topic 7 wave_interference(latest)

Topic 4.1 Waves, Interference and Optics

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UEEP1033 Oscillations and Waves

Topic 7:

Interference and Diffraction


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• When a wavefront encounters an aperture in an opaque barrier, the barrier suppresses all propagation of the wave except through the aperture

• Following Huygen’s principle, the points on the wavefront across the aperture act as sources of secondary wavelets

• When the width of the aperture is comparable with the wavelength, the aperture acts like a point source and the outgoing wavefronts are semicircular

Huygen’s Principle


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• Ignores most of each secondary wavelet and only retaining the portions common to the envelope

• As a result, Huygens’s principle by itself is unable to account for the details of the diffraction process

• The difficulty was resolved by Fresnel with his addition of the concept of interference

Huygens’s Principle


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Augustin Jean Fresnel

• 1818, Fresnel brought together the ideas of Huygens and Young and by making some arbitrary assumptions about the amplitude and phases of Huygens’ secondary sources

• Fresnel able to calculate the distribution of light in diffraction patterns with excellent accuracy by allowing the various wavelet to mutually interfere

Huygens-Fresnel Principle


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Huygens-Fresnel Principle

Every unobstructed point of a wavefront, at given instant, serves as a source of spherical secondary

wavelets (with the same frequency as that of the primary

wave)The amplitude of the optical field at any point

beyond is the superposition of all these wavelets (considering their amplitudes and relative phases)


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Christian Huygens


Each point on the wavefront of a disturbance were considered to be a new source of a “secondary” spherical disturbance, then the wavefront at a later

instant could be found by constructing the “envelope” of the secondary wavelets”


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Huygens’s PrincipleEvery point on a propagation wavefront serves as the source of spherical secondary wavelets, such

that the wavefront at some later time is the envelope of these wavelets

Plane wave Spherical wave


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Plane waveSpherical wave

Every point on a propagation wavefront serves as the source of spherical

secondary wavelets

the wavefront at some later time is the envelope of these wavelets


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ri Law of Reflection

Law of Refraction (Snell’s law)

ttii nn sinsin

Interface

Incident medium ni

Refractingmedium ni

Surface normal


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Law of ReflectionWhen a ray of light is reflected at an interface dividing two uniform media, the reflected ray remains within the plane of incidence, and the angle of reflection equals the angle of incidence. The plane of incidence includes the incident ray and the normal to the point of incidence

Law of Refraction (Snell’s law)When a ray of light is refracted at an interface dividing two uniform media, the transmitted ray remains within the plane of incidence and the sine of the angle of refraction is directly proportional to the sine of the angle of incidence


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Huygens’ construction to prove the law of reflection

Narrow, parallel ray of light

Plane of interface XY

Angle of incidence

Angle of reflection


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• Since points along the plane wavefront do not arrive at the interface simultaneously, allowance is made for these differences in constructing the wavelets that determine the reflected wavefront

• If the interface XY were not present, the Huygens construction would produce the wavefront GI at the instant ray CF reached the interface at I

• The intrusion of the reflecting surface, means that during the same time interval required for ray CF to progress from F to I, ray BE has progressed from E to J and then a distance equivalent to JH after reflection


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• Wavelet of radius JN = JH centered at J is drawn above the reflecting surface

• Wavelet of radius DG is drawn centered at D to represent the propagation after reflection of the lower part of the light

• The new wavefront, which must now be tangent to these wavelets at points M and N, and include the point I, is shown as KI in the figure

• A representative reflected ray is DL, shown perpendicular to the reflected wavefront

• The normal PD drawn for this ray is used to define angles of incidence and reflection for the light


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The Law of Refraction

Use Huygen’s principle to derive the law of refraction

The refraction of a plane wave at an air-glass interface

Figures show three successive stages of the refraction of several wavefronts at a plane interface between air (medium 1) and glass (medium 2)

1 = wavelength in medium 1v1 = speed of light in medium 1v2 = speed of light in medium 2 < v1 1 = angle of incidence


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As the wave moves into the glass, a Huygens wavelet at point e will expand to pass through point c, at a distance of 1 from point e.

