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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

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Interference, Diffraction & Polarization

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Interference diffraction polarization l.jpg

Interference, Diffraction & Polarization

PHY232 – Spring 2007

Jon Pumplin

http://www.pa.msu.edu/~pumplin/PHY232

(Ppt courtesy of Remco Zegers)


Light as waves l.jpg

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 

PHY232 - Remco Zegers - interference, diffraction & polarization


Interference l.jpg

interference

constructive interference

destructive interference

at any point in time one can construct the total amplitude

by adding the individual components

PHY232 - Remco Zegers - interference, diffraction & polarization


Slide4 l.jpg

+

=

destructive interference

waves ½ out of phase

demo: interference

Interference III

+

=

constructive interference

waves in phase

PHY232 - Remco Zegers - interference, diffraction & polarization


Slide5 l.jpg

r1=r2

r1

r2

positive constructive interference

negative constructive interference

destructive interference

if r2-r1 = n then constructive interference occurs

if r2-r1 = (n+½) then destructive interference occurs

Interference in spherical waves

maximum of wave

minimum of wave

PHY232 - Remco Zegers - interference, diffraction & polarization


Slide6 l.jpg

PHY232 - Remco Zegers - interference, diffraction & polarization


Light as waves7 l.jpg

light as waves

it works the same for light waves, sound waves,

and *small* water waves

PHY232 - Remco Zegers - interference, diffraction & polarization


Double slit experiment l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Slide9 l.jpg

Young’s interference experiment

there is a path difference: depending on its size the waves

coming from S1 or S2 are in or out of phase

PHY232 - Remco Zegers - interference, diffraction & polarization


Young s interference experiment l.jpg

Young’s 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, m=0,1,2,3…

m = 0: zeroth order, m=1: 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 = [m+1/2], m=0,1,2,3…)

path difference

demo

PHY232 - Remco Zegers - interference, diffraction & polarization


Slide11 l.jpg

distance between bright spots

tan=y/L

L

if  is small, then sin     tan 

so: d sin = m, m=0,1,2,3… converts to

dy/L = m

difference between maximum m and maximum m+1:

ym+1-ym= (m+1)L/d-mL/d= L/d

ym=mL/d

demo

PHY232 - Remco Zegers - interference, diffraction & polarization


Question l.jpg

distance is equal

so 1/2 difference:

destructive int.

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Question13 l.jpg

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 m=3

  • =sin-1(m/d)=sin-1(3x600x10-9/0.01)=0.01030

  • b) Ym = mL/d

  • m=0: y0 =0

  • m=3: y3 = 3x600x10-9x5/0.01 = 9x10-4 m = 0.9 mm

PHY232 - Remco Zegers - interference, diffraction & polarization


Other ways of causing interference l.jpg

other ways of causing interference

  • remember

equivalent to:

n1>n2

n1<n2

1

2

2

1

PHY232 - Remco Zegers - interference, diffraction & polarization


Phase changes at boundaries l.jpg

phase changes at boundaries

If a light ray travels from medium 1 to medium 2 with n1<n2,

the phase of the light ray will change by 1/2. This will not

happen if n1>n2.

n1>n2

1

2

1

2

n1<n2

1/2 phase change

no phase change

In a medium with index of refraction n, the wavelength

changes (relative to vacuum) to /n

PHY232 - Remco Zegers - interference, diffraction & polarization


Thin film interference l.jpg

thin film interference

n=1

The two reflected rays can

interfere. To analyze this system,

4 steps are needed:

n=1.5

n=1

  • Is there phase inversion at the top surface?

  • Is there phase inversion at the bottom surface

  • What are the conditions for constructive/destructive interference?

  • what should the thickness d be for 3) to happen?

PHY232 - Remco Zegers - interference, diffraction & polarization


Thin film analysis l.jpg

thin film analysis

  • top surface?

  • bottom surface?

  • conditions?

  • d?

1

2

n=1

n=1.5

  • top surface: n1<n2 so phase inversion 1/2

  • bottom surface: n1>n2 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:

  • 2d=m (difference in path length=integer number of wavelengths)

  • due to phase inversion at top surface: 2d=(m+1/2)

  • since the ray travels through film: 2d=(m+1/2)film =(m+1/2)/nfilm

  • destructive: 2d=mfilm =m/nfilm

n=1

PHY232 - Remco Zegers - interference, diffraction & polarization


Slide18 l.jpg

Note

The interference is different for light of different

wavelengths

PHY232 - Remco Zegers - interference, diffraction & polarization


Question19 l.jpg

n1<n2 in both cases

question

  • Phase inversion will occur at

  • top surface

  • bottom surface

  • top and bottom surface

  • neither surface

na=1

nb=1.5

nc=2

  • constructive interference will occur if:

  • 2d=(m+1/2)/nb

  • 2d=m/nb

  • 2d=(m+1/2)/nc

  • 2d=m/nc

note: if destructive 2d=(m+1/2)/nb

this is used e.g. on sunglasses to

reduce reflections

PHY232 - Remco Zegers - interference, diffraction & polarization


Another case l.jpg

another case

1

2

The air gap in between the plates has varying thickness.

