Chapter 24:Wave Optics Conditions for Interference Homework assignment : 10,22,23,33,42,55 ...

Chapter 24:Wave OpticsConditions for Interference

Homework assignment : 10,22,23,33,42,55

Conditions for interference

Adding two waves: wave 1 and wave 2

Conditions for interference (cont’d)

• The sources must be coherent, which means the waves they emit must maintain a constant phase with respect to each other.• The waves must have identical wavelengths.

• Two sources producing two traveling waves are needed to create interference.

An old style method to produce two coherent sources

Conditions for Interference

Two slits and interference

Conditions for Interference

Two monochromatic sources of the same frequency and with anyConstant phase relation are said to be coherent.

Young’s double-slit experiment

Young’s Double-Slit Experiment

• Red dots represent spots where the two waves are in phase and constructive interference occurs. This produces bright spots.

• Blue circles represent spots where the two waves are out of phase and destructive interference occurs. This produces dark spots.

Young’s double-slit experiment (cont’d)


In phase

Bright image

In phase

Bright image

Out of phase

Dark image



sin12 drr ,...)2,1,0( sin mmd bright

Path difference :

Constructive interference:

m : order number m=0 zeroth-order, m=+-1 first-order…

Destructive interference : ,...)2,1,0( )2

1(sin mmd bright



1 if sintan LLy

,...)2,1,0( mmd

Lybright

Constructive interference:

Destructive interference: ,...)2,1,0( )2

1( mm

d

Lydark

Example 24.1: Measuring the wavelength of a light source• A screen is separated from a double-slit source by 1.20 m. The distance between the two slits is 0.0300 mm. The second-order bright fringe (m=2) is measured to be 4.50 cm from the centerline. Determine: (a) the wavelength of the light

(b) the distance between the adjacent bright fringes.

m 1063.5 72 mL

dy

cm 25.2)1(1 d

Lm

d

Lm

d

Lyyy mm


Lloyd’s mirror• In the arrangement of a mirror and the screen on the right, at a point on the screen the light waves have two possible paths.

Change of Phase Due to Reflection

• At point P’, the path difference of two paths is zero: Therefore we expect a bright fringe. But we see a dark fringe there.• Similarly on the screen the pattern of bright and dark fringes is reversed.

An EM wave undergoes a phase change of 180o or upon reflectionfrom a medium that has an index of refraction higher than the one inwhich the wave is traveling.

Phase change upon reflection

Change of Phase Due to Reflection

An EM wave undergoes a phase change of 180o upon reflection froma medium that has an index of refraction higher than the one in whichthe wave is traveling.

Colorful patterns

Interference in Thin Film

Interference in thin film


• An EM wave traveling from a medium of index of refraction n1

toward a medium of index of refraction n2 undergoes a phase change on reflection when n2>n1. There is no phase change in the reflected light wave if n1>n2.• The wavelength of light n in a medium with index of refraction n is n=/n where is the wavelength of light in vacuum.Ray 1: A phase change of at surface A with respect to the incident wave (n>nair=1)

Ray 2: No phase change at surface B with respect to the incident wave (n>nair=1)

pha

se d

iff. b

y

due

to r

efle

ctio

n

Phase difference Ray 1 : Ray 2 : 2 x 2t/n

)/2(212 nt

.)0,1,2,3,..(m )2( m

1,2,3,...),0(m )2)(2/1( m

constructive

destructive

Interference in thin film (cont’d)


Interference between ray 1 and ray 2

,...)2,1,0( )2

1(2 mmt n

,...)2,1,0( 2 mmt n

constructive

destructive

)2

1(2 mnt

mnt 2

thicker dueto gravity

Newton’s Rings


• Another example of interference: A planoconvex lens on top of a flat glass surface

Blue ray : phase change at the interface between air and a flat glass.Red ray : no phase change due to reflection.

pha

se d

iff. b

y

due

to r

efle

ctio

n

Interference between red and blue rayconstructive

destructive

RdRdRRdRr

RddRRr

2)/21(1)/1(1

)(

2

22

0,1,2,...)(m )2/)(2/1( mdm

0,1,2,...)(m )2/( mdm

Non-reflective coating for optical lens


,...)2,1,0( 2 mmt n

)2

1(2 mnt

MgF2 n = 1.38

Glass, n = 1.50

Air, n = 1.00

1 2

t

Thin layer of MgF2 with index intermediate between air and glass;there are two phase shifts of in this case, as shown.Phase shifts of reflected rays:

Ray 1: Ray 2: + (2/n)(2t)

For no reflection: (n) (2t) =

ntt

n 42

/

2

Numbers: for no reflection at = 550 nm, t = 100 nm.

