Chapter Seven
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2-
Physical
Optics
Chapter Seven (Diffraction)
1-7 Diffraction
When light passes through a single slit whose width is comparable with or less
than the wavelength of the light, it spreads out after passing through the slit.
The bending of waves around the edges of an aperture is called diffraction.
However, contrary to what one might expect, if the emerging beam falls on a
screen, instead of a getting a broad image of the slit, something more interesting
occurs. A series of light and dark bands, similar to an interference pattern, is
obtained. The pattern obtained is called a diffraction pattern. It consists of an intense
central band (the central maximum) flanked by a series of narrower, less intense
bands (called secondary maxima) as shown in figure (1).
Figure (1): diffraction pattern
Figure (2) illustrates the passage of plane wave with wavelength λ through an
aperture of diameter b, we notes the followings:
a) When the aperture is large compared to the wavelength (λ<< b), the waves do not
bend round the edges and continue in a straight-line path (figure 2-a).
b) When the aperture is small (λ ≈ b), the bending round the edges is noticeable rays
and waves spread out after passing through the aperture (figure 2-b).
c) When the aperture is very small (λ >> b), the waves spread over the surface
behind the aperture and the aperture behaves as a point source emitting spherical
waves (figure 2-c).
The aperture acts as an independent source of waves which propagate in all
directions. The diffraction effect is observable quite close to the aperture when the
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size of the aperture is very small. When the aperture is large, diffraction effect is
observed at greater distances from the aperture.
In general diffraction of waves becomes noticeable only when the aperture
width is nearly the same size or smaller than the wavelength
b
b
b
b
(a
(b
(c
Figure (2): A plane waves of wavelength λ passing through
a small opening of diameter b
Differences between Interference and Diffraction
The main differences between interference and diffraction are as follows:
Interference
Diffraction
1. Interference is the result of
1. Diffraction is the result of interaction of
interaction of light coming from
light coming from different parts of the
different wave fronts originating
same wave front.
from the source.
2. Interference fringes may or may 2. Diffraction fringes are not of the same
not be of the same width.
width.
3. Regions of minimum intensity are 3. Regions of minimum intensity are not
perfectly dark.
perfectly dark.
4.The intensity of all positions of maxima
4. All bright bands are of same
are vary
intensity.
.
5- in interference ,we must have two 5- in diffraction, single slit is enough to
slits to produce fringes
produce fringe
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2-7 Fresnel and Fraunhofer Diffraction
The diffraction phenomena are broadly classified into two types: Fresnel
diffraction and Fraunhoffer diffraction.
 Fresnel diffraction:
In this type of diffraction,
1 The source of light and the screen are at finite distances from the aperture (or
slit) (Figure 3-a).
2 Observation of Fresnel diffraction phenomenon does not require any lenses.
3 The incident wave front is curved (spherical).
4 The phase of secondary wavelengths is not the same at all points in the plane
of the slit.
5 The resultant amplitude at any point of the screen is obtained by the mutual
interference of secondary wavelengths from different elements of unblocked
portions of wave front.
6 This type is more complex to treat mathematically because the wave fronts
here are divergent instead of plane.
 Fraunhofer diffraction:
In this type of diffraction,
1 The source of light and the screen are at infinite distances from the slit.
2 The conditions required for Fraunhofer diffraction are achieved using two
convex lenses, one to make the light from the source parallel and the other to
focus the light after diffraction on to the screen (Figure 3-b).
3 The diffraction is thus produced by the interference between parallel rays.
4 The incident wave front as such is plane wave and the secondary wavelets,
which originate from the unblocked portions of the wave front, are in the same
phase at every point in the plane of the slit.
5 This type is simple to treat mathematically because the rays are parallel.
L1
L2
Figure (3): Conditions for (a) Fresnel diffraction and (b) Fraunhofer diffraction
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3-7 Diffraction by a Single Silt
Fraunhofer diffraction pattern can be easily observed in practice and its simple
to treat mathematically therefore we will deal with this type of diffraction.
Fraunhofer experimental arrangement for obtaining the diffraction pattern of a single
slit is shown in figure (4).
Figure (4): Experimental arrangement for obtaining the diffraction pattern of a
single slit; Fraunhofer diffraction
A slit is a rectangular aperture of length large compared to its width (b) and
perpendicular to the plane of the page, and illuminated by parallel monochromatic
light from the source, at the principal focus of the lens L1. The light focused by
another lens L2 on a screen or photographic plate P at its principal focus will form a
diffraction pattern, as indicated schematically. It is important to notice that the width
of the central maximum is twice as great as that of the fainter side maxima. That this
effect comes under the heading of diffraction.
