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Diffraction

Types of Fresnel’s Diffraction

1 Zone Plate

A zone plate is a specially constructed transparent plate hav-

ing concentric circles with their radii proportional to the square

root of natural numbers i.e.,

rn=√nwhere n=1,2,3,....

The alternate zones are painted blck such that light is ob-

structed through them. Such a plate behaves like a convex

lens and produces an image of a source of light on a screen

placed at suitable distance.

1.1 Construction

Concentric circles are drawn on white paper such that the radii

are proportional to the square root of the natural numbers. The

alternate zones (odd) are covered with black ink and a reduced

photograph is taken. The drawing appears as (a) and the neg-

ative of the photograph will be as (b). In the developed neg-

ative, the odd zones are transparent to incident light and even

zones are opaque.

1.2 Theory

Consider a point source O emiting ligh of wavelength λ. Let

PH1be the imaginary transparent plate perpendicular to the

plane of paper. Let OP ⊥PHnand OP is produced to I where

the intensity of light will be calculated. Here, P is the centre

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and radii are equal to.

PH1=r1,PH2=r2,PH3=r3....PHn=rn

Circles are constructed in such a way that

OH1I−OPI =λ

OH2I−OPI =2λ

........

OHnI−OPI =nλ

2...(1)

Let OP = u and PI = v

Now, OH2

n=OP2+H2

or, OH2

n=u2+r2

or, OHn=pu2+r2

n=u1+r2

u21/2

=u1+1

u2

∴OHn=u+r2

Similarly, IHn=v+r2

so, equation (1) becomes

u+r2

2u+v+r2

2v−(u+v) = nλ

textor,r2

21

u+1

v

or, r2

n=nλuv

u+v...(2)

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Here, rn∝√n

So, for a given object and image distance, the focal length of

the zone plate is

f=1

u+1

v=u+v

uv =nλ

[from (2)]

∴fn=r2

nλ

The area A of nth zone is (πr2

n−πr2

n−1)

=πnλuv

u+v−(n−1)λuv

u+v

=πλuv

u+v

or, A=πλ f=const.

Hence, for a given object and image distance, area of zone

remains same.

1.3 Comparison of Zone plate and convex lens

1. Both zone plate and convex lens form a real image of the

object and the equations connecting the conjugate distances

are similar.

2. The focal length of both depends on the wavelength λand

hence suffer from chromatic aberration.

3. In case of zone plate, image is formed by diffraction phe-

nomenon. In case of a convex lens, the image is fomed due

to refraction of light.

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4. Zone plate has multiple foci on either side of plate and thus

intensity of image will be less. Convex lens has one focus

and hence intensity of image will be more.

5. The path difference of image point in zone plate is λbut in

convex lens is zero.

6. Zone plate can be used over wide range of wavelengths

from microwave to x-rays but convex glass lens can only

be used in visible region.

2 Diffraction produced by circular aperture

Let AB be a small circular aperture of radius r and S is a point

source of monochromatic light. O is the centre of aperture and

QOQ’ be a section of a spherical wavefront incident on the

aperture. Consider P be a point on the screen such that SOP is

perpendicular to the circular aperture AB and the screen at P.

The aperture will cast a shadow on the screen outise M and N.

The aperture is considered to have half period zones contructed

with centre at O. The no. of half period zones depends upon

the distance between the screen and the aperture.

Let a be the distance of aperture from the source and b be the

distance of screen from the aperture. Let δbe the path differ-

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ence for the wave reaching P along the paths SAP and SOP.

Then, δ= (SA +AP)−(SO +OP)

= (a2+r2)1/2+ (b2+r2)1/2−(a+b)

=a(1+r2

a2)1/2+b(1+r2

b2)1/2−(a+b)

=a(1+r2

2a2) + b(1+r2

2b2)−(a+b)

=r2

21

a+1

b

∴1

a+1

b=2δ

r2...(1)

If the position of the screen is such that n half period zones

are constructed, then

path diff ,δ=nλ

or, 2δ=nλ

substituting in equation (1) becomes

1

a+1

b=nλ

r2...(2)

If the source is at inﬁnite distance, then a=∞and

b=n|lambda

or, n=r2

λb

or, n=πr2

πλb

πr2is the area of aperture AB and πλbis the area of half period

zones constructed at a distance b from the wavefront. If nis

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odd we get maximum at P an if nis even we get a minmum at

2.1 Intensity at a point away from the centre

Let abe the disance of aperture from the source and bbe the

distance of screen from the aperture. Let P′be any point on

the screen such that PP′=x.