(destructive interference)

Interference in a wedge-shaped film


A pair of glass slides 10.0 cm longand with n=1.52 are separated onone end by a hair, forming a triangularwedge of air as illustrated. Whencoherent light from a helium-neonlaser with wavelength 633 nm isincident on the film from above, 15.0dark fringes per cm are observed.How thick is the hair?For destructive interference with nair=1: 2/mt If d is the distance from one dark bandto the next, the x-coord. of the m-th fringe is: mdx s)(cm/fringe 1067.6)fringes/cm 0.15( 2-1 d

Using similar triangles : LDdmdmxt /)2/()/()2/(/

m 1075.42

5d

LD

What is diffraction?

Diffraction

Diffraction is bending of waves around obstacles. Diffractionalso occurs when waves from a large number of sources interfere.

Diffraction Single slit

Bending on the corner

Diffraction at slits Single slit

•

•

Diffraction and Huygens principleEvery 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!

Diffraction at slits


1

1

11

sin4

a

= path difference

= path difference

When a/2 sin = /2 every ray originating in top half of slit interferes destructively with the corresponding ray originating in bottom half.

When a/4 sin = /2 every ray originating in top quarter of slit interferes destructively with the corresponding ray originating in the second quarter.

a2

a

sin2

a

4

a

a


Condition for quarters of slit to destructively interfere

Condition for halves of slit to destructively interfere a

sin

a

2sin

Condition for sixths of slit to destructively interferea

3sin

This formula locates

minima (for dark fringes)!!

All together… ,...)3,2,1( sin ma

mdark

Diffraction Grating Diffraction grating• Combination of diffraction and interference• Consists of a large number of equally spaced parallel slits.• Can be made by scratching parallel lines on a glass plate with a precision machine technique.

Each slit causes diffraction. Each slit acts as a source of waves in phase. Path diff. between two adjacent slits at P: sind

d=spacing of slits If slits are made to satisfy :

waves from slits are in phase.,...)2,1,0( sin mmd bright

bright are the positions of bright maxima.

Diffraction Grating Diffraction grating (cont’d)

Pattern of light intensity Diffraction grating spectrometer

By measuring the positions of maxima, using the telescope in a diffractiongrating spectrometer, the wavelength of the incident waves can bedetermined.

Polarization

EM wave

BE

tkxEjtxB

tkxEitxE

)cos(ˆ),(

)cos(ˆ),(

max

max

Polarization (defined by the direction of )E

Linear polarization Circular polarization

vector inx-dir.vector iny-dir.

If the E-field is in the same directionall the time, the EM wave is linearly polerized.

Polarization (cont’d)

Polarization (defined by the direction of )E

Circular polarization

Polarization (cont’d) Polarization by reflection

p p

b

an

bn

plane of incidenceWhen the angle of incident coincides withthe polarizing angle or Brewster’s angle,the reflected light is 100% polarized.

p

pbpbbbpa nnnn cos)90sin(sinsin

a

bp n

ntan Brewsters’ law of the polarizing angle

Polarization (cont’d)

Polarization by absorption and polarizing filters

Polaroid: polarizes light through selective absorption by oriented molecules

These molecules have a preferred axis (transmission axis) along which electronscan move easily. All lights with E parallel(perpendicular) to the axis is transmitted(absorbed).

• 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:

20

2 cosIIEI

Malus’s Law

Malus’s law

• When linearly polarized light passes through a polarizer whose axis

of transmission make an angle with the direction of the E-field vector (or polarization), the intensity of the light follows the law:

02

2

1IIEI

Malus’s Law

Malus’s law (cont’d)

2max cosII

Malus’ law for polarizedlight passing through ananalyzer/polarizer:

This is true only whenthe incident light islinearly polarized.

Unpolarized light on linear polarizer

• Most light comes from electrons accelerating in random directions and is unpolarized.

• Averaging over all directions, intensity of transmitted light reduces due to reduction in E

Malus’s Law

Example:Malus’s law

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

1) Intensity of unpolarized light incident on linear polarizer is reduced by half . I1= I0 / 2

2) Now ignore the 2nd polarizer for now.

I = I0I1

I3

3) Light transmitted through first polarizer is vertically polarized. Angle between it and 3rd polarizer is =90º . I3= I1 cos2(90º) = 0

Malus’s Law

Example: Malus’s law

2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is =45º. I2= I1 cos2 (45º)=0.5I1=0.25 I0

3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is =45º. I3= I2 cos2(45º)

I2= I1cos2(45)

=0.125 I0

I1= 0.5 I0

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

1) Now put 2nd polarizer in.

Malus’s Law

I1 I2

Example: Malus’s law

Malus’s Law

Controlled Polarization

• Electrical voltage on a liquid crystal diode– Turns on and off

polarizing filter effect– Used in LCD display

LCD screen

Chapter 24:Wave Optics Conditions for Interference Homework assignment : 10,22,23,33,42,55 ...

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