Wave optics

Wave optics second course

Assist Prof. DR.Mariam Mohamed Abud

1

HUYGENS' PRINCIPLE

Huygens nearly three centuries ago proposed the rule that each point on a

wavefront may be regarded as a new source of waves (see chapter one)

Huygens’ theory is essentially based on a geometrical construction which

allows us to determine the shape of the wavefront at any time. He results are

follows:

1) When waves pass through an aperture or past the edge of an obstacle, they

always spread to some extent into the region which is not directly exposed to

the oncoming waves. This phenomenon is called diffraction.

2) Huygens explained many of the known propagation characteristics of light,

According to Huygens ’ Principle, each point of a wave front is a source of

secondary disturbance, and the wavelets emanating from these points spread out in

all directions with the speed of the wave. The envelope of these wavelets gives

the shape of the new wave front.

Huygens’ Principle, along with the fact that the secondary wavelets mutually

interfere, is known as the Huygens–Fresnel principle.

(8-2)Interference in the phenomena of superposition of two coherence

waves in the region of superposition, light waves occurs whenever two or

more waves overlap at a given point.



2

Conditions for sustained interference of light waves

1. The sources are coherent, which means that the waves they emit must

maintain a constant phase with respect to one another.

coherence

incoherence waves

2. The waves have identical wavelengths and frequency, the interfering

waves equal amplitudes

3. The two waves is interference must be propagated at the same line.

4. Two sources must be narrow

We can write the separate displacements as follows: we can write

the separate displacements as follows:

(1)

According to the principle of superposition, the resultant

displacement Y is merely the sum of y1 and y2, and we have:

(2)

When the expression for the sine of the difference of two angles is

used, this can be written

(3)



3

We are justified in setting:

(4)

The Amplitude of the wave is conserved by each wave period A the

law of cosines gives

(5)

Phase difference φ between two waves are given by ∅ = 𝛼1 − 𝛼2

(8-2) Theory of interference

According figure (1) has two sources such as young’s experiment

Figure (1): Construction for analyzing the double-slit interference pattern. A bright fringe, or

intensity maximum, is observed at O.

Where the wave travels the distance L in a medium, then

Phase difference= 2𝜋

𝜆∗ 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒



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∅ =2𝜋∆

𝜆=

2𝜋𝑛𝐿

𝜆 (7)

∆ is the optical path difference=n L

Where ∆ is the optical path difference, and E the superposition of two

sinusoidal waves having the same frequency

ER = EA+EB

When two waves have a different phase of electric field can be given by:

The Phase difference produces a resultant amplitude by two waves we use

the law of cosine

ER2 = E1

2 + E22 – 2 E1 E

2cos (𝜋 − ∅) (8)

Although the waves are in phase as the slits, the phase difference

∅ = 𝑟2 − 𝑟1 = 𝑑 sin 𝜃 At point P.

A path difference of λ (condition of constructive interference) ,according

phase difference 2𝜋 rad , we get the ratio following:

∆

𝜆=

∅

2𝜋

∅ = 2𝜋

𝜆 ∆=

2𝜋

𝜆 𝑑 sin 𝜃

we can obtain the magnitude of the resultant electric field at point P :



5

(10)

Interference

Ep=a1+a2 if coherence source Ep= E+E=2E this is maimum

amplitude where a1=a2=E

The intensity is squre of amplitude .

When the phase difference ∅ = 0 and I1=I2=I0,

Ir= A2= (a1+a2)2

Ir=I1+I2 =2IO the interference produced at this points is known as

constructive interference (As stationary bright band)

We have the maximum amount of light: Imax= 4I0

The following features about monochromatic wave

1- The wave has a single frequency definite (𝑓 =𝑤

2𝜋)in optics ,the waves

have a single frequency and wavelength called monochromatic.



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2- the electric vector E-vibration are coordinate y-z plane .therfore the

wave is plane polarized (linearly polarized)

3- phase difference and coherence :

∅ =2𝜋𝐿

𝜆 (11)

(8-2-1)INTENSITY DISTRIBUTION OF THE DOUBLE-SLIT

INTERFERENCE PATTERN

The bright fringes in the interference pattern do not have sharp

edges.

The equations developed give the location of only the centers of the

bright and dark fringes.

▪ We can calculate the distribution of light intensity associated with

the double-slit interference pattern.

▪ Assumptions:

▪ The two slits represent coherent sources of sinusoidal waves.

▪ The waves from the slits have the same angular frequency, ω.

▪ The waves have a constant phase difference, φ.

▪ The total magnitude of the electric field at any point on the

screen is the superposition of the two waves.

▪



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

Producing coherence sources

Light from a monochromatic source is allowed to pass through a narrow slit.

•The light from the single slit is allowed to fall on a screen containing two narrow slits.

•The first slit is needed to insure the light comes from a tiny region of the source which is coherent.



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Currently, it is much more common to use a laser as a coherent source.

•The laser produces an intense, coherent, monochromatic beam over a width of several millimeters.

•The laser light can be used to illuminate multiple slits directly. The intensity of a wave is proportional to the square of the resultant electric

field magnitude at that point

Therefore express the light intensity at point P as

(12)

Most light-detecting instruments measure time-averaged light intensity, and

the time-averaged value of sin2(w t +Φ/2) over one cycle is Therefore, we

can write the average light intensity at point P as

(13)

(8-2-2) intensity Distribution

The intensity light wave is given by square of its amplitude

Ir= I1+I2 +2√𝐼1𝐼2 cos ∅

At point that max and min, when I1=I2=I0 ,we get:

Ir= I0 +I0 + 2I0cos∅

= 2I0 (1+cos ∅)

Since the identify 1+cos∅ = 2 cos2 (1

2∅) , 𝑤𝑒 𝑔𝑒𝑡:

𝐼 = 4𝐼0𝑐𝑜𝑠2(1

2∅) (14)



9

Figure ( 2 ): Light intensity versus d sin for a double-slit interference pattern when the screen is

far from the slits (L >>d)

(15)



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(8-3) PRACTICAL (EXPERIMENTAL) COHERENCE PRODUCES

I. INTERFERENCE BY DIVISION OF WAVE FRONT:

1- YOUNG’S DOUBLE-SLIT EXPERIMENT

2-FRESNEL'S BIPRISM

3-FRESNEL'S MIRROR

4- LLOYD’S MIRROR

II. INTERFERENCE BY DVSION OF AMPLITUDE

1-MICHELSON INTERFEROMETER

2-TWEEMAN&GREEN INTERFERMETER

3-JAMIN INTERFEROMETER

4-MACH-ZEHNDER INTERFEROMETER

5-RAYLEGH'S REFREFOMETER

III.INTERFEROMETER BY A MULTPLE REFLECTON

1-NEWTON'S RINGS

2-TWO PARALLEL THIN FILM

In constructive interference the amplitude of the resultant wave is greater

than that of either individual wave.