The time interval required for this expansion is that distance divided by the speed of the wavelet = 1/v1

In the same time interval, a Huygens wavelet at point h will expand to pass through point g, at the reduced speed v2 and with wavelength 2, i.e. the time interval = 2/v2

2

2

1

1

vv

2

1

2

1

v

v


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According to Huygens’ principle, the refracted wavefront must be tangent to an arc of radius 2 centered on h, say at point g

the refracted wavefront must also be tangent to an arc of radius 1 centered on e, say at point c

2 = angle of refraction

h c

e

h c

g

hc1

1sin

hc2

2sin

2

1

2

1

2

1

sin

sin

v

v


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Define: refraction index for a medium

c = speed of lightv = speed of light in the medium

Speed of light in any medium depends on the index of refraction of the medium

11 v

cn e.g.

22 v

cn

v

cn

1

2

2

1

2

1

2

1

/

/

sin

sin

n

n

nc

nc

v

v

2211 sinsin nn


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The wavelength of light in any medium depends on the index of refraction of the medium

Let a certain monochromatic light:Medium refraction index wavelength

speed vacuum 1 c medium n

n v2

1

2

1

v

v

From slide-8:c

vn

The greater the index of refraction of a medium, the smaller the wavelength of light in that medium

nn


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Frequency Between Media

• As light travels from one medium to another, its frequency does not change.

– Both the wave speed and the wavelength do change.

– The wavefronts do not pile up, nor are they created or destroyed at the boundary, so ƒ must stay the same.


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nn

vf

Frequency of the light in a medium with index of refraction n

fv

fc

n

ncfn

/

/

f = frequency of the light in vacuum

The frequency of the light in the medium is the same as it is in vacuum


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The fact that the wavelength of light depends on the index of refraction is important in situations involving the interference of light waves

Example: Two light rays travel through two media having different indexes of refraction

• Two light rays have identical wavelength and are initially in phase in air (n 1)

• One of the waves travels through medium 1 of index of refraction n1 and length L

• The other travels through medium 2 of index of refraction n2 and the same length L


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• When the waves leave the two media, they will have the same wavelength – their wavelength in air

• However, because their wavelengths differed in the two media, the two waves may no longer be in phase

The phase difference between two light waves can change if the waves travel through different materials having different indexes of refraction

How the light waves will interfere if they reach some common point?


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Number N1 of wavelengths in the length L of medium 1

11 / nn wavelength in medium 1:

1

11

LnLN

n

wavelength in medium 2: 22 / nn

2

22

LnLN

n

)( 1212 nnL

NN

Phase difference between the waves

21 nn


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Example: phase difference = 45.6 wavelengths

• i.e. taking the initially in-phase waves and shifting one of them by 45.6 wavelengths

• A shift of an integers number of wavelengths (such as 45) would put the waves back in phase

• Only the decimal fraction (such as 0.6) that is important• i.e. phase difference of 45.6 wavelengths 0.6 wavelengths

• Phase difference = 0.5 wavelength puts two waves exactly out of phase

• If the two waves had equal amplitudes and were to reach some common point, they would then undergo fully destructive interference, producing darkness at that point


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• With the phase difference = 0 or 1wavelengths, they would undergo fully constructive interference, resulting brightness at that common point

• In this example, the phase difference = 0.6 wavelengths is an intermediate situation, but closer to destructive interference, and the wave would produces a dimly illuminated common point


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Example:

= 550 nm

Two light waves have equal amplitudes and re in phase before entering media 1 and 2

Medium 1 = air (n1 1)

Medium 2 = transparent plastic (n2 1.60, L = 2.60 m)

Phase difference of the emerging waves:

o

9

6

1212

1020 rad17.8

swavelength84.2

)00.160.1(10550

1060.2

)(

nnL

NN


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Effective phase difference = 0.84 wavelengths = 5.3 rad 300o

• 0.84 wavelengths is between 0.5 wavelength and 1.0 wavelength, but closer to 1.0 wavelength.

• Thus, the waves would produce intermediate interference that is closer to fully constructive interference,

• i.e. they would produce a relatively bright spot at some common point.