Now we have the geometrical construction for investigating the intensity in the
single-slit diffraction pattern as shown in figure (5).
r1
r2
b
(b)
(a)
Figure (5): Geometrical construction for single-slit diffraction
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Let ds be an element of width of the wave front in the plane of the slit, at a
distance s from the center O. The parts of each secondary wave which travel normal
to the plane of the slit will be focused at Po, while those which travel at any angle 𝜃
will reach P. Considering first the wavelet emitted by the element ds situated at the
origin, its amplitude will be directly proportional to the length ds and inversely
proportional to the distance r1.
According to Huygens’s principle, each portion of the slit acts as a source of
light waves. Hence, light from one portion of the slit can interfere with light from
another portion, and the resultant light intensity on a viewing screen depends on the
direction . Based on this analysis, we recognize that a diffraction pattern is actually
an interference pattern, in which the different sources of light are different portions
of the single slit.
We then look at two sources of Huygen’s wavelets located at the top and the
middle of the slit, separated by distance b/2 [see figure (6)]. We will analyze these
two wavelets, but our analysis will be the same for any two wavelets from slits
separated by the same distance.
𝒃
The path difference between rays 1 and 3 or between rays 2 and 4 is [ 𝟐𝒔𝒊𝒏𝜽]
Therefore, waves from the upper half of the slit interfere destructively with waves
from the lower half.
b/2
b
b/2
𝑏
𝑠𝑖𝑛𝜃
2
Figure (6): Paths of light rays that encounter a narrow slit of width b
Therefore we have:
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𝑏
2
𝑠𝑖𝑛𝜃 =
𝜆
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……………………………………………..…………(1)
2
When 𝑠𝑖𝑛𝜃 =
𝜆
𝑏
………………………………..……………..(2)
Conditions of obtaining bright fringes and dark fringes:
b sinθ = m λ
b sinθ=(m+½)λ
the condition for obtaining a dark fringe ….…(3)
the condition for obtaining a bright fringe ……(4)
Where 𝑚 = ±1, ±2, … ….
where b :is the slit width
Broad central bright fringe is observed; this fringe is flanked by much weaker
bright fringes alternating with dark fringes. The positions of two minima on each
side of the central maximum are labeled as shown in figure (7).
b
b
b
b
b
D
Figure (7): Intensity distribution for a Fraunhofer diffraction pattern from b
From figure (8) we see that the path difference is just the slit spacing times the
sin of the angle:
(r1 − r2 ) = b sinθ …………………………….………………(5)
2
And, we use the same small angle approximations as before:
𝑡𝑎𝑛𝜃 ≈ 𝑠𝑖𝑛𝜃 =
𝑦
𝐷
………………………………………………(6)
the phase difference is given by:
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δ=
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2π b y
2π
(r1 − r2 ) =
λ
λ 2D
….……………………………….……(7)
The condition for a dark band is that the two sources are out of phase, so:
δ=
2π
(r1 − r2 ) = 2π b y =
λ
λ 2D
π ….…………………………………(8)
Solving for y, we get:
𝑦𝑚𝑖𝑛 =
𝐷𝜆
𝑏
So in general,
𝒚𝒎𝒊𝒏 =
𝒎𝑫𝝀
𝒃
, 𝒎 = ±𝟏, ±𝟐, ±𝟑, … …
…………………(9)
(a)
b
D
Po
b/2
b/2
b
θ′1
θ1 θ1
θ1
θ1 θ1
ym =mλD/b
b
Figure (8): Paths of light rays that encounter a narrow slit of width b
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Intensity of Single-Slit Diffraction Patterns
The light intensity at point P [in figure (5)] is the resultant of all the incremental
electric field magnitudes from zones of width ds. We may represent its amplitude
as:
𝐴 = 𝐴𝑜
𝛽=
𝑠𝑖𝑛𝛽
……………………………………………….….(10)
𝛽
(𝜋𝑏 𝑠𝑖𝑛𝜃)
1
𝑘𝑏 𝑠𝑖𝑛𝜃 =
2
𝜆
And 𝐴𝑜 =
𝑎𝑏
….……………………..….…….(11)
…………………………………….………(12)
𝑟
The quantity β is a convenient variable, which signifies one-half the phase
difference between the contributions coming from opposite edges of the slit. The
intensity on the screen is then:
𝐼 ≈ 𝐴2 = 𝐴2𝑜
𝑠𝑖𝑛 2 𝛽
𝛽2
………………………………..…………..(13)
If the light makes an angle i, then it is necessary to replace the above expression for
β by the more general expression:
𝛽 = 𝜋𝑏(𝑠𝑖𝑛𝑖+𝑠𝑖𝑛𝜃) ……………………………………..………(14)
𝜆
The minima intensity occur when β is a multiple of π, i.e. at
𝛽 = 𝜋 , 2𝜋 , 3𝜋 , 4𝜋, … . . = 𝑚𝜋
m=1, 2, 3, …….