In destructive interference the amplitude of the resultant wave is less than

that of either individual wave.

All interference associated with light waves arises when the electromagnetic

fields that constitute the individual waves combine.



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(8-3-1) Interference by Division of Wave front

1-YOUNG’S DOUBLE-SLIT EXPERIMENT

Thomas Young first demonstrated interference in light waves from two

sources in 1801.the described of details the experiment as following:

•Light is incident on a screen with a narrow slit S0,

•The light waves emerging from this slit arrive at a second screen that

contains two narrow, parallel slits, S1and S2

•The narrow slits, S1and S2 act as real sources of waves.

•The waves emerging from the slits originate from the same wave front and

therefore are always in phase. As shown in fig (3).

Figure (3): (a) Geometric construction for describing Young’s double-slit experiment

(not to scale). (b) The path difference between the two rays is



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∅ = 𝑟2 − 𝑟1 = 𝑑𝑠𝑖𝑛𝜃 . it is essential that L >>d.

The incident waves (planar wave) on the slit So from this slit arrive at a

second barrier that contains two parallel slits S1 and S2. These two slits serve

as a pair of coherent light sources because waves emerging from them

originate from the same wave front and therefore maintain a constant phase

relationship. The light from S1 and S2 produces on a viewing screen a

visible pattern of bright and dark parallel bands called fringes .When the

light from the two slits combines destructively at any location on the screen,

a dark fringe results an interference pattern produced by two coherent

vibrating sources in a water tank.

Figure (4) : the two slits produced a pair coherence sources and form max bright fringes

on screen.



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The planar waves clashed on slit So, it is emerging from this slit arrive at a

second barrier that contains two parallel slits S1 and S2. These two slits serve

as a pair of coherent light sources because waves emerging from them

originate from the same wave front and therefore maintain a constant phase

relationship. The light from S1 and S2 produces on a viewing screen a

visible pattern of bright and dark parallel bands called fringes .When the

light from the two slits combines destructively at any location on the screen,

a dark fringe results an interference pattern produced by two coherent

vibrating sources in a water tank.

Results interference pattern

The light from the two slits form a visible pattern on a screen.

•The pattern consists of a series of bright and dark parallel bands called

fringes.

•Constructive interference occurs where a bright fringe appears.

Path difference ∅ =𝑟2 − 𝑟1 = 𝑑𝑠𝑖𝑛𝜃 = m λ (15)

•Destructive interference results in a dark fringe. The condition for destructive interference

(16)

It is useful to obtain expressions for the positions of the bright and dark

fringes measured vertically from O to P. In addition to our assumption that

we assume that These can be valid assumptions because in practice

L is often of the order of 1 m, d a fraction of a millimeter, and a fraction of a

micrometer for visible light. Under these conditions, is small



14

(17)

The positions of the bright fringes measured from O are given by the

expression

(18)

(19)

Young’s double-slit experiment provides a method for measuring

wavelength of the light.

Figure (5): Young’s double-slit experiment set up

φ



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▪ This experiment gave the wave model of light a great deal of

credibility.

▪ It was inconceivable that particles of light could cancel each other

in a way that would explain the dark fringes.

▪ ∅ = 𝒅 𝒔𝒊𝒏𝜽 = 𝒅 𝑿

𝑳 (20)

Example 1: A pair of screens are placed 13.7m apart. A third order fringe is

seen on the screen 2.50cm from the central fringe. If the slits were cut

0.0960 cm apart, determine the wavelength of this light. Roughly what

color is it?

Just to make sure you’ve got all the numbers from the question matched

with the correct variables…

L = 13.7 m

n = 3

x = 2.50cm = 0.0250 m

d = 0.0960cm = 9.60e-4 m

It’s probably a yellow light being used given the wavelength we've measured.

If a white light is used in the double slit experiment, the different colours will

be split up on the viewing screen according to their wavelengths.

• The violet end of the spectrum (with the shortest wavelengths) is

closer to the central fringe, with the other colours being further away

in order.

There is also a version of the formula where you measure the angle between

the central fringe and whatever fringe you are measuring.



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Example 2: For a single slit experiment apparatus like the one described

above, determine how far from the central fringe the first order violet (λ =

350nm) and red (λ = 700nm) color’s will appear if the screen is 10 m away

and the slit is 0.050 cm wide.

We need to solve the formula for “x”, the distance from the central fringe.

For the violet light

For the red light

1. Fresnel’s Biprism:

Some objected Young’s Experiment result and said that: “the bright fringes

were probably due to some complicated modification of the light by the



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edges of the slits and not interference” Fresnel did new experiments to prove

that they were interference. S is the real source, and the other S’s are virtual.

Figure (8) Diagram of Fresnel's biprism experiment

To find d: the distance between sources or virtual images sources

When a: distance between the slit and biprism

𝛼 = 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑖𝑝𝑟𝑖𝑠𝑚

𝑑 = 2𝛿𝑥𝑎 = 2𝑎(𝑛 − 1)𝛼 (21)

𝛿: 𝑡ℎ𝑒 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑏𝑦 𝑒𝑎𝑐ℎ ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑠𝑚 = 𝛿 = (𝑛 − 1)𝛼

In fig 4 the central point between be region fringe of maximum intensity

occurs there.

The fringes width of dark or bright fringe is 𝛽 =𝜆𝐷

𝑑 (22)

Where D: distance between source and screen

The wavelength is given by:

𝜆 =𝛽 𝑑

𝑎 + 𝐶

A

a

D

B



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D= a+c

Example 3 :a) calculate the separation between the coherent sources formed by biprism whose inclined faces make angles of 2 with its base ,the slit being 0.1 m away from the biprism (n=1.5)?

