected light is partially polarized. - if the angle between the reflected ray and the

refracted ray is exactly 900 the reflected light

is completely polarized - the above condition is met if for the angle of

incidence the equation tan?n2/n1 - the angle ?tan-1(n2/n1) is called the Brewster

angle - the polarization of the reflected light is

(mostly) parallel to the surface of reflection

n1

n2

question

- Because of reflection from sunlight of the glass

window, the curtain behind the glass is hard to

see. If I would wear polaroid sunglasses that

allow polarized light through, I would be able

to see the curtain much better. - a) horizontally
- b) vertically

sunglasses

wearing sunglasses will help reducing glare

(reflection) from flat surfaces (highway/water)

without with sunglasses

polarization by scattering

- certain molecules tend to polarize light when

struck by it since the electrons in the molecules

act as little antennas that can only oscillate in

a certain direction

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