𝛽=
1
2
𝑚𝜋 =
𝑘𝑏 𝑠𝑖𝑛𝜃
1 2𝜋
( ) 𝑏 𝑠𝑖𝑛𝜃
2 𝜆
∴bsinθ =m λ ,
(𝑚 = ±1, ±2, ±3, … . )
Width of Central Peak
With the relative dimensions shown in figure (8-b), the geometrical shadow of the
slit would considerably widen the central maximum as drawn. Just as was true with
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Young's experiment, when the screen is at infinity, the relations become simpler.
Then the two angles 𝜃1 and 𝜃1′ in Fig.(8-b) become exactly equal, i.e., the two lines
are perpendicular to each other, and λ = b sin θ1 for the first minimum corresponding
to β = π.
Then the positions of the first minima are given by:
sinθ1 =
mλ
, m = ±1
b
In practice θ1 is usually very small angle then we have (sinθ1 = θ1), Then:
𝛌
……………………….…………………………..(15)
𝛉𝟏 = 𝐛
‫يزكرمال ةبدهلل يوازال ضرعال فصن‬: Ө
This relation shows that the dimensions of the pattern vary with λ and b.
The width of the central peak is
λ
−λ
b
b
θ1 − θ−1 = −
2λ
= b ………………………………………..(16)
The linear width of the pattern on a screen will be proportional to the slit-screen
distance, which is the focal length f of a lens placed close to the slit. The linear
distance d between successive minima corresponding to the angular separation θ1 =
λ / b is thus:
𝒅 = 𝒇𝝀
……………………………………………………….(17)
𝒃
‫ةبدهلل يطً خال ضرعال فصن‬: d
The angular and linear width of the fringe is shown in figure (9)
b
Figure (9): Angular and linear width of the fringe for diffraction pattern
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Also, as a approaches the wavelength, there are no minima, and the diffraction
pattern is just one broad peak [see figure (10)].
b =λ
b =5λ
b =8λ
Figure (10): Variation of diffraction pattern by the variation of the slits width
The narrower the slit, the broader the diffraction pattern
The width of the pattern increases in proportion to the wavelength, so that for red
light it is roughly twice as wide as for violet light, the slit width, etc., being the same. If
white light is used, the central maximum is white in the middle but reddish on its outer
edge, shading into a purple and other impure colors farther out. The angular width of the
pattern for a given wavelength is inversely proportional to the slit width b, so that as
b is made larger, the pattern shrinks rapidly to a smaller scale.
Example (1): Light of wavelength 580 nm is incident on a slit having a width of
0.300 mm. The viewing screen is 2.00 m from the slit. Find the
positions of the first dark fringes and the width of the central bright
fringe.
Solution:
𝜆
5.80 × 10−7𝑚
𝑠𝑖𝑛𝜃 = ± = ±
= ±1.93 × 10−3
𝑏
0.300 × 10−3𝑚
𝑦1
𝑠𝑖𝑛𝜃 ≈
𝐷
𝜆
𝑦1 ≈ 𝐷𝑠𝑖𝑛𝜃 = ±𝐷 = 2 × (±1.93 × 10−3) = ±3.87 × 10−3𝑚
𝑏
The positive and negative signs correspond to the dark fringes on either side of
the central bright fringe.
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Hence, the width of the central bright fringe is equal to
2|𝑦1| = 7.74 × 10−3𝑚 = 7.74 𝑚𝑚
Example (2): You pass 633-nm light through a narrow slit and observe the
diffraction pattern on a screen 6.0 m away. The distance at the screen
between the center and the first minima on either side is 32 mm long.
How wide is the slit?
Solution:
We have
𝑦𝑚𝑖𝑛 =
=
m=1, D= 6m , λ=633nm , b= 32
𝑚𝐷𝜆
𝑏
(6000𝑚𝑚)(0.000633𝑚𝑚)
32𝑚𝑚
D
= 0.24 𝑚𝑚
Example (3) : Light from a red laser passes through a single slit to form a diffraction
pattern on a distant screen. If the width of the slit is increased by a
factor of two, what happens to the width of the central maximum on
the screen?
Solution:
The central maximum occurs between θ = 0 and θ as determined by the location
of the 1st minimum in the diffraction pattern:
b sinθ = m λ
Let 𝑚 = ∓1 and assume that θ is small.