𝛼 = 2𝑜 = 𝜋

180𝑥 2 𝑟𝑎𝑑𝑖𝑎𝑛𝑠

𝑑 = 2𝑥0.1𝑥(1.5 − 1)𝑥𝜋

180 𝑥2 = 0.0034

Example 4: In a Fresnel’s bi-prism experiment, the refracting angle of prism

were 1.5 o and refractive index of the glass was 1.5.with the angle slit 5 cm

from the bi-prism and using light of wavelength 580nm,fringes were formed

on a screen 1m from the single slit ,calculate the fringes width?

Solution

Example 3: In a Fresnel’s bi-prism experiment, the refracting angle of prism

were 1.5 o and refractive index of the glass was 1.5.with the angle slit 5 cm

from the bi-prism and using light of wavelength 580nm,fringes were formed

on a screen 1m from the single slit ,calculate the fringes width?

Solution

𝑑 = 2𝛿𝑥𝑎 = 2𝑎(𝑛 − 1)𝛼

d=2x5(1.5-1)(1.5x 3.14/180)=0.1308=0.131 cm

The fringe width is

∆𝑥 =𝜆 𝐷

𝑑=

580 𝑥 10−7𝑐𝑚 𝑥 100 𝑐𝑚

0.131 𝑐𝑚= 0.044 𝑐𝑚



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Figure (9): Interference and diffraction fringes produced in the Fresnel biprism

3. Fresnel’s Mirrors:

Here, Fresnel used two mirrors. Light from a slit S is reflected (not

refracted) from two plane mirrors slightly inclined to each other. The

mirrors produce two virtual images of S. They act exactly like two

sources formed by the biprism.

Figure (10): Fresnel's mirrors.

L

a b



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Light from a slit is reflected in two plane mirrors slightly inclined to each

other. The mirrors produce two virtual images of the slit, a narrow slit S is

illuminated by a monochromatic light wavelength 𝜆

It is placed parallel to the line of intersection of mirror surface, one portion

of cylinder wavelength is reflected from the first mirror and another portion

of the wave front is reflected from the second mirror. The virtual sources S1

and S2 are coherence sources which produced from the source S. the fringes

from (ae) region are equal width. The distance between virtual sources is d

Fringes width ∆𝑥 =𝜆𝐿

𝑑=

𝜆(𝑎+𝑏)

𝑑 (23)

∆𝑆1𝑂𝑆2 = 2𝜃

Then the distance between two virtual slits 𝐴𝑟𝑐 𝑆1𝑆2 = 𝑑 = 𝑎 ∗ 2𝜃 Then the eq.23 becomes:

∆𝑥 =𝜆𝐿

𝑑=

𝜆(𝑎 + 𝑏)

2𝑎𝜃

Compared between the fringes by Fresnel’s biprism and Mirrors:

Both cases are similar, but the double mirror are narrower than Fresnel’s biprism fringes.

4. Lloyd’s Mirrors:

This method is used to single long mirror about 30 cm. The

interference is produced by the light reflected from the mirror and the

light coming directly from the source (the slit 𝑆1)

Figure (11): Set up Lloyd’s Mirrors

O



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A cylindrical wave front coming from a narrow slit S1 falls on the

mirror which reflects a portion of incident wave front, giving rise to a

virtual image of the slit s’.

The slit s1 and s2 represent at two coherent sources. The point O

equidistant from s and s’

An arrangement for producing an interference pattern with a single light

source

•Waves reach point O either by a direct path or by reflection.

•The reflected ray can be treated as a ray from the source S’ behind the

mirror.

O is central (zero-order) fringes. With white light the central point o is

expected white, but in practice it is dark. The phase change leads to a

path difference of λ/2 and here destructive interference occurs.

To determine of wavelength𝜆, the fringes width is:

∆𝑥 =𝜆𝐿

𝑑

𝑚𝑒𝑎𝑠𝑢𝑟𝑚𝑖𝑛𝑔 ∆𝑥, 𝐿 𝑎𝑛𝑑 𝑑.

Comparison between the fringes produced by biprism and

Lloyd’s mirror:

1. In biprism the complete set of fringes is obtained. In Lloyd’s

mirror a few fringes on one side of the central fringe are noted, the

central fringe which being invisible

2. In biprism the zero-order (central fringe) is bright, but the central

fringe of Lloyd’s mirror is dark.

3. The central fringe is less sharp in biprism than in Lloyd’s mirror.

O



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Example (5):

In young’s experiment, interference bands are produced on screen placed 1.5

m from two slits 0.15 mm apart and illuminated by the light of wavelength

6000Å. Find

a) Fringe width

b) Change in fringe width if the screen is taken away from the slits by 50

cm?

Solution

a) ∆𝑥 =𝜆𝐿

𝑑=

6000𝑥10−10 𝑚𝑥1.5 𝑚

0.15𝑥10−3𝑚

∆𝑥 = 6 𝑚𝑚

b) ∆𝑥 =𝜆𝐿

𝑑=

6000𝑥10−10 𝑚𝑥50 𝑥10−2 𝑚

0.15𝑥10−3𝑚= 2mm

Example (6):

A light falls on two slits ( 2 mm) apart and produces on a screen 1 m away

from the fourth –order bright line 1 mm from the center of the pattern .what

is the wavelength of the light used?

Solution:

N=4 order , d=2mm

D=1 m

sin 𝜃 = tan 𝜃 = 𝑦

𝐿

𝑑 sin 𝜃 = 𝑚𝜆

𝜆 = 𝑑 ( 𝑦

𝐿)(

1

𝑛)



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𝜆 = (2 𝑥 10−3)(10−3(1

4)

𝜆 = (1

2) 𝑥 10−6

𝜆 = (0.5𝑥 10−6 = 500 𝑛𝑚

Example (7): yellow light pass through two slits and interference pattern

is observed on a screen

1) The bright fringes will increase in width if the yellow light is replaced

blue

2) The bright fringes will increase in width if the distance between slit

minimized

3) The intensity of light decreases if it is far from the central fringes

4) The intensity of light is constant if it is far from the central fringe which

is the correct statement?