Ray 1 is not inverted (n1>n2)

Ray 2 is inverted (n1<n2)

where the two glasses touch: no path length difference:

dark fringe.

if 2t=(m+1/2) constructive interference

if 2t=m destructive interference.

PHY232 - Remco Zegers - interference, diffraction & polarization


Question21 l.jpg

question

Given h=1x10-5 m

30 bright fringes are seen,

with a dark fringe at the left

and the right.

What is the wavelength of

the light?

2t=m destructive interference.

m goes from 0 (left) to 30 (right).

=2t/m=2h/m=2x1x10-5/30=6.67x10-7 m=667 nm

PHY232 - Remco Zegers - interference, diffraction & polarization


Newton s rings l.jpg

newton’s rings

demo

spacing not equal

PHY232 - Remco Zegers - interference, diffraction & polarization


Quiz extra credit l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Question24 l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Diffraction l.jpg

diffraction

In Young’s 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

PHY232 - Remco Zegers - interference, diffraction & polarization


Slide26 l.jpg

PHY232 - Remco Zegers - interference, diffraction & polarization


Interference pattern from a single slit l.jpg

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 m=1,2,3…

In between the minima, are maxima: sin = (m+1/2)/a m=1,2,3…

AND sin=0 or =0

PHY232 - Remco Zegers - interference, diffraction & polarization


Slit width l.jpg

slit width

a

a

if >a sin=/a > 1

Not possible, so no

patterns

if <<a

sin=m/a is very small

diffraction hardly seen

<a : interference

pattern is seen

PHY232 - Remco Zegers - interference, diffraction & polarization


The diffraction pattern l.jpg

the diffraction pattern

The intensity is not uniform:

I=I0sin2()/2 =a(sin)/ 

a

a

a

a

a

a

PHY232 - Remco Zegers - interference, diffraction & polarization


Question30 l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Diffraction from a single hair l.jpg

diffraction from a single hair

instead of an slit, we can also use an inverse

image, for example a hair!

demo

PHY232 - Remco Zegers - interference, diffraction & polarization


Double slit interference revisited l.jpg

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, m=1,2,3…

maxima asin=(m+1/2), m=1,2,3…

and =0

a: width of individual slit

due to 2-slit

interference

maxima dsin=m, m=0,1,2,3…

minima dsin=(m+1/2), m=0,1,2,3…

d: distance between two slits

PHY232 - Remco Zegers - interference, diffraction & polarization


Double slit experiment33 l.jpg

double-slit experiment

a

d

if >d, each slit acts as a single

source of light and we get

a more or less prefect double-slit

interference spectrum

if <d the interference spectrum

is folded with the diffraction

pattern.

PHY232 - Remco Zegers - interference, diffraction & polarization


Question34 l.jpg

7th

7th

question

A person has a double slit plate. He measures the distance between the two slits to be d=1 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 /a=7/d and a=d/7=1/7 mm

PHY232 - Remco Zegers - interference, diffraction & polarization


Diffraction grating l.jpg

diffraction grating

consider a grating with

many slits, each separated by

a distance d. Assume that for

each slit >d. We saw that for 2 slits

maxima appear if:

d sin = m, m=0,1,2,3…

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

d=1/(300 lines/mm)=0.0033 mm

PHY232 - Remco Zegers - interference, diffraction & polarization


Separating colors l.jpg

separating colors

d sin = m, m=0,1,2,3… for maxima (same as for double slit)

so  = sin-1(m/d) depends on , the wavelength.

cd’s can act as a diffraction grating

(DVD’s work even better because

their tracks are more closely spaced.)

PHY232 - Remco Zegers - interference, diffraction & polarization


Question37 l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Question38 l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Polarization l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Polarized light l.jpg

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. I~Emax2

PHY232 - Remco Zegers - interference, diffraction & polarization


How to polarize l.jpg

How to polarize?

  • absorption

  • reflection

  • scattering

PHY232 - Remco Zegers - interference, diffraction & polarization


Polarization by absorption l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Polarizers and intensity l.jpg

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.

I=I0cos2=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 I=I0cos2

PHY232 - Remco Zegers - interference, diffraction & polarization


Question44 l.jpg

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, I1=I0/2. After passing through

the second polarizer, I2=I1cos230=0.75I1=0.375I0

PHY232 - Remco Zegers - interference, diffraction & polarization


Polarization by reflection l.jpg

polarization by reflection

n1

  • 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

n2

PHY232 - Remco Zegers - interference, diffraction & polarization


Question46 l.jpg

horizontal

vertical

direction of

polarization

of reflected

light

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

PHY232 - Remco Zegers - interference, diffraction & polarization


Sunglasses l.jpg

sunglasses

wearing sunglasses will help reducing glare (reflection)

from flat surfaces (highway/water)

withoutwith sunglasses

PHY232 - Remco Zegers - interference, diffraction & polarization


Polarization by scattering l.jpg

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

PHY232 - Remco Zegers - interference, diffraction & polarization


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