Unit -II Diffraction of Light

“Diffraction is the bending of light around an obstacle such as the edge of a slit”.

We know that waves bend round the obstacles placed in their path. For example, water waves escaping
through a small hole spread out in all directions as if they have originated at the hole.

Similarly sound, form of wave motion, is found to pass round the obstacles of moderate dimensions.

“The amount of bending, however, depends upon the size of the obstacle and wavelength of wave”.
It was argued that light, which is a wave motion, should also bend round the obstacles. A minute
investigation reveals that light suffers some deviation from its straight path in passing close to the edge
of opaque obstacles and narrow slits. The deviation is extremely small when the wavelength of light
waves is small in comparison to the dimensions of the obstacle or aperture. But, when the size of aperture
is comparable with the wavelength of light, this deviation becomes much pronounced.

For example, when light waves diverging from narrow slit S which is illuminated by a monochromatic
source O, pass an obstacle AB with straight edge A parallel to the slit, the geometrical shadow on the
screen is never sharp. A small portion of light bends around the edge into geometrical shadow. Outside
the shadow parallel to its edge several bright and comparatively dark bands are observed.

Thus, when light falls on obstacles or small apertures whose size is comparable with the wavelength
of light, there is a departure from straight lines propagation, the light bends round the corners of
the obstacles or apertures and enters in the geometrical shadow. This bending of light is called
diffraction.

It was found that diffraction produces bright and dark fringes known as diffraction hands or fringes.The
correct interpretation of diffraction phenomenon was provided by Fresnel. According to Fresnel, the
diffraction phenomenon is due to mutual interference of secondary wavelets originating from various
points of the wavefront which are not blocked off by the obstacle. Fresnel, thus, applied Huygens’s
principle of secondary wavelets in conjunction with the principle of interference and calculated the
position of fringes. It was observed that the calculated results were in good agreement with the observed
diffraction pattern.

It should be remembered that the diffraction effects are observed only when a portion of the wavefront is
cut off by some obstacle.

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Types of Diffractions
To study diffraction phenomenon, we need (i) source, (ii) aperture and (iii) screen or telescope.
Diffraction phenomenon can be divided into following two general classes:

(1) Fraunhofer's diffraction: In this class of diffraction, source and the screen or telescope (through which
the image is viewed) are placed at infinity or effectively at infinity (using lenses) from aperture. In this
case the wavefront which is incident on the aperture or obstacle is plane.

(2) Fresnel's diffraction: In this class of diffraction, source and screen are placed at finite distances from
the aperture or obstacle having sharp edges. In this case no lenses are used for making the rays parallel or
convergent. The incident wavefronts are either spherical or cylindrical.
Difference between Fresnel's and Fraunhofer's diffractions
Fresnel’s diffraction Fraunhofer’s diffraction
i)The source and the screen are at finite i)The source and screen are at infinite distances
distances from the aperture or obstacle. from the aperture.
ii)No lenses are required to observe the ii)The convex lenses are necessary to observe
diffraction pattern. the diffraction pattern.
iii)The incident wavefront is either iii)The wavefront is plane.
Spherical or cylindrical.
iv)Diffraction effects can be studied in
iv) Diffraction patterns are studied in the
any direction.
direction of propagation of light.

Fraunhofer diffraction at Single slit

Figure represents a section AB of a narrow slit of width e perpendicular to the plane of the paper.

Let a plane wave front WW’ of monochromatic light of wavelength λ propagating normally to the slit be
incident on it. Let the diffracted light be focused by means of a convex lens on a screen placed in the focal
plane of the lens.

According to Huygens-Fresnel, every point of the wavefront in the plane of the slit is a source of
secondary spherical wavelets, which spread out to the right in all directions.

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The secondary wavelets travelling normally to the slit, i.e., along the direction OP, are brought to focus
at Po by the lens. Thus, Po is a bright central image.

The secondary wavelets travelling at an angle θ with the normal are focused at a point P₁ on the screen.
The point P₁ is of the minimum intensity or maximum intensity depending upon the path difference
between the secondary waves originating from the corresponding points of the wavefront.