Solution:

d:distance between slit is smaller than the distance between slit and

screen, so the angle is very small



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sin 𝜃 = tan 𝜃 = 𝑦

𝐿

𝑑 sin 𝜃 = 𝑚𝜆

𝜆 = 𝑑 ( 𝑦

𝐿)(

1

𝑛)

n: no. of bright lines is proportional inversely to the wavelength

Yellow wavelength (small frequency) is larger than blue light.

Blue light is changed; the wavelength λ is decrease, so the number of

bright lines is increase.

This statement is correct

2) The bright fringes will increase in width if the distance between slit

minimized

d sin 𝜃 = tan 𝜃 = 𝑦

𝐿



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If The bright fringes will increase in width if the distance between slit

minimized, the number of bright lines (n) decreses.

𝑛𝜆 = 𝑑 ( 𝑦

𝐿)

The bright fringes will increase in width if the distance between slit

minimized

This statement is incorrect

3) The intensity of light decreases if it is far from the central fringes

This statement is correct

Intensity related to the light level. Intensity is proportional inversely to

the distance. If the distance is increased the intensity is decrease(the

light dimmer)

4)The intensity of light is constant if it is far from the central fringe which

is the correct statement?

This statement is incorrect

Example (8):

Suppose in double slit arrangement, d=0.150 mm, L=120 cm, the

wavelength =833 nm, and the width fringe y=2 cm

a) What is the path difference δ for the rays from the two slits arriving at

point P

b) Express this path difference in terms of λ?



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c) Does point P corresponds to a maximum or minimum, is intermediate

condition?

Solution:

a) φ = d sin θ bright = d 𝒚

𝑳= 𝟏. 𝟓 𝒙𝟏𝟎−𝟒𝒎

𝟐𝒙 𝟏𝟎−𝟐𝒎

𝟏.𝟐= 𝟐. 𝟓 𝒙 𝟏𝟎−𝟔𝒎

b) from a) we get

𝜑

𝜆=

2.5𝑥 10−6𝑚

8.33𝑥10−7𝑚 ≈ 3

Path difference

𝜑 = 3𝜆

c) at point P the path difference φ = ∓ 𝑚𝜆 𝑖𝑠 𝑎the multiple

wavelength , so the intensity y at point p is maximum

COHERENT SOURCES

The two incident light beams must have the same frequency and be in phase

at the two slits (or at least have a definite time-independent phase relation);

two such beams are said to be coherent sources. The result is that the

difference in phase between any pair of points in the two sources always

remains constant, and so the interference fringes are stationary. Such as

Young's experiment and in Fresnel's mirrors and biprism, the two sources Sl

and S2 always have a point-to-point correspondence of phase, since they are

both derived from the same source any interference experiment with light

that the sources must have this point-to-point phase relation, and sources that

have this relation are called coherent sources.

While special arrangements are necessary for producing coherent sources of

light, the same is not true of microwaves, which are radio waves of a few

centimeters wavelength. If in Young's experiment the source slit S (Fig. 12)

is made too wide or the angle between the rays which leave it too large, the



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double slit no longer represents two coherent sources and the interference

fringes disappear

Interference fringes with white light and monochromatic

If the slit illuminated by white light in the in Young's experiment the source

slit S is used and Fresnel’s biprism. The interference pattern consist of a

central white light around it on both sides a few coloured fringes. The

central fringe will be dark and others will be colors. With white light, fringes

are observed only when the path difference is small.

(8-3-2 ). INTERFERENCE BY DVSION OF AMPLITUDE

1-MICHELSON’s INTERFEROMETER

Principle:

Two beams are interference formed by division of amplitude.

Construction of Michelson and works:

The light source S is monochromatic(known (as with a laser beam), have

incident beam (a plane waver fronts on light silvered Mirror pass by striking

glass which has inclined 45 degree.

The two plane-parallel plates of glass Gland G2, sometime used one glass

G.to get interference the mirrors M1 and M2 have perpendicular each other,

with a finely threaded screw adjustment.

Figure (12): diagram of young’s Experiment



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Figure (13): Construction of Michelson

The beam is divided into two beams

1) from reflection

2) from transmission

in this process the amplitudes have divide represents a two sources

Under these conditions, the interference pattern is a series of bright and dark

parallel fringes as described in Example figure (14(b). The tilting screws on

M2 are turned until one pair of images falls exactly on the other, the

interference fringes should appear. When they first appear, the

fringes will not be clear unless the eye is focused on or near the back mirror

M1 so the observer should look constantly at this mirror while searching for

the fringes.

As M1 is moved, the fringe pattern shifts. For example, if a dark fringe

appears in the field of view (corresponding to destructive interference) and

M1 is then moved a distance λ/4 toward M, the path difference changes by

λ/2 (twice the separation between M1 see fig (14(c) and what was a dark

fringe now becomes a bright Fringe.

As M1 is moved an additional distance λ /4 toward M, the bright fringe

becomes a dark fringe. Thus, the fringe pattern shifts by one-half fringe each

time M1 is moved a distance λ /4. The wavelength of light is then measured



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by counting the number of fringe shifts for a given displacement of M1. If

the wavelength is accurately mirror displacements can be measured to within

a fraction of the wavelength.

(a) (b) (c )

Figure (14) : a) A pair of images, b) arrangement two image and

Circular fringes are produced with laser light when the mirrors M1 and

M2 are exactly perpendicular to each other.(c) As M1 is moved, the

fringe pattern shifts. For example, if a dark fringe appears in the field

of view as forward or backward.

Optical path

The two waves will interfere constructively or destructively as per the following conditions of

Path difference ∆:

∆ =2𝑚𝜆

2= 𝑚𝜆 𝑎𝑡 𝑚𝑎𝑥. 𝑖. 𝑒 𝑏𝑟𝑖𝑔ℎ𝑡 𝑓𝑟𝑖𝑛𝑔𝑒𝑠 , where m is integer

𝑚 = ∓1, ∓2, ∓3 … … … … . . ∓𝑚

∆= (2𝑚 + 1)𝜆/2

at min. or dark fringes ,when m is also integer

(24)



30

Figure (15) Schematic of the Michelson interferometer.

Types of fringes

1) Circular fringes

Circular fringes are produced with monochromatic light when the

mirrors M1 and M2 are exactly perpendicular to each other.