𝜃 ≈
𝜆
𝑏
From the previous picture, θ only determines the half-width of the maximum. If a is
doubled, the width of the maximum is halved.
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4-7 Diffraction Grating
A diffraction grating is an array of a large number of slits having the same
width and equal spacing. We symbolize the number of slits by N each of width (a)
and separated from the next by a distance d as shown in figure (11).
Figure (11): Diffraction grating
The condition for the principal maxima is:
dsinθ =m λ
Where m= 0, 1, 2, 3,........
m is the order number of the diffraction pattern.
The mth order of, the diffraction pattern is:
R = Nm
Example (4): The wavelengths of the visible spectrum are approximately 380 nm
(violet) to 750 nm (red). (a) Find the angular limits of the first-order
visible spectrum produced by a plane grating with 600 slits per
millimeter when white light falls normally on the grating. (b) Do the
first order and second order spectra overlap? What about the 2nd
and 3rd orders?
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Solution:
(a) distance between slits is
1𝑚𝑚
𝑑=
= 1.67 × 10−6𝑚
600 𝑠𝑙𝑖𝑡𝑠
Violet light for 1st order occurs at:
𝜃 = 𝑠𝑖𝑛−1(𝜆/𝑑) = 𝑠𝑖𝑛−1(3.8 × 10−7 /1.67 × 10−6) = 13.2𝑜
Red light for 1storder occurs at:
𝜃 = 𝑠𝑖𝑛−1(𝜆/𝑑) = 𝑠𝑖𝑛−1(7.5 × 10−7 /1.67 × 10−6) = 26.7𝑜
(b)recalculate for m = 2 and m = 3
The 2nd-order spectrum extends from 27.1- 63.9° while the 3rd order is from
43-90.
5-7 Rectangular Aperture
In the preceding sections the intensity function for a slit was derived by summing
the effects of the spherical wavelets originating from a linear section of the wave
front by a plane perpendicular to the length of the slit. Nothing was said about the
contributions from parts of the wave front out of this plane. A more thorough
mathematical investigation, involving a double integration over both dimensions of
the wave front, shows, however, that the above result is correct when the slit is very
long compared to its width. The complete treatment gives, for a slit of width b and
length l, the following expression for the intensity:
2
2
𝐼 ≈ 𝑏2𝑙2 𝑠𝑖𝑛𝛽 2 𝛽 𝑠𝑖𝑛𝛾2 𝛾
Where 𝛽 =
(𝜋𝑏 𝑠𝑖𝑛𝜃)
𝜆
………………………....……….(18)
as before, and 𝛾 =
(𝜋/𝑠𝑖𝑛𝞨)
𝜆
…… …..(19)
The angles θ and Ω are measured from the normal to the aperture at its center,
in planes through the normal parallel to the sides b and l, respectively.
2
𝑠𝑖𝑛 𝛾
Now for a slit having l very large, the factor ( 𝛾2 ) in Eq.(18) is zero.
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6-7 Resolving power
Resolving power: its ability of an optical instrument to produce separate images of
objects very close together. Using the laws of geometrical optics, one designs a
telescope or a microscope to give an image of a point source which is as small as
possible.
‫(اهناب روشومالو بوسوركيمالو بوكسلتالك يرصب زاهج يال‬resolving power ) ‫ ٌلحتال ةردق فرعت‬. ‫اهضعب نم ادج‬
‫ةبيرقال ماسجالل ةدرفنمو ةحضاو ةروص نيوكت ىلع زاهجال كلذ ةيٌلباق‬
1-6-7 Resolving power with a Rectangular Aperture
Figure (12) shows two plano-convex lenses (equivalent to a single double
convex lens) limited by a rectangular aperture of vertical dimension b. Two narrow
slit sources S1 and S2 perpendicular to the plane of the figure form real images S'1
and S'2 on a screen.
Figure (12): Diffraction images of two slit sources formed by a rectangular aperture
Each image consists of a single-slit diffraction pattern for which the intensity
distribution is plotted in a vertical direction. The angular separation α of the central
maxima is equal to the angular separation of the sources, and with the value shown
in the figure is adequate to give separate images. The condition illustrated is that in
which each principal maximum falls exactly on the second minimum of the adjacent
pattern. This is the smallest possible value of α which will give zero intensity
between the two strong maxima in the resultant pattern. The angular separation from
the center to the second minimum in either pattern then corresponds to β = 2π or,
𝑠𝑖𝑛𝜃 ≈ 𝜃 =
2𝜆
= 2𝜃1 ………………….…………..(20)
𝑏
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As α made smaller than this, and the two images move closer together, the intensity
between the maxima will rise, until finally no minimum remains at the center.