Th path difference between the secondary waves from A and

B and reaching P′can be given by,

δ=BP′−AP′

=pb2+ (x+r)2−pb2+ (x−r)2

=b1+(x+r)2

2b2−b1+(x−r)2

2b2

2b(x+r)2−(x−r)2

2b(4xr)

∴δ=2rx

b...(1)

The point P′will be dark if path difference is

δ=2nλ

or, 2nλ

2=2rx

or, xn=nλb

2r...(2)

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where xngives the radius of nth dark ring as screen.

Similarly, the point P′will be bright if path diff is

δ= (2n+1)λ

or, (2n+1)λ

2=2rx

or, xn= (2n+1)bλ

4r...(3)

where xngives the radius of nth bright ring.

2.2 Application

The objective of a telescope consists of an achromatic convex

lens and a circular aperture is ﬁxed infront of the lens. Let the

diameter of the aperture be D=2r. While viewing distance

objects the incident wavefront is plane and the diffraction pat-

tern consist of a bright centre surrounded by dark and bright

rings of gradually decreasing intensity.

The radii of the dark ring is

xn=nλb

The radii of 1st dark ring is x1=λb

2r.

For an incident plane wavefront, b=f, the focal length of

objective.

∴x1=λf

However, according to Airy’s theory, x1=1.22λf

The value of x1measures the distance of the 1st secondary

minima from the central bright maximum. It is interesting to

note that the size of the central image depends upon λ,fand

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3 Diffraction due to a straight edge

Let S be a narrow slit illuminated by a source of monochro-

matic light of wavelength λ. The length of the slit is perpen-

dicular to the plane of the paper. AD is the straight edge and

the length of the edge is parallel to the length of the slit. P is

a point of the screen and SAP is perpendicular to the screen.

Below the point P is the geometrical shadow and above P is

the illuminated portion.

Let SA = aand AP = b

Let P′be a point in the screen such that PP′=x, the path

difference at P′is

δ=AP′−BP′

= (b2+x2)1/2−(SP′−SB)

=b1+x2

2b2−(p(a+b)2+x2−a)

=b+x2

2b−(a+b)1+x2

2(a+b)2+a

=b+x2

2b−a−b−x2

2(a+b)+a

=x2

21

b−1

a+b

∴δ=x2

2.a

(a+b)b

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Now, P′will be maximum if δ= (2n+1)λ

∴(2n+1)λ

2=ax2

2b(a+b)

or, xn=r(2n+1)(a+b)bλ

a=c√2n+1

where c=q(a+b)bλ

a= constant

This provides position of nth maxima in diffraction pattern.

So, position of successive maxima are

x1=Cfor n = 0

x2=√3Cfor n = 1

x3=√5Cfor n = 2

x4=√7Cfor n = 3

Therefore, the distance between successive maxima will be

x2−x1= (√3−1)C=0.73C

x3−x2= (√5−√3)C=0.50C

x4−x3= (√7−√5)C=0.43C

Thus, the separation of successive maxima are in decreasing

order. As we go above on the screen, the fringes comes closer.

Now, P′will be minimum if δ=2nλ

∴2nλ

2=ax2

2b(a+b)

or, xn=r2n(a+b)bλ

a=c√2n

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where c=q(a+b)bλ

a= constant and xnis the distance of the

nth dark band from P. The intensity distribution of diffracted

beam is shown here.

3.1 Determination of wavelength of light

For nth maxima

n=(2n+1)(a+b)bλ

For (n+m)th maxima

n+m=(2n+2m+1)(a+b)bλ

on subtracting

n+m−x2

n=(a+b)bλ×2m

∴λ=x2

n+m−x2

2mb(a+b)λa

knowing all quantities, wavelength can be determined easily.

4 Fraunhofer’s Diffraction at a single slit

Consider a parallel beam of light incident normally ona slit

AB of width ′d′as shown in ﬁgure below. After passing through

the slit, the beam is focused on to the screen MN by means of

a lens L. A bright fringe is formed at centre O of the screen

and alternate dark and bright bands on either sides of central

bright spot at O. These bands are secondary minima and max-

ima respectively.

It is observed that

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i) The secondary maxima lies mid way between the secondary

minima.

ii) The intensity of secondary maxima decreases with distance

from O.

iii) The width of central maxima is double than that of a sec-

ondary maxima.

Let the duffracted ray meet at point P on the screen. The path

difference between wavelets from A and B is given by

BN =BP −Ap =dsin θ

If the path difference is equal to nλ, secondary minima will be

produced at P

i.e.,dsin θn=nλ

or, sinθn=nλ

d...(1)

where n=0,1,2,...

If the path diffeence is equal to (2n+1)λ

2, secondary maxima

will be produced at P

i.e.,dsin θn= (2n+1)λ

or, sinθn= (2n+1)λ

2d...(2)

where n=0,1,2,...