The ones used in most kinds of measurement with the interferometer. Their

origin can be understood by reference to the diagram in Fig. 10. M2'

Is the virtual image of M2 and it is parallel to M1. For simplicity, light

source L is at the observer's position. L1 and L2 are the virtual images of L

formed by M1 and M2', and are coherent. Let d be the distance between M1

and M2', therefore the distance between L1 and L2 is 2d.



31

θ: the angle between the incident beam originated at P and the reflected

beams from M1 and M2'. Then path difference between light beams from

points Pand P" is 2d cos θ.

A maximum (bright fringe) will be formed when 2dcosθ = m . For a fixed

value of n, and d, the value of θ is a constant, and the contour of the

maximum point becomes a ring. The Centre of the ring is in line with the

observer and perpendicular to the mirror plane. Each circular ring

corresponds to a particular value of θ. Hence the fringes are known as

fringes of equal inclination. Fig (11 a) Fringes of this kind, where parallel

beams are brought to interference with a

phase difference determined by the angle of inclination θ, are often referred

to as fringes of equal inclination.

This type may remain visible over very large path differences

2𝑑 cos 𝜃 = 𝑚𝜆 (25)

Figure (16) form of circular fringes (Concentric circular fringes) as known as

fringes of equal inclination

Since for a given m, λ.,and d the angle θ is constant, the maxima will lie in

the form of circles about the foot of the perpendicular from the eye to the

p P’’

p 2d

θ P”

2d cos θ

L1 L2



32

mirrors. By expanding the cosine, it can be shown from Eq. (25) that the

radii of the rings are proportional to the square roots of integers, as in the

case of Newton's rings . The intensity distribution across the rings follows

Eq. (13), in which the phase difference is given by

∅ = 2𝜋

𝜆 2𝑑 cos 𝜃 (26)

2) Curved fringes (fringes of equal thickness):

When M1 and virtual image M2' are inclined to each other, the film enclosed

is wedge shaped. Then curved fringes can be observed as shown in Fig. 11.

These are also known as fringes of equal thickness

Figure 17: form fringes of equal thickness

The difference between two fringes has explain in fig.18



33

Figure 18: a) difference fringes between a) fringes of equal inclination

B) fringes of equal thickness and c. d) moved M2’ formed circular fringes e) form

straight fringes.

a b

c

d

e

https://i.stack.imgur.com/ktu36.jpg



34

3) Straight line fringes:

When M1 and virtual image M2' intersect, straight line fringes are obtained

around the point of intersection (see Fig. 13). The path

Figure 19: shape of Straight line fringes

Figure 20: Appearance of the various types of fringes observed in the Michelson

interferometer. Upper row, circular fringes. Lower row, localized fringes. Path

difference increases outward, in both directions, from the center.

4) White light fringes: if white light is used, the central fringe will be

dark and other will be coloured.fringes are observed only when the

path difference is small. These fringes are important because they are

used to locate the position of zero path difference.



35

Figure 21:Micheslon’ interferometer using white light

a)wit compensating plate b) with beam splitter

Example 9: Assume your laser has a wavelength of 560 nm, and your

mirrors start at equal distances from the beam splitter. You shift your

movable mirror by 1.5876 mm.

a) What’s the path difference your two mirrors now?

b) It the center of the image on your detector dark, or bright?

Solution-

Steps

1) The unit is nm , convert to nm 1587.6 nm

2) The light beam must travel to the movable mirror and back,

So multiple the distance that the mirror shifted by 2

2x1587.6= 3175.2 nm

The path difference is 3175.2 nm



36

This no. an odd integer multiple of (λ/2) (in which case it would cause

destructive interference and result in darkness in the center), or an

integer multiple of λ (which would cause constructive interference

and a result in a bright center)

Step 2:

3175.2 nm/560 nm =5.67 ≈ 𝟔

5.67 is an integer, so it’s constructive interference resulting in a bright

center on the image at the detector.

Note: Limiting path difference as determined by the length of wave packets.

VISIBILITY OF THE FRINGES

There are three principal types of measurement that can be made with the

interferometer:

(I) width and fine structure of spectrum lines, (2) lengths or displacements

in terms of wavelengths of light, and (3) refractive indices

In Michelson’s interferometer, the intensity is given by:

𝐼 = 4𝐼0𝑐𝑜𝑠2(1

2∅) (14)

Since the phase path ∅ =2𝜋

𝜆 (2𝑑 cos 𝜃)

Where d: distance between M1 & M2’.the intensity is max. When ∅ is an

integral multiple of 2𝜋 . The intensity is zero when ∅ is an odd multiple of 𝜋

Michelson’s interferometer measured the visibility, defined as

𝑉 =𝐼𝑚𝑎𝑥− 𝐼𝑚𝑖𝑛

𝐼𝑚𝑎𝑥+ 𝐼𝑚𝑖𝑛 (27)

Where Imax and Imin are the intensities at the maxima and minima of the

fringe pattern. The more slowly V decreases with increasing path difference,

the sharper the line.



37

Example (10): which case will the visibility be 1? What does a visibility of

one mean?

Imax= 4I, I min = 0

𝑉 =𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛

𝐼𝑚𝑎𝑥 + 𝐼𝑚𝑖𝑛

𝑉 =4𝐼 − 0

4𝐼 + 0 = 1

A visibility of 1 means that fringes are most visible (i.e. .this is the highest

possible contrast)

Uses of Michelson's Interferometer

1. Determination of wavelength of monochromatic light

2d1= m1 λ

2d2=m2 λ

∆𝒅 = 𝒅𝟐− 𝒅𝟏 = (𝐦𝟐 − 𝐦𝟏) .𝝀

𝟐

N=m2- m1

𝛌 =𝟐(𝒅𝟐−𝒅𝟏)

𝑵=

𝟐𝒅

𝑵 (28)

d2 -d1: no. of fringes counting the center of the field of view.

(Conversely, if x is increased, the fringe pattern will expand.)