General Mathematical Theory

In order to find out intensity at P₁ , draw a perpendicular AC on BR, the path difference
between secondary wavelets from A and B in direction θ = BC = AB sin θ = e sin θ, and the corresponding
2𝜋
phase difference = 𝑒 𝑠𝑖𝑛 𝜃
𝜆

Let us consider that the width of the slit is divided into n-equal parts and the amplitude of the wave from
each part is a (because width of each part is same). The phase difference between consecutive waves from
1 1 2𝜋
these parts would be (𝑇𝑜𝑡𝑎𝑙 𝑝ℎ𝑎𝑠𝑒 ) = ( 𝑒 𝑠𝑖𝑛 𝜃) = 𝑑 (say)
𝑛 𝑛 𝜆

Using the method of vector addition of amplitudes, the resultant amplitude R is given by
𝑛ⅆ
𝑎𝑠𝑖𝑛
2
𝑅 = ⅆ
𝑠𝑖𝑛
2

𝑠𝑖𝑛(𝜋𝑒 𝑠𝑖𝑛𝜃/𝜆)
= 𝑎 𝑠𝑖𝑛(𝜋𝑒 𝑠𝑖𝑛𝜃/𝑛𝜆)

𝑠𝑖𝑛 𝛼
= 𝑎 where 𝛼 = 𝜋𝑒 𝑠𝑖𝑛𝜃/𝜆
𝑠𝑖𝑛(𝛼/𝑛)

𝑠𝑖𝑛 𝛼
= 𝑎 𝑠𝑖𝑛(𝛼/𝑛)

𝑠𝑖𝑛 𝛼
= 𝑎 ( ∵ 𝛼/𝑛 is very small 𝑠𝑖𝑛( 𝛼/𝑛) ≈ 𝛼/𝑛 )
𝛼/𝑛

𝑠𝑖𝑛 𝛼
= 𝑛𝑎 𝛼
𝑠𝑖𝑛 𝛼
= 𝐴 𝛼
𝑠𝑖𝑛 𝛼
Thus, Resultant amplitude is given by 𝑅= 𝐴 --------( 1 )
𝛼

When 𝑛 → ∞, 𝑎 → 0 , but product 𝑛 𝑎 = 𝐴 ( 𝑟𝑒𝑚𝑎𝑖𝑛𝑠 𝑓𝑖𝑛𝑖𝑡𝑒)

𝑠𝑖𝑛 𝛼 2
Now , the Intensity is given by 𝐼 = 𝑅2 = 𝐴2 ( )
𝛼

𝑠𝑖𝑛 𝛼 2
𝐼 = 𝐼0 ( ) ---------( 2 )
𝛼

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Intensity distribution
The expression for resultant amplitude R can be written in ascending powers of 𝛼 as
𝐴 𝛼3 𝛼5 𝛼7
𝑅= [𝛼 − + − +⋯ ]
𝛼 3! 5! 7!
𝛼2 𝛼4 𝛼6
= 𝐴 [1 − + − +⋯ ]
3! 5! 7!

If the negative terms vanish, the value of R will be maximum, i,e., 𝛼 = 0

𝜋𝑒 𝑠𝑖𝑛𝜃
α= =0 or sinθ = 0
𝜆

θ = 0 ------------( 3 )
Now, maximum value of R is A and intensity is proportional to 𝐴2 .
The condition θ = 0 means that this maximum is formed by those secondary wavelets which travel
normally to the slit. The maximum is known as principal maximum.
2) Maximum intensity positions
The intensity will be maximum when 𝑠𝑖𝑛 𝛼 = 0. The values of α which satisfy this equation are 𝛼 =
±𝜋, ±2𝜋, ±3𝜋, ±4𝜋, ………….. etc = ±𝑚 𝜋
𝜋𝑒 𝑠𝑖𝑛𝜃
⇒ = ±𝑚 𝜋
𝜆
⇒ 𝑒 𝑠𝑖𝑛𝜃 = ±𝑚 𝜆 where m = 1,2,3,………etc
In this way we obtain he points of minimum intensity on either side the principal maximum.