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Fermat’s Principle

• The ray of light traveled the path of least time from A to B

• If light travels more slowly in the second medium, light bends at the interface so as to take a path that favors a shorter time in the second medium, thereby minimizing the overall transit time from A to B

Construction to prove the law of refraction from Fermat’s principle


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InterferenceYoung’s Double-Slit Experiment


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• Mathematically, we are required to minimize the total time:

ti v

OB

v

AOt

22 xaAO 22 )( xcbOB

ti v

xcb

v

xat

2222 )(


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0)( 2222

xcbv

xc

xav

x

dx

dt

ti

• minimize the total time by setting dt / dx = 0

22sin

xa

xi

• From diagram:

22 )(sin

xcb

xct

0sinsin

t

t

i

i

vvdx

dt

0/

sin

/

sin

t

t

i

i

ncnc ttii nn sinsin


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Interference

two waves are out of phase

destructive interference

two waves are in phase

constructive interference

amplitude of their superposition is zero

amplitude of the superposition (ψ1 + ψ2) = 2A

A is the amplitude of the individual waves


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Figure (a)• Two monochromatic waves ψ1 and ψ2 at a

particular point in space where the path difference from their common source is equal to an integral number of wavelengths

• There is constructive interference and their superposition (ψ1 + ψ2) has an amplitude that is equal to 2A where A is the amplitude of the individual waves.

Figure (b)• The two waves ψ1 and ψ2 where the path

difference is equal to an odd number of half wavelengths

• There is destructive interference and the amplitude of their superposition is zero

Interference


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Light source

ApertureObservatio

n plane

Screen

Arrangement used for observing

diffraction of light

Corpuscular Theoryshadow behind the

screen should be well defined, with sharp

borders

Observations• The transition from light to shadow was gradual rather than

abrupt

• Presence of bright and dark fringes

extending far into the geometrical shadow of the

screen


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Young’s Double-Slit Experiment

L >> a

d

d = slits separation

d


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• A monochromatic plane wave of wavelength λ is incident upon an opaque barrier containing two slits S1 and S2

• Each of these slits acts as a source of secondary wavelets according to Huygen’s Principle and the disturbance beyond the barrier is the superposition of all the wavelets spreading out from the two slits

• These slits are very narrow but have a long length in the direction normal to the page, making this a two-dimensional problem

• The resultant amplitude at point P is due to the superposition of secondary wavelets from the two slits



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• Since these secondary wavelets are driven by the same incident wave there is a well defined phase relationship between them

• This condition is called coherence and implies a systematic phase relationship between the secondary wavelets when they are superposed at some distant point P

• It is this phase relationship that gives rise to the interference pattern, which is observed on a screen a distance L beyond the barrier



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The secondary wavelets from S1 and S2 arriving at an arbitrary point P on the screen, at a distance x from the point O that coincides with the mid-point of the two slits

Distances: S1P = l1 S2P = l2 Since L >> d it can be assumed that the secondary wavelets arriving at P have the same amplitude A

The superposition of the wavelets at P gives the resultant amplitude:


)cos()cos( 21 kltkltAR

ω = angular frequencyk = wave number

(5)

d= slits separation


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This result can be rewritten as:

Since L >> d, the lines from S1 and S2 to P can be assumed to be parallel and also to make the same angle θ with respect to the horizontal axis


2/)(cos[]2/)(cos2 1212 llkllktAR

The line joining P to the mid-point of the slits makes an angle θ with respect to the horizontal axis

21 cos/ lLl

cos/212 Lll

(6)



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When the two slits are separated by many wavelengths, θ is very small and cos θ 1. Hence, we can write the resultant amplitude as:


)2/cos()cos(2 lkkLtAR

= path difference of the secondary wavelets

The intensity I at point P = R2

12 lll

)2/(cos)(cos4 222 lkkLtAI

This equation describes the instantaneous intensity at PThe variation of the intensity with time is described by the cos2(ωt − kL) term

(7)

(8)


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• The frequency of oscillation of visible light is of the order of 1015 Hz, which is far too high for the human eye and any laboratory apparatus to follow.

• What we observe is a time average of the intensity• Since the time average of cos2(ωt − kL) over many

cycles = 1/2

the time average of the intensity is given by:


)2/(cos20 lkII

20 2AI = intensity observed at a maximum of the interference pattern

described how the intensity varies with l)2/(cos2 lk

(9)


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I = maximum whenever l = n (n = 0,±1, ±2, …)I = 0 whenever l = (n + ½)


From figure on slide-25: l a sin θSubstituting for l in Equation (9), we obtain:

(10) )2/sin(cos)( 20 kdII

When θ is small so that sinθ θ, we can write:

)/(cos)(

)2/(cos)(2

0

20

dII

kdII

(11)

/2where kd = slits separation


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If there were no interference, the intensity would be uniform and equal to Io/2 as indicated by the horizontal dashed line


Light intensity I (θ) vs angle θ


L/dseparation of the bright fringes


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Intensity maxima: .....,2,1,0, nd

n

.....,2,1,0, nd

LnLx

(12)

