Interference, Diffraction & Polarization. interference, diffraction & polarization 2 light as waves ...

$: Interference, Diffraction & Polarization. interference, diffraction & polarization 2 light as waves so far, light has been treated as if it travels.$
Interference, Diffraction & Polarization

interference, diffraction & polarization 2

light as waves

so far, light has been treated as if it travels in straight linesray diagramsrefraction, 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 amplitudeby adding the individual components


Interference III

constructive interferencewaves in phase

demo: interference

+

=

destructive interferencewaves ½ out of phase

+

=


Interference in spherical wavesmaximum of wave minimum of wave

positive constructive interference

negative constructive interference

destructive interference

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

r1

r2

r1=r2


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 isno phase shift between the two sources

•the two sources of light must have identical wave lengths


Young’s interference experiment

there is a path difference: depending on its size the wavescoming from S1 or S2 are in or out of phase


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


distance between bright spots

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-mL/d= L/dym=mL/d

tan=y/L

L

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

distance is equalso 1/2 difference:destructive int.


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?

a) use d sin = m with m=3 =sin-1(m/d)=sin-1(3x600x10-9/0.01)=0.01030

b) Ym = mL/d m=0: y0 =0 m=3: y3 = 3x600x10-9x5/0.01 = 9x10-4 m = 0.9 mm


other ways of causing interference

remember

equivalent to:

1 2n1<n2

n1>n2

1 2


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.

1 2

n1<n2

1/2 phase change

n1>n2

1 2

no phase changeIn a medium with index of refraction n, the wavelengthchanges (relative to vacuum) to /n


thin film interference

n=1

n=1.5

n=1

The two reflected rays caninterfere. To analyze this system,4 steps are needed:

1. Is there phase inversion at the top surface?2. Is there phase inversion at the bottom surface3. What are the conditions for

constructive/destructive interference?4. what should the thickness d be for 3) to

happen?


n=1

n=1.5

n=1

thin film analysis1. top surface?2. bottom surface?3. conditions?4. d?

1. top surface: n1<n2 so phase inversion 1/2 2. bottom surface: n1>n2 so no phase inversion3. conditions:

1. constructive: ray 1 and 2 must be in phase2. destructive: ray 1 and 2 must be out of phase by 1/2

4. 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=mfilm =m/nfilm

1 2


Note

The interference is different for light of differentwavelengths


question

na=1

nb=1.5

nc=2

Phase inversion will occur ata) top surfaceb) bottom surfacec) top and bottom surfaced) neither surface

n1<n2 in both cases

constructive interference will occur if:a) 2d=(m+1/2)/nb

b) 2d=m/nb

c) 2d=(m+1/2)/nc

d) 2d=m/nc

note: if destructive 2d=(m+1/2)/nb

this is used e.g. on sunglasses to reduce reflections


another case

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 interferenceif 2t=m destructive interference.

1 2


question

Given h=1x10-5 m30 bright fringes are seen,with a dark fringe at the leftand the right.What is the wavelength ofthe 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


newton’s rings

spacing not equal

demo


quiz (extra credit)

Two beams of coherent light travel different paths arrivingat point P. If constructive interference occurs at point P,the two beams must:a) travel paths that differ by a whole number of wavelengthsb)travel paths that differ by an odd number of halfwavelengths


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?a) the very narrow slits required would generate

different wavelength, thereby washing out the

interference patternb) the two slits would not emit coherent lightc) the fringes would be too close togetherd) in no direction could a path difference as large as

one wavelength be obtained


diffraction

In Young’s experiment, two slits were used to producean 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/smallopenings.

In fact, every point in the slit openingacts as the source of a new wave front


interference pattern from a single slit

pick two points, 1 and 2, one inthe top top half of the slit, one in the bottom half of the slit.Light from these two points interferesdestructively if:x=(a/2)sin=/2 so sin=/a

we could also have divided up the slitinto 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


slit width

if >a sin=/a > 1 Not possible, so nopatterns

if <<a sin=m/a is very smalldiffraction hardly seen

<a : interference pattern is seen

aa


the diffraction pattern

The intensity is not uniform:

I=I0sin2()/2 =a(sin)/

a a a a a a


question

light with a wavelength of 500 nm is used to illuminatea slit of 5m. 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 inverseimage, 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 witha diffraction pattern from each of the slit

due to diffraction

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

minima asin=m, m=1,2,3…maxima asin=(m+1/2), m=1,2,3… and =0a: width of individual slit


double-slit experimenta

if >d, each slit acts as a singlesource of light and we geta more or less prefect double-slitinterference spectrum

if <d the interference spectrumis folded with the diffraction pattern.

d


7th7th

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 andthus 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=/asin must be equal for both, so /a=7/d and a=d/7=1/7 mm

question


diffraction grating

d

consider a grating withmany slits, each separated bya distance d. Assume that foreach 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.

Diffraction gratings can be madeby scratching lines on glass andare often used to analyze light

instead of giving d, one usuallygives the number of slits perunit distance: e.g. 300 lines/mmd=1/(300 lines/mm)=0.0033 mm


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.)


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 anglebetween the central maximum and the fourth ordermaximum 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 mm = 4, sin(47.2)=0.734so = 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

propagation into screen

E-vectors could pointanywhere: unpolarized


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~Emax

2


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 intensitypolarization

axisdirection of E-vector

If E-field is parallelto polarization axis,all light passes

If E-field makes an angle pol. axisonly the componentparallel to the pol. axispasses: E0cosSo I=I0cos2

For unpolarized light, on average, the E-fieldhas an angle of 450 withthe polarizer.I=I0cos2=I0cos2(45)=I0/2


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


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

n2

n1


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

horizontalverticaldirection of polarizationof reflected light


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

Interference, Diffraction & Polarization. interference, diffraction & polarization 2 light as waves ...

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