3) Secondary maxima
In addition to principal maximum at 𝛼 = 0, there are weak secondary maxima between equally spaced
minima. The positions can be obtained with the rule of finding maxima and minima of a given function
in calculus. Differentiating the expression of I with respect to 𝛼 and equating to zero, we have

ⅆ𝐼 ⅆ 𝑠𝑖𝑛 𝛼 2
= ⅆ𝛼 (𝐴2 ) =0
ⅆ𝛼 𝛼

2 𝑠𝑖𝑛 𝛼 (𝛼 𝑐𝑜𝑠 𝛼−𝑠𝑖𝑛𝛼 )

⇒ 𝐴2 . =0
𝛼 𝛼2

Here, either 𝑠𝑖𝑛 𝛼 = 0 or (𝛼 𝑐𝑜𝑠 𝛼 − 𝑠𝑖𝑛 𝛼 ) = 0

The equation 𝛼 = 0 gives the value of 𝛼 ( except 0) for which the intensity is zero on the screen.
Hence, the positions of maxima are given by the roots of the equation

𝛼 𝑐𝑜𝑠 𝛼 − 𝑠𝑖𝑛 𝛼 ( or ) 𝑡𝑎𝑛 𝛼 = 𝛼 --------( 4 )

The values of a satisfying the above equation are obtained graphically by plotting the curves y = 𝛼 and
y= tan 𝛼 on the same graph.

The points of intersection of two curves give the values of 𝛼 which satisfy eq(2). The plots of
y = 𝛼 and y = tan 𝛼 are shown in figure.

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 3𝜋 5𝜋 7𝜋
The points of intersections are 𝛼 = 0, ± ,± ,±
2 2 2

or more exactly to 𝛼 = 0, ±1.430𝜋, ±2.462𝜋, ±3.471𝜋 etc

Substituting the approximate values of 𝛼 in eq(2) , we get the intensities in various maxima
𝐼0 = 𝐴2 ( 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑎)
𝑠𝑖𝑛(3𝜋/2) 2 𝐴2
𝐼1 = 𝐴2 ( ) = approx ( 1st Subsidiary maxima)
3𝜋/2) 22

𝑠𝑖𝑛(5𝜋/2) 2 𝐴2
𝐼2 = 𝐴2 ( ) = 𝑎𝑝𝑝𝑟𝑜𝑥 ( 2nd Subsidiary maxima)
5𝜋/2) 62

and so on.
From the expressions of 𝐼0 , 𝐼1 , 𝐼2 it is evident that most of the incident light is concentrated in the
principal maximum.

Intensity distribution graph

A graph showing the variation of intensity with 𝛼 is shown in figure.

The diffraction pattern consists of a central principal maximum occurring in the direction of incident
rays. There are subsidiary maxima of decreasing intensity on either sides of it at positions
3𝜋 5𝜋 7𝜋
𝛼 = 0, ± ,± ,± and so on.
2 2 2

Between subsidiary maxima, there are minima at positions 𝛼 = 0, ±𝜋, ±2𝜋 , ±3𝜋,

Note: It should be noted that subsidiary maxima do not fall exactly mid-way between two minima, but
they are displaced towards the centre of the pattern, of course, the displacement decreases as the order
of maximum increases.

Note: Fraunhofer used the first grating consisting of a large number of parallel wires placed very
closely side by side at regular intervals. The diameters of the win were of the order of 0.05 mm and
their spacing, varied from 0.0533 mm to 0.687 mm.
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Fresnel’s half period zones
Fresnel calculated the resultant intensity at any point due to wavefront by dividing it into a number of
zones which are called as Fresnel’s half period zones.

Let ABCD be a monochromatic plane wavefront of

wavelength λ proceeding in the direction of arrow and P is a
point where the resultant intensity to be calculated. Fresnel
divided the wavefront into a number of zones.

Let PO be the normal to the wavefront (PO = p say). With P as

the centre and radii equal to p+λ/2 p+2λ/2...... p+nλ/2, spheres
are drawn. The plane ABCD cuts these spheres in concentric circles with center O and radii OM1 , OM2
OM3……, OMn. The area of the first innermost circle is the first half-period zone.