(13)

(14)

(15)

The bright fringes occur at distances from the point O given by:

Minimum intensity occur when:

The distance between adjacent bright fringes is:

.....,2,1,0,2

1

n

d

Lnx

d

Lxx nn

1



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Point source of light is illuminating an opaque object, casting a shadow where the edge of the shadow fades gradually over a short distance and made up of bright and dark bands, the diffraction fringes. Shadow fades gradually

>> Bright and Dark Bands

= Diffraction Fringes

Diffraction


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Francesco Grimaldi in 1665 first accurate report

description of deviation of light from rectilinear propagation (diffraction)

The effect is a general characteristics of wave phenomena occurring whenever a portion of a wavefront is

obstructed in some way

Diffraction


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Plane wavefronts approach a barrier with an opening or an obstruction, which both the opening and the obstruction are large compared to the wavelength

Opening(size = a)

Obstruction (size = a)

wavelength, a >>


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• If the size of the opening or obstruction becomes comparable to the wavelength

• The waves is not allowed to propagate freely through the opening or past the obstruction

• But experiences some retardation of some parts of the wavefront

• The wave proceed to "bend through" or around the opening or obstruction

• The wave experiences significant curvature upon emerging from the opening or the obstruction

curvaturea


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As the barrier or opening size gets smaller, the wavefront experiences more and more

curvature

More curvature

Diffraction

a


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Fraunhofer and Fresnel Diffraction


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Observationscreen


S

Lens

Plane waves

Opaque shield , with a singlesmall aperture of width a is

being illuminated by plane wave of wavelength from a distant

point source S

Case-1observation screen is very

close to

Image of aperture is projected onto the screen


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Observationscreen


S

Lens

Plane waves Case-2

observation screen is moved farther away from

Image of aperture become increasingly more structured as the

fringes become prominent

Fresnel or Near-Field Diffraction


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S

Lens

Plane waves Case-3

observation screen is at very great distance away from

Projected pattern will have spread out considerably, bearing a little or

no resemblance to the actual aperture

Observationscreen

Thereafter moving the screen away from the aperture change

only the size of the pattern and not its shape

Fraunhofer or Far-Field Diffraction


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S

Lens

Plane waves Case-4

If at that point, the wavelength of the incoming radiation is reduce

Observationscreen

the pattern would revert back to the Fresnel case

If were decreased even more, so that → 0The fringes would disappear, and the image

would take on the limiting shape of the aperture


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If a point source S and the observation screen are very far

from

S

Lens

Plane waves

Observationscreen

Fraunhofer Diffraction

If a point source S and the observation screen are

too near Fresnel Diffraction


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S

Lens

Plane waves

Observationscreen

Fraunhofer Diffractiona

R R

R is the smaller of the two distances from S to and to

2aR

d = slit width


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Practical realization of the Fraunhofer condition

F1 F2


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Diffraction

• Any obstacle in the path of the wave affects the way it spreads out; the wave appears to ‘bend’ around the obstacle

• Similarly, the wave spreads out beyond any aperture that it meets. such bending or spreading of the wave is called diffraction

• The effects of diffraction are evident in the shadow of an object that is illuminated by a point source. The edges of the shadow are not sharp but are blurred due to the bending of the light at the edges of the object

• The degree of spreading of a wave after passing through an aperture depends on the ratio of the wavelength λ of the wave to the size d of the aperture

• The angular width of the spreading is approximately equal to λ/d; the bigger this ratio, the greater is the spreading


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The Mechanism of Diffraction• Diffraction arises because of the way in which waves propagate as

described by the Huygens-Fresnel Principle

• The propagation of a wave can be visualized by considering every point on a wavefront as a point source for a secondary radial wave

• The subsequent propagation and addition of all these radial waves form the new wavefront

• When waves are added together, their sum is determined by the relative phases as well as the amplitudes of the individual waves, an effect which is often known as wave interference

• The summed amplitude of the waves can have any value between zero and the sum of the individual amplitudes

• Hence, diffraction patterns usually have a series of maxima and minima


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• A monochromatic plane wave is incident upon an opaque barrier containing a single slit

• Replace the relatively wide slit by an increasing number of narrow subslits

• Each point in the subslits acts as a point source for a secondary radial wave

• When waves are added together, their sum is determined by the relative phases and the amplitudes of the individual waves, an effect which is often known as wave interference

• The summed amplitude of the waves can have any value between zero and the sum of the individual amplitudes

• Hence, diffraction patterns usually have a series of maxima and minima

Single Slit Diffraction


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One can find the second dark fringes above and below the central axis as the first dark fringes were found, except that we now divide the slit into four zones of equal widths a/4, as shown in Fig. 36-6a.