N: no. of fringes



38

Michelson tested the lines from various sources and concluded that a certain

red line in the spectrum of cadmium was the most satisfactory .The standard

meter in Paris with the wavelengths of intense red, green, And blue lines of

cadmium by Michelson and Benoit

Red line λ= 6438.4722 Å or λ = 643.84 nm

Green line λ= 5085.8240 Å or λ = 508.5824 nm

Blue line λ = 4799.9107 Å or λ= 479.9991 nm

Example 11: A red laser (He-Ne) of wavelength 632.8 nm is used in

Michelson’s interferometer. While kept the mirror M1 is fixed and M2 is

moved. The fringes were found to move past a fixed cross-hair in the

observer .find the distance of the mirror M2 is moved for single fringe to

move past the origin line.



39

Solution:

∆𝒅 = 𝒅𝟐− 𝒅𝟏 = (𝐦𝟐 − 𝐦𝟏) .𝝀

𝟐

2(∆𝑑) = 𝑚𝜆

∆𝑑 = 𝑚𝜆

2= 1𝑥

632.8𝑛𝑚

2= 316.4 𝑛𝑚 = 0.3146 𝜇𝑚

Example 12: In an interferometer using laser produced 200 fringes cross

the field of view when the mirror is moved through 0.0589 mm. calculate the

wavelength of light used?

Solution

𝛌 =𝟐(𝒅𝟐−𝒅𝟏)

𝑵=

𝟐𝒅

𝑵=

𝟐𝒙 𝟎.𝟎𝟓𝟖𝟗 𝒙𝟏𝟎−𝟑𝒎

𝟐𝟎𝟎= 𝟓𝟖𝟗 𝒏𝒎

2- Determination of different in wavelength between two neighbor lines

or two waves

Let the source of light emit close wavelength 𝛌1 and λ2 such as sodium

lamp, the apparatus is adjusted to form circular rings. the arrangement of the

position of mirror M1 is moved and reached when a bright fringes of one set

falls on the bright fringe of the other an fringes are again distinct.

2d=m1λ1=m2 λ2 (29)

If condition λ1>λ2, since m2=m1+1, subst. in eq. 29



40

2d= m1λ1 = (m1+1) λ2

𝑚1 =𝜆2

𝜆1 − 𝜆2

∴ 2𝑑 =𝜆1𝜆2

𝜆1 − 𝜆2

𝜆1 − 𝜆2 = 𝜆1𝜆2

2𝑑

If used λ as a mean wavelength of two wavelength λ1and λ2, the small

difference between them is∶

∆𝜆 =𝜆2

𝑚𝑒𝑎𝑛

2𝑥 (30)

3-determination of the refractive index of gases

The path difference introduce between the two interfering beam is 2(n-1) L

Where n: Refractive index of gas

L: length of the tube. If m fringes cross the center of the field of view thus:

2(n-1)L=mλ

𝑛 =𝑚𝜆

2𝐿+ 1 (31)

Example (13):

In an experiment for determine the refractive index of gas using Michelson

interferometer a shift of 140 fringes is observed. When all the gas is

removed from the tube. If the wavelength of light used is 5460Å and the

length of the tube is 20 cm, calculate the refractive index of the gas?

Solution



41

𝑛 =𝑚𝜆

2𝐿+ 1

𝑛 =140𝑥5460𝑥10−10𝑚

2𝑥0.2+ 1 = 1.00019

(8-4) Index of refractive by interference Methods

If a thickness t of a substance having an index of refraction n is introduced

into the path of one of the interfering beams in the interferometer, the optical

path in this beam is increased because of the fact that light travels more

slowly in the substance and consequently has a shorter wavelength. The

optical path ∆= 𝑛 𝑡 is now nt through the medium, whereas it was

practically t through the corresponding thickness of air (n = 1). Thus the

increase in optical path due to insertion of the substance is (n - l)t.t This will

introduce (n - l)t). Extra waves in the path of one beam; so if we call ∆𝑚 the

number of fringes by which the fringe system is displaced when the

substance is placed in the beam, we have

Interferometry

Instrument used by the principle of interference of light is called

interferometer this instrument designed used to determine the refractive

index of each solid ,liquid and solutions transparent substances .

Refractive index of Solid state using 1) Michelson’s form of interferometer

2) Twyman & Green interferometer.

3) James and 4) Rayleigh is used to determine the refractive index of gases

and is known as refract meter

2. Twyman and Green interferometer

The Twyman Green interferometer is a variant of the Michelson

interferometer. It is used industrially for the interferometry control of non-

plane optical surfaces and mirrors or lens objectives. Figure 22 presents the



42

Twyman-Green device. It instrument for testing the distortion of optical

parts such as prisms and lenses. The piece to be tested is located in one of

the two beam paths. A mirror behind it, chosen that the reflected beams is

plane waves from the other arm of the interferometer by another lens, at the

focus of which the eye is placed. If the prism or lens is optically perfect, so

that the returning waves are strictly plane, the field will appear uniformly

illuminated. Any local variation of the optical path will, however, produce

fringes in the corresponding part of the field, which are essentially the

contour lines of the distorted wave front.

Figure (22): Twyman Green Interferometer



43

3) Jamin’s interferometer

The Jamin interferometer allows very exact measurements of

the refractive index and of gases; a transparent pressure chamber can be

positioned in the instrument. The phase shift due to changes in pressure is

quite easy to measure.

Figure (23) The Jamin interferometer.

G1&G2 is two thickness parallel plates with one mirror face. S is broader

source of monochromatic light. Two Similar evacuated tubes T1 and T2 of

equal length are placed in the two parallel beams. To calculate its refractive

index (n)

Gas is slowly admitted to tube T2. If the number of fringes ∆m crossing the

field is counted while the gas reaches the desired pressure and temperature,

the value of n can be found by applying the optical path is now nt

Through the medium, whereas it was practically t through the corresponding

thickness of air (n = 1). Thus the increase in optical path due to insertion of

the substance is (n - 1) t.

Extended

source

https://en.wikipedia.org/wiki/Refractive_index

https://en.wikipedia.org/wiki/Gas

https://en.wikipedia.org/wiki/Phase_shift



44

The number of fringes by which the fringe system is displaced when the

Substance is placed in the beam, we have (𝑛 − 1)𝑡 = ∆𝑚 ∗ 𝜆

In principle a measurement of ∆𝑚, t, and λ thus gives a determination of n.