Similarly, the areas enclosed between first and second circles, second and third circles,.….(n-1)th and nth
circles are known as 2nd , 3rd, ….. nth half-period zones respectively.

It is assumed that a resultant wave starts from each zone. The successive zones differ in phase by 𝞹 or
by a half period (T/2), called as half-period zones.

1. Amplitude due to a zone

The amplitude of disturbance at P due to the wave from a zone varies:
(1) directly as the area of the zone,
(ii) inversely as the average distance of the cone from P,
(iii) directly as f(𝜃𝑛 ), where f(𝜃𝑛 ) is a function of the 𝜃𝑛 . 𝜃𝑛 is a measure of the obliquity of
the nth zone.
(i) The area of the first half period zone
Let us calculate the area of the half period zone. We have
𝜆2
𝜋(O𝑀1 )2 = 𝜋[ (𝑀1 𝑃)2 − (O𝑃)2 ] = 𝜋[ (p + λ/2 )2 − 𝑝2 ] = 𝜋 [ 𝑝𝜆 + ] ≃ 𝜋 𝑝𝜆
2

The radius OM1 of the first half period zone is √𝑝𝜆

The radius OM2 of the second half period zone is 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
1⁄
OM2 = [(𝑀2 𝑃)2 − (O𝑃)2 ] 2

1⁄
= [(p + λ )2 − 𝑝2 ] 2

= √2𝑝𝜆

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∴ The area of the second half period zone = 𝜋[ (𝑂𝑀2 )2 − (O𝑀1 )2 ]
= 𝜋 [ 2𝑝𝜆 − 𝑝𝜆 ]
= 𝜋 𝑝𝜆
The area of second half period is equal to 𝜋 𝑝𝜆, i.e., approximately same with first zone
Therefor, all zones have approximately equal area.
1⁄
In general, the radius of nth zone (𝑂𝑀𝑛 )2 = [(𝑀𝑛 𝑃)2 − (O𝑃)2 ] 2

= 𝜋[ (p + n λ/2 )2 − 𝑝2 ]
λ
= 𝜋[ 𝑝2 + (n λ/2 )2 +2pn 2 − 𝑝2 ]

𝑟𝑛 2 ≃ n p λ

𝒓𝒏 ≃ √𝒏 𝒑 𝝀

ii)Average distance of the zone

Let us consider the average distance of nth zone from P
𝑛𝜆 (𝑛−1)𝜆
( 𝑝+ 2 )+ {𝑝+ 2
}
=
2
𝜆
= 𝑝 + (2𝑛 − 1)
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iii)Obliquity factor f(𝜽𝒏 )
𝜃𝑛 angle between the normal to the zone and the line joining the zone to P.
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑧𝑜𝑛𝑒
The amplitude of nth zone ∝ 𝑓(𝜃𝑛 )
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡ℎ𝑒 𝑧𝑜𝑛𝑒
𝜆
𝜋 [ 𝑝+(2𝑛−1) 4 ]𝜆
∝ 𝜆 𝑓(𝜃𝑛 )
𝑝+(2𝑛−1)
4

∝ 𝜋 𝝀 f(𝜃𝑛 )
f(𝜃𝑛 ) = 1 for 𝜃𝑛 = 0 and f(𝜃𝑛 ) = 0 for 𝜃𝑛 = 900
As the order of the zone increases 𝑓(𝜃𝑛 ) decreases. Thus, the amplitude of wave from a zone at P
decreases as n increases.
3) Resultant amplitude
Let 𝑅1 , 𝑅2 , 𝑅3 , 𝑅4 , ……….(in figure ) represent the amplitudes of the waves at P due to
the secondary waves from the first, second, third ……….etc half period zones respectively.

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 As discussed above, the amplitude decreases, as 𝑛 increases, because obliquity, hence, 𝑅1 ,
𝑅2 , 𝑅3 …are of continuously decreasing order 𝑅1 is slightly grater than 𝑅2 , 𝑅2 is slightly greater than
𝑅3 and so on. As the amplitudes are gradually decreasing magnitude, the amplitude of any vibration at P
due to any zone can be approximately taken as the mean of the amplitudes due to the zones preceding and
𝑅1 +𝑅3
succeeding it. Hence, 𝑅2 = . The successive amplitudes are shown in reverse directions as there is
2

a phase difference of 𝜋 between any two consecutive zones.