In general,


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Example, Single Slit Diffraction Pattern with White Light:


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36.4: Intensity in Single-Slit Diffraction Pattern, Qualitatively:


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Fig. 36-8 The relative intensity in single-slit diffraction for three values of the ratio a/l. The wider the slit is, the narrower is the central diffraction maximum.

The intensity pattern is: where

For intensity minimum,


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36.5: Intensity in Single-Slit Diffraction Pattern, Quantitatively:

From the geometry, f is also the angle between the two radii marked R. The dashed line in the figure, which bisects f, forms two congruent right triangles.


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Example, Intensities of the Maximum in a Single Slit Interference Pattern:


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36.6: Diffraction by a Circular Aperture:


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36.6: Diffraction by a Circular Aperture, Resolvability:

Fig. 36-11 At the top, the images of two point sources (stars) formed by a converging lens. At the bottom, representations of the image intensities. In (a) the angular separation ofthe sources is too small for them to be distinguished, in (b) they can be marginally distinguished, and in (c) they are clearly distinguished. Rayleigh’s criterion is satisfied in (b), with the central maximum of one diffraction pattern coinciding with the first minimum of the other.

Two objects that are barely resolvable when the angular separation is given by:


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Example, Pointillistic paintings use the diffraction of your eye:


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Example, Rayleigh’s criterion for resolving two distant objects:


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Fig. 36-15 (a) The intensity plot to be expected in a double-slit interference experiment with vanishingly narrow slits. (b) The intensity plot for diffraction by a typical slit of width a (not vanishingly narrow). (c) The intensity plot to be expected for two slits of width a. The curve of (b) acts as an envelope, limiting the intensity of the double-slit fringes in (a). Note that the first minima of the diffraction pattern of (b) eliminate the double-slit fringes that would occur near 12° in (c).

The intensity of a double slit pattern is:


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Example, Double slit experiment, with diffraction of each slit included:


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Example, Double slit experiment, with diffraction of each slit included, cont. :


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Diffraction GratingDefinition

A repetitive array of diffracting elements that has the effect of producing periodic alterations in the phase, amplitude, or

both of an emergent wave

An idealized grating consisting of only

five slits

Opaque surface with narrow parallel grooves

e.g. made by ruling or scratching parallel notches into the surface of a flat, clean glass plate

Each of the scratches serves as a source of scattered light, and together they form a regular array of parallel line sources


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Diffraction Grating

Grating Equation: d sinm = m

m = specify the order of the various principal maxima

The intensity plot produced by a diffraction grating consists of narrow peaks, here label with their order number m

The corresponding bright fringes seen on the screen are called lines

The maxima are very narrow and they separated by relatively wide dark region

d = grating spacing (spacing between rulings or slits)

N rulings occupy a total width w, then d = w/N


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36.8: Diffraction Gratings:


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36.8: Diffraction Gratings, Width of the Lines:


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Diffraction Grating

Application: Grating Spectroscope

collimator

Plane waveDiffraction grating

telescope

Visible emission lines of cadmium

Visible emission lines from hydrogen

The lines are farther apart at greater angles


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36.9: Gratings, Dispersion and Resolving Power:

A grating spreads apart the diffraction lines associated with the various wavelengths.

This spreading, called dispersion, is defined as

Here Dq is the angular separation of two lines whose wavelengths differ by Dl.

Also,

To resolve lines whose wavelengths are close together, the line should also be as narrow as possible. The resolving power R, of the grating is defined as

It turns out that


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Gratings, Dispersion and Resolving Power, proofs:

The expression for the locations of the lines in the diffraction pattern of a grating is:

Also, If Dq is to be the smallest angle that will permit the two lines to be resolved, it must (by Rayleigh’s criterion) be equal to the half-width of each line, which is given by :


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36.9: Gratings, Dispersion and Resolving Power Compared:


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topic 7 wave_interference(latest)

Science

waves topic

waves law of reflection

waves huygens construction

ueep1033 oscillations

plane wavefront

waves augustin jean

propagation wavefront

reflected wavefront