The main source beam will broke into two parallel beams passing through

the two tubes. The recombines in G2 causes an interference fringes known

as Brewster’s fringes where

𝑛2 − 1

𝑛2 + 2= (𝑛 − 1)

𝑛 + 1

𝑛2 + 2= 𝑐𝑜𝑛𝑠𝑡.∗ 𝜌

This is a special case of the Lorenz-Lorentz* law, according to which

𝜌: 𝑔𝑎𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦

4- Mach-Zehnder interferometer

Figure 25: the Mach-Zehnder interferometer.

The Mach-Zehnder interferometer is suitable only for studying slight

changes of refractive index. Which like Michelson’ interferometer

except that it have a large separation in beams path.



45

5- Rayleigh's refractometer

In Rayleigh's refractometer (Fig. 26) monochromatic light from a

linear source S is made parallel by a lens L1, L2 are two converging

lenses and split into two beams by a fairly wide double slit. After

passing through two exactly similar tubes T1, T2 and the

compensating plates C1 &C2, these are brought to interfere by the lens

L2. This form of refractometer is often used to measure slight

differences in refractive index of liquids and solutions.

Figure 27: Rayleigh's refractometer

III.INTERFEROMETER BY A MULTPLE REFLECTON

1-NEWTON'S RINGS

2-TWO PARALLEL THIN FILM

1) Newton's rings This method for determining the wavelength of light was proposed by Sir

Isaac Newton in his book Optics, published in 1717. The experimental

arrangement is shown in Figure 1.

A plano-convex lens of large radius of curvature R is placed on a plane glass

plate with its curved surface downwards and is illuminated from above with a

parallel beam of monochromatic light. Some of the light is reflected from the

upper



46

surface of the glass plate and some from the lower surface of the lens;

interference thus occurs by division of amplitude, the fringes being localised

in the air gap between the lens and plate.

At any point a distance r from the axis of the lens the path difference will be

2h, where h is the distance between the lens and the plate at that point (See

Figure 2). whenever the thickness of the air film satisfies the condition R2= rm2+(R –t) 2



47

rm represents the radius of the mth dark ring;the thickness of the air film

(where the mth dark ring is formed) is t. the optical path difference between

the two waves is very nearly equal to 2nt, where n is the refractive index of



48

the film and t is the thickness of the film. Thus, whenever the thickness of

the air film

satisfies the condition

If a liquid of refractive index n is introduced between the

lens and the glass plate, the radii of the dark rings are given by

(because it is reflected from a medium of lower

refractive index). Hence,

the optical path difference

The interference fringes are circular because the system is symmetrical about

the centre of the lens. The radius of any ring is given by:

(2R - t)t = r2 r2 = 2Rt – t2



49

But t2 is small compared with 2Rt and so: 2Rt = r2 The path difference (2t) is therefore r2/R

A phase change of 𝜋 occurs when the light reflects from the top surface of

the plate but not at the lower surface of the lens, and therefore:

For a bright ring viewed by reflection: (2m + 1)λ/2 = rm2/R

For a dark ring viewed by reflection: mλ = rm2/R

Where m = 0, 1, 2, 3, etc and rm is the radius of the mth ring. If a graph is plotted of r2 against m for the dark

rings a straight line should be produced with a gradient given by:

(rm

2 - r12)/(m - 1) = λR

where r1 and rm are the radii of the first and mth rings respectively. (See Figure 3).

When doing the experiment it is much easier (and more accurate) to measure the diameter of the rings and then calculate their radius. A dark

central spot should be obtained when viewed by reflection.

The rings can be viewed by transmission by putting the microscope below the plate, and if this is done

the equations for bright and dark rings should be interchanged as two phase changes will occur,

producing an effective phase difference of 2𝜋. A

bright central spot should be obtained.

If white light is used a few coloured rings will be seen due to the different wavelengths of the different colours of light.

.



50

2) Interference in thin films

The beautiful colours in soap films and in oil floating

on water are due to interference by reflection. We

will first consider the interference due to a parallel-

sided film as shown in Figure 1. Interference can be

seen at O by division of amplitude, some light being

reflected from each side of the film.

Part of the light beam travels a distance BEP = 2BE

while the other travels a distance BM.

The path difference is therefore = 2nBE – BM

Now the distance 2BE is in the thin film which has a

refractive index n and so the equivalent path length

in a vacuum is 2nBE

So path difference = 2nBE – nBN = n(BE + EF - BN)

= nNF = 2nt cos r

where n is the refractive index of the film of

thickness t.

Path difference in a thin film = 2nt cosr

For a soap film in air a phase change will occur at the first face and so a maximum will

occur when the path difference is equal to an odd number of half wavelengths.

Therefore:

Path difference in a thin film = 2nt cosr = (m + ½)λ

where m = 0, 1, 2, etc.

For an oil film on water the phase changes will depend on the refractive index of the oil.

If we now consider the soap film and normal incidence (i.e. cos r = 1) it is easy to see

why thick films reflect light but do not show coloured effects and appear transparent and

why very thin films appear 'black' i.e. do not reflect light. The idea of the next set of

calculation is to see which wavelengths will 'fit' into the film and so give a maximum. In

other words to find out which colours will be seen. Consider a soap film of refractive

index 1.33 illuminated by white light.

For t =10-7m and m = 0 λ = 5.32x 10-7m = 532 nm (green)

while for m=1 λ = 1.77x10-7 m = 177 nm (ultraviolet)

Therefore only one colour is visible at this thickness at normal incidence.

For t = 10-6 m and m= 0 λ = 5.32 x 10-6 m = 5320 nm (infrared)

m = 4 λ = 5.91 x 10-7 m = 591 nm (yellow)



51

m = 5 λ = 4.75 x 10-7 m = 475 nm (blue)

All higher values of m give wavelengths in the ultraviolet region, so only two colours are

visible.

For t= 10-5 m and with m= 0 λ = 5.32 x 10-5m (infrared)

with m=36 λ = 7.28x10-7m (red)

with m=59 λ= 4.52x10- 7m (violet)

All higher values of m give wavelengths in the ultraviolet region. Therefore values of m

between 36 and 59 will "fit", there are 24 wavelengths that give maxima and so the

appearance is very nearly white.

For thicker films still many more wavelengths can 'fit in', and so all thicker films appear

to reflect white light.