The resultant amplitude at P at any instant is given by

𝑅 = 𝑅1 −𝑅2 + 𝑅3 − 𝑅4 … … … … . . 𝑅𝑛 if n is odd
𝑅 = 𝑅1 −𝑅2 + 𝑅3 − 𝑅4 … … … … . . −𝑅𝑛 if n is even
Considering n as odd
𝑅1 𝑅1 𝑅3 𝑅3 𝑅5
𝑅 = +[ −𝑅1 + ]+[ −𝑅4 + ] + ⋯ …. if n is odd
2 2 2 2 2

𝑅1 𝑅𝑛 𝑅1 +𝑅3
= + [∵ 𝑅2 = ]
2 2 2

Considering n as even
𝑅1 𝑅𝑛−1
R= + − 𝑅𝑛
2 2

As 𝑅 → ∞, 𝑅𝑛 and 𝑅𝑛−1 tend to zero as amplitudes are gradually diminishing. Therefore, the resultant
𝑅
amplitude at P due to whole wavefront = 21 .

𝑹𝟏 𝟐 𝑹𝟏 𝟐
The resultant intensity I ∝ ( ) ∝
𝟐 𝟒

Explanation of rectilinear propagation of light

Suppose a small circular opaque obstacle is placed at O which covers only the first half-period
zone of the wave. The resultant intensity at P due to the exposed wavefront is evidently proportional to
𝑹𝟏 𝟐
( ) . Similarly, if the size of the circular obstacle increases and successively covers the first two, first
𝟐

𝑹𝟑 𝟐
three, etc. half-period zones, the resultant intensity at P becomes respectively proportional to ( ) ,
𝟐

𝑹 𝟐
( 𝟒 ) , etc. Thus, the illumination at P gradually diminishes and ultimately becomes too small when the
𝟐
size of the obstacle is large enough to intercept an appreciable number of half-period zones. As the sizes
of these half-period zones are very small, a tiny obstacle is sufficient to cover a large number of half-
period zones by which the light from the source is practically cut off. This fact is interpreted as the
rectilinear propagation of light.

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***Zone Plate
The correctness of Fresnel's method of dividing a plane wavefront into half period zones, which
forms the basis of the proof of the rectilinear propagation of light, can be verified by means of an optical
device known as zone plate.
1. Construction of zone plate
To construct a zone plate, a large number of concentric circles with radii proportional to the
square root of natural numbers [√1, √2, √3, ]are drawn on a sheet of white paper. The alternate zones are
painted black.
A highly reduced size photograph of this pattern is taken on a thin glass plate. The resulting glass negative
is known as zone plate.

In the negative, the zones which were painted black in

the original drawing appear transparent while others
appear black. In this way two types of zone plates may
be constructed.

When the central zone is transparent, the zone plate is called positive and when the central zone is opaque
it is called negative. Both types of zone plates are shown in figure.
2. Action of a zone plate
Let XY represent the section of a zone plate perpendicular to the plane of the paper as shown in figure.
S is a point source of monochromatic light giving out spherical
waves of wavelength x. P is the position of the screen for a
bright image. OM₁,OM2 OM3 ….OMn (𝑟1 , 𝑟2 ,...... 𝑟𝑛 ) are the
radii of the first, second, ....nth half period zones. Let 𝑎 be the
distance of the source from the zone plate and 𝑏, the distance
of screen from the zone plate. The position of the screen in such
that there is an increasing path difference of λ/2 from one zone
to the next zone,
𝜆 𝜆
i.e., 𝑆𝑀1 + 𝑀1 𝑃 = 𝑆𝑂 + 𝑂𝑃 + 2 = 𝑎 + 𝑏 + 2
2𝜆 2𝜆
𝑆𝑀2 + 𝑀2 𝑃 = 𝑆𝑂 + 𝑂𝑃 + = 𝑎+𝑏+
2 2