4.1 Width of Central Maxima

The width of the central maxima is deﬁned as the distance

between the ﬁrst minima on two sides of central maxima such

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that

sinθ=±λ

For small angle, sinθ≈θ

∴θ=±λ

Thus angular width of the central maxima, 2θ=2λ

d...(3)

If y be the half width of central maxima and D be distance

between screen and lens, we can write

tanθ=y

For small angle, tanθ≈θ

∴θ=y

or, 2θ=2y

d...(4)

From (3) and (4), we have

2λ

d=2y

or, y=λD

Therefore, width of central maxima, β=2y=2λD

The intensity istribution on thescreen due to the diffraction at

a single slit is shown below.

It is seen that the width of central maxia si directly propor-

tional to wavelength λand inversely proportional to width of

the slit d. Thus, increasing the width of the slit, the secondary

maxima becomes narrower. Thus, distinct diffraction pattern

will be obtained only in the case of narrow slit.

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5 Fraunhofer Diffracton ata double slit

Consider AB and CD as two rectangular slits parallel to one

another and perpendicular to the plane of the paper. The width

of each slit is ′a′and the width of the opaque portion is ′b′; L

is the collecting lens and MN is a screen placed perpendicular

tot he plane of paper.

Let a plane wave front be incident as the surface of XY. All

the secondary wavelets, traveling in a direction parallel to OP

comes to focus at P.Thus, P corresponds tot he position of cen-

tral bright maxima.

Here, diffraction is considered in two parts.

i) The interference phenomemon due to the secondary waves

originating fromthe corresponding points of the two slits and,

ii) The diffraction pattern due to the secondary waves from the

two slits individually.

5.1 Interference Maxima and Minima

Consider the secondary waves traveling in a direction inclined

at an angle θwith the initial direction.

We have,

sinθ=CN

AC =CN

a+b

or, CN = (a+b)sin θ

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Interference maxima will be formed at P′if

CN = (2n+1)λ

or, (a+b)sinθn= (2n+1)λ

2λn=1,2,....

or, sinθn=(2n+1)λ

2(a+b)

It gives dark interference fringe.

Similarly, interference maxima will be formed at P′if

CN =nλ

or, (a+b)sinθn=nλn=1,2,....

or, sinθn=nλ

a+b

Th angular separation between any two consecutive minima is

given by

θ2−θ1≈sinθ2−sintheta1

=5λ

2(a+b)−3λ

2(a+b)

=λ

a+b

Similary, angular separation between any two consecutive maxia

is given by

θ2−θ1=λ

a+b

Thus,angular separation is inversely proportional to the dis-

tance between two slits (a+b).

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5.2 Diffraction Maxima and Minima

Consider the secondary waves traveling inthe direction inclined

at an angle θwith the initial direction.

If path difference, BM =λ, then θwill be give the direction of

diffraction mnima at P′.

The path difference corresponding to the upper half and lower

half of the wavefront AB is equal to λ

2. The effect at P′due to

wavefront incident on AB is zero.

Similarly, the effect at P′due to the wavefront on CD is zero.

In general, for diffraction minima

asinθn=nλn=1,2,3.....

Intensity distribution of the diffraction pattern due to double

slit is shown below:

5.3 Difference between single slit and double slit

i. It consist of central

bright maxima with sec-

ondary inia and maxia of

gradually creasing intensity

i. It consist of equally

spaced interference max-

ima and minima within

central maxima

ii. Intensity of central max-

ima is less

ii. Intensity of central max-

ima is four times than sin-

gle slit.

6 Fraunhofer Diffraction at a Circular Aperture

Consider a circular aperture AB of diameter d. A plane wave-

front is incident on the circular aperture. The secondary waves

traveling in the direction CO comes to focus at P. Therefore,

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P corresponds to the position of central maximum. Now, con-

sider the secondary waves traveling in a direction inclined at

an angle θwith the direction CP,that meet at P′on the screen.

Let PP′=y.

Here, path difference is BD =dsinθ

If the path difference is equal to integral multiple of λ

i.e.,dsin θn=nλ..(1)

the point P′will have minimum intensity.

Similarly the point P′will have maximum intensity if

dsinθn= (2n+1)λ

2n=1,2,3....

If P′is a point of minimum intensity, all the points on the

screen at the same distance from P such as P′and lying on a

circle of radius y will be of minimum intensity. Thus, diffrac-

tion pattern due to a circular aperture consist of a central bright

disc called the Airy’s disc, surrounded by alternate dark and

bright concentric rings called Airy’s ring. The intensity of the

dark ring is zero and that of the bright rings decreases gradu-

ally outwards from P.