For very thin films the distance travelled inside the film is insignificant and so the two

reflected waves are almost exactly out of phase with each other (due to the phase

change at one surface); they interfere destructively and the film appears 'black'. For this

reason when a soap film goes black it is about to burst.

Example (1) A film of oil 0.0005 mm thick and of refractive index 1.42 lies on a pool of water. Which colour will be missing from the spectrum when a point on the film is viewed at 400 to the vertical? Solution Destructive interference occurs: m = 2ntcos r

m = 2 x 1.42 x 5 x 10-7 x cos r

Therefore: m = 2 x 1.42 x 5 x 10-7 x 0.89

m = 1 1.26 x 10-6 m (infrared) m = 2 6.33 x 10-7 m (orange) m = 3 4.22 x 10-7 m (ultraviolet). Thus the only colour missing from the visible part of the spectrum will be an orange line of wavelength 6.33 x 10-7 m.

The colours in a soap film can be observed clearly by

projecting them on to a screen. A wire ring with a soap film

formed across it should be mounted vertically in the beam of

a high-intensity light source in such a way that the light is

focused just behind the film. A further lens may then be used

to project the colours formed on to a screen. As the water

runs from the film it gets progressively narrower at the top

and turns black when it is about to break. Beautiful effects

may be obtained by blowing gently on the film.



52

Problems interferences

Q1-How far must the movable mirror of a Michelson interferometer be

displaced for 2500 fringes of the red cadmium line to cross the center of the

field of view?

Q2- If the mirror of a Michelson interferometer is moved 1.0 mm, how

many fringes of the blue cadmium line will be counted crossing the field of

view?

Q3: A double-slit source with slit separation 0.2 mm is located 1.2 m from a

screen. The distance between successive bright fringes on the screen is

measured to be 3.30 mm. What is the wavelength of the light?

Answer: 550nm

Q4: In young’s Experiment, the distance between two slit is 0.8 mm and the

distance of the screen from the slits is 1.2 m. If the fringe width is 0.75 mm,

calculate the wavelength of light? Answer 500 nm

Q5: In young’s double slit experiment the distance between the slits is 1 mm

and fringe width is 0.6 mm when a light of certain wavelength is used. When

the screen is moved through 0.25 m the fringe width increases by 0.15 mm?

What is the wavelength of light used?

Q6: In young’s experiment, interference bands are produced on the screen

placed 1.5 m from two slits apart and illuminated by the light of

wavelength 6000 Å. Find (1) the fringe width (2) change the fringe width

taken away from the slit by 50 cm.

Q7: two slit 3 mm apart, 1 meter from the screen .if produced the sixth –

order bright line 1 mm from the center of the pattern , what is the

wavelength of the light used?

Q8: laser light illuminates two narrow parallel slits and the interference

pattern produced is observed on a wall 4 m away. The distance between the

two slits is 0.5 mm and the angular position of the fourth maximum is 0.3˚.

Find the wavelength λ of the light?



53

Q9: In a biprism experiment, the width of a fringe is 0.75 mm when the eye-

piece is at a distance of one meter from the slits. The eye-piece is now

moved away from the slits by 50 cm.Find the fringes width

Answer: new fringe width is 1.125 mm

Q10: In a biprism experiment, the width of a fringe is 0.12 mm when the

eye-piece is at a distance of 100 cm from the slits. The eye-piece is now

moved away from the slits by 25 cm.Find the fringes width

Answer: new fringe width is 0.09 mm

Chapter nine Diffraction is special case from interference, to explain the diffraction

Define of Diffraction:

The bending of waves around the edges of an obstacle

When a beam of light passes through a narrow slit, it spreads out to a certain

extent into the region of the geometrical shadow.

Fresnel and Fraunhofer diffraction

We can define two distinct types of diffraction:

(a) Fresnel diffraction is produced when light from a point source meets an

obstacle, the waves are spherical and the pattern observed is a fringed image

of the object.

(b) Fraunhofer diffraction occurs with plane wave-fronts with the object

effectively at infinity. The pattern is in a particular direction and is a fringed

image of the source.

Fresnel diffraction

Fresnel diffraction can be observed with the minimum of apparatus but the

mathematics are complex. We will therefore only treat it experimentally

here.

If a razor blade is placed between the observer and a point source of

monochromatic light, dark and bright diffraction fringes can be seen in the

edges of the shadow. The same effects can be produced with a pinhead,

when a spot of light will be seen in the center of the shadow.



54

Fraunhofer diffraction - Single slit

the general condition for a minimum for a single slit is:

mλ = a sin θ

where m = 1, 2, 3, 4 and so on

For a circular aperture and a small angle the formula has to be modified to

give:

Smallest resolvable angle (φ) = 1.2λ/a

You can see that higher resolution is possible with large apertures or with

short- wavelength radiation such as ultraviolet light, X-rays or even

electrons.

This is why ultra violet light or electrons are used in microscopes where a very high magnification is needed.

The eye can resolve fine detail rather better if the lighting is not too strong,

so that the pupil will have a large aperture.



55

The problem with very large optical telescope mirrors has been overcome to

some extent by the use of multiple mirror telescopes. These instruments use

a number of smaller mirrors mounted to give the same light-gathering power

and resolving power as a very large single mirror

Fresnel diffraction Fraunhofer diffraction

1- Either the source and the screen

or both are finite distance from the

diffraction element

1- both of light source and

screen are an infinite

distance from diffraction

element

2-We are dealing with spherical

wave

2-We are dealing with plane wave

3- lenses are not required 3-lenses are required

A diffraction grating consists of a large number of equally spaced, identical

slits. The condition for intensity maxima in the interference pattern of a



56

diffraction grating for normal incidence i s

Where d is the spacing between adjacent slits and m is the order number of

the diffraction pattern. The mth order of, the diffraction pattern is

Where N is the number of lines in the grating that are illuminated.

And difference between interference and diffraction

Interference Diffraction

1.is results of interaction of light

coming from different wave front

originating from the source

1. is results of interaction of light

coming from different parts of the

same wave front .

2.interference fringes are the same

width or may not be the same width

2. diffraction fringes are not the

same width

3. All bright bands are of same

intensity

3. the difference max. of varying

intensities with max. intensity for

central maximum.

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