…. …. …. ……. …… ……….. ………

…. …. …. ……. …… ……….. ……….
𝑛𝜆 𝑛𝜆
𝑆𝑀𝑛 + 𝑀𝑛 𝑃 = 𝑆𝑂 + 𝑂𝑃 + =𝑎+𝑏+ --------( 1 )
2 2

From the figure, (𝑆𝑀𝑛 )2 = (𝑆𝑂)2 + (𝑂𝑀𝑛 )2

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 (𝑆𝑀𝑛 )2 = 𝑎2 + 𝑟𝑛 2
1⁄
= [ 𝑎2 + 𝑟𝑛 2 ] 2

1⁄
𝑟𝑛 2 2
= 𝑎[ 1+ ]
𝑎2

𝑟𝑛 2
= 𝑎+ ------------------( 2 )
2𝑎

𝑟𝑛 2
Similarly, 𝑀𝑛 𝑃 = 𝑏 + ------------------( 3 )
2𝑏

Substituting these values in equ(1), we have

𝑟𝑛 2 𝑟𝑛 2 𝑛𝜆
𝑎+ +𝑏+ =𝑎+𝑏+
2𝑎 2𝑏 2
1 1
⇒ 𝑟𝑛2 (𝑎 + 𝑏) = 𝑛𝜆 -------------( 4 )
−1 1
Applying sign convention, 𝑟𝑛2 ( 𝑎 + 𝑏) = 𝑛𝜆
𝟏 𝟏 𝒏𝝀 1 𝑟𝑛 2
−𝒂=𝑟 2 = , where 𝑓𝑛 = ---------( 5 )
𝒃 𝑛 𝑓𝑛 𝑛𝜆

1 1 1
.Equation(5) is similar to a lens formula, −𝑢=
𝑣 𝑓

v = b as image distance, u = a as object distance 𝑓 = 𝑓𝑛 as the focal length of the lens.

𝒓𝒏 𝟐
Now, comparing the two, we have 𝒇𝒏 = ---------------( 6 )
𝒏𝝀

Eq(6) gives the focal length of zone plate. Thus, the zone plate behaves like a convergent lens.
Comparison between a zone plate and a convex lens
Similarities:
i) Both form real image of an object on the side other than the object. The distances of the object and
image are connected together by similar formulae in both the cases.
ii) Focal length for both depends upon λ and hence both shows chromatic aberration.
𝑟𝑛 2
For zone plate, 𝑓𝑛 = 𝑛𝜆
1 1 1
For convex lens, = ( μ − 1 )[𝑅 − 𝑅 ]
𝑓 1 2
Dissimilarities:
i) In case of convex lens, the rays are brought to focus by refraction while in case of a zone plate, the
image is formed by diffraction.
ii)In case of convex lens, all the rays reaching the image point have the same optical path while in a
zone plate, waves reaching the image point through two alternate zone differ in path by λ.

iii) The image due to a convex lens is more intense than due to a zone plate.

iv) For a convex lens 𝑓𝑣 < 𝑓𝑟 while for zone plate 𝑓𝑟 < 𝑓𝑣 .
v) A convex lens has only one focus on one side of it, while a zone plate has multiple foci at distances

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𝒓𝒏 𝟐 𝒓 𝟐 𝒓 𝟐 𝒓 𝟐
, 𝟑 𝒏𝒏𝝀 , 𝟓𝒏𝝀
𝒏 𝒏
, … … . . ( 𝟐𝒎+𝟏)𝒏𝝀 . The intensity diminishes as the focal length decreases.
𝒏𝝀

Sample questions
1) What is zone plate? Explain the working of zone plate with necessary theory.
2) Describe the Fresnel’s half period zones with necessary theory and determine the
area of half period zones.
3) What is diffraction of light? Explain the Fraunhofer diffraction due to a single slit.
4) Differentiate between Fresnel and Fraunhofer diffractions. (5Marks)
5) Define zone plate and compare with a convex lens. (5 Marks)

-----------o O o -----------

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