6.1 Radius of Airy’s disc

If the circular aperture is near to the lens L and screen is at

large distance from L.

sinθ≈θ=y

f...(1)

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For 1st secondary minima,

dsinθ=λ

or, sinθ=λ

d...(2)

∴λ

d=y

∴y=λf

d...(3)

where yis the radius of Airy’s disc.

But, according to Airy’s, the actual radius of 1st dark ring is

given by

y=1.22 fλ

d...(4)

Thus, the radius of central disc decreases with the increase in

diameter of the aperture.

7 Plane Transmission Grating

A diffraction grating is an arrangement of large number of

equally spaced narrow parallel slits ruled on a plane galss or

metal plates. It can be constructed by drawing a large numbers

of straight and equidistant lines or scratches over a thin glass

plate by means of a ﬁne sharp diamond point. Such surface

acts as a transimssion grating.

Let XY be the grating surface and RS be the screen. Let a be

the thickness of slit and b be the thickness of the opaque por-

tion. Consider a beam of monochromatic light of wavelength

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λis incident normally as the grating surface.

At O, the wavelets from all slits interfere constructively pro-

ducing a maximum. Thus, a central maxima is formed. For

any point P, let the light diffracted at an angle θwith the direc-

tion of incident light. Then, the path difference between the

rays from A and C is

CN = (a+b)sin θ

The point P will be maximum if

path difference , (a+b)sinθn=nλ

where n=0,1,2,...

7.1 Maximum numbers of order

We have,

(a+b)sinθ=nλ

or, n=(a+b)sinθn

the maximum value of sinθis at θ=90◦

∴nmax =(a+b)sin90

or, nmax =a+b

If (a+b)<2λthen nmax <2.

i.e., only 1st order will be obtained.

7.2 Dispersive Power of Grating

If the incident light consists of two nearly equal wavelengths λ

and (λ+dλ)of incident light, the beam gets dispersed and the

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angle of diffraction for different wavelengths will be different.

Then, we have,

(a+b)sinθ=nλ...(1)

and (a+b)sin(θ+dθ) = n(λ+dλ)...(2)

where θand (θ+dθ)are the angles of diffraction correspond-

ing to these wavelength.

The dispersive power of the grating is the rate of change of an-

gle of diffraction with the wavelength of light. It is expressed

as dθ

dλ.

We have, (a+b)sin θ=nλ

differentiating both sides, we get

(a+b)cosθdθ=ndλ

or, dθ

dλ=n

(a+b)cosθ

Thus, the dispersive power is

dθ

dλ=n

(a+b)cosθ

There is more dispersion for red than violet colour of light.

8 Resolving Power of an Optical Instrument

The resolving power of an optical instrument is the ability to

produce distinctly separate images of two points objects very

close together. It depends on the size of the slit through which

objects are viewd and on the wavelength of light used. The

images of two objects are said to be resolvedd when the central

maxima of the diffraction pattern of one object just coincides

with the 1st minima of the other.

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The resolving power of a grating is deﬁned as the ratio of the

wavelength of any spectral line to the difference in wavelength

between this lines and neighbouring lines such that the two

lines appear to be just resolved.

Thus, resolving power of a grating = λ

dλ

Here, at P1

(a+b)sinθ=nλ...(1)

and at P2

(a+b)sin(θ+dθ) = n(λ+dλ)...(2)

The two lines will appear just resolvd if the angle of diffrac-

tion (θ+dθ)also corresponds to direction of 1st minima. This

is possible if the extra path difference introduced is λ

Nwhere

N=1

a+bis the number of lines on the grating surface.

∴(a+b)sin(θ+dθ) = nλ+λ

from (1) and (2)

n(λ+dλ) = nλ+λ

or, ndλ=λ

or, λ

dλ=nN

This gives the resolving power of grating.

9 Resolving Power of Microscope and Telescope

The resolving power of the microscope is its ability to form

separate image of two points objects lying close together. It

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is determined by the least distance between two points objects

which can be distinguished. Thus, distance ’d’ is resolving

power and is given by

d=λ

2µsinθ

where λis the wavelength of light used to illuminate the object

and µis refractive indexs of the medium between the object

and the objective. The angle θis the half angle of the cone

of light from the object point. The term µsin θis called the

numerical aperture of the objective.

The resolving power of a telescope is the reciprocal of the

smallest angular separation between two distant objects where

image are separated in the telescope. This is given by

dθ=1.22λ

where dθis the angle substented by the point object at the ob-

jective, λis the wavelength of light used and d is the diameter

of the telescope objective. It is seen that a telescope with a

large aperture objective gives a high resolving power.

10 Numerical

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