If the two sources are far enough apart to keep their central maxima from
overlapping, as shown in Figure (13-a), their images can be distinguished and are
said to be resolved. If the sources are close together, however, as shown in Figure
(13-d), the two central maxima overlap, and the images are not resolved. In
determining whether two images are resolved, the following condition is often used:
Figure (13) showing the resultant curve (heavy line) for four different values of α.
In each case the resultant pattern has been obtained by merely adding the intensities
due to the separate patterns (dotted and light curves.
Figure (13): Diffraction images of two slit sources (a) and (b) well resolved; (c)just
resolved; (d) not resolved.
When the central maximum of one image falls on the first minimum of the other
image, the images are said to be just resolved. This limiting condition of resolution
is known as Rayleigh’s criterion. Which is given by:
𝜆
𝛼 = 𝜃1 = 𝑏
The angle θ1 is sometimes called the resolving power of the aperture b . The ability
to resolve increases as θ1 becomes smaller.
ٌ ‫وهةلي‬
: ‫ط ٌ تسمةحتفل يليار طرش بسح راسمال قرف‬
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𝛽=
𝜋=
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1
𝑘𝑏 𝑠𝑖𝑛𝜃
2
and
𝛽 = 𝜋 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑤𝑎𝑣𝑒 𝑛𝑜.
𝑘 = 2𝜋
𝜆
1 2𝜋
( ) 𝑏 𝑠𝑖𝑛𝜃
2 𝜆
∴bsinθ = λ ,
………………(21)
path difference for a rectangular aperture
At small angle sinθ=dθ , since dθ is limit of resolution
∴ 𝑏𝑑𝜃 = 𝜆 ………………………………………………………(22)
According to Rayleigh’s criterion, this expression gives the smallest angular
separation for which the two images are resolved. Because in most situations,
sin θ = θ is small, and we can use the approximation . Therefore, the limiting angle
of resolution for a slit of width a is:
𝜽𝒎𝒊𝒏 =
𝝀
………………………………….…………….…….(23)
𝒃
𝜃𝑚𝑖𝑛 : is the minimum angle of resolution expressed in radians
Example (5): Light with a wavelength of 511 nm forms a diffraction pattern after
passing through a single slit of width 2.20×10-6m. Find the angle
associated with (a) the first and (b) the second dark fringe above the
central bright fringe.
Solution:
(a) First Dark Fringe, m=1
Since b sinθ=mλ
2.2×10-6 sinθ = (1) (511×10-9)
θ=13.40
(c)
Second Dark Fringe, m=2
b sinθ = mλ
2.2×10-6 sinθ= (2)(511×10-9)
θ =27.70
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2-6-7 Chromatic resolving power of a prism
The chromatic resolving power of a prism is invariably stated for the case in
which parallel rays of light are incident on the prism, in which the prism is oriented
at the angle of minimum deviation at wavelength , and in which the complete
height of the prism is utilized. The corresponding resolving power R1 deduced on
the basis of Rayleigh's criterion is:
R = / = bdn/
……………………………………(24)
where n: is the refractive index of the prism for the wavelength 
b: is the maximum thickness of the prism traversed by light rays.
dn/d : called the dispersion of the prism.
d: base length of the prism.
3-6-7 Resolving Power a Circular Aperture
When light from a point source passes through a small circular aperture, it does
not produce a bright dot as an image, but rather a diffuse circular disc known as
Airy's disc surrounded by much fainter concentric circular rings as shown in figure
(14).
Figure (14): Diffraction images circular aperture .
This example of diffraction is of great importance because the eye and many
optical instruments have circular apertures. If this smearing of the image of the point
source is larger that produced by the aberrations of the system, the imaging process
is said to be diffraction-limited, and that is the best that can be done with that size
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aperture. This limitation on the resolution of images is quantified in terms of the
Rayleigh criterion so that the limiting resolution of a system can be calculated.
The intensity distribution is very much the same as that which obtained by the
single-slit pattern. the dimensions of the pattern are ,however ,appreciably different
from those in single-slit pattern for a slit of width equal to the diameter of the
circular aperture.
For the single-slit pattern, the angular separation of the minima from the center :
Sinθ≈ 𝜃= 𝑚𝜆/ …………………………………………………………(25)
where b=diameter of the aperture
For a circular aperture and a small angle the formula for resolving power is given
by:
θR = 1.2λ/b ,
path difference for a circular aperture
……..(26)
4-6-7 Resolving Power of a Telescope
A telescope is an instrument used to view objects in remote areas. The angular
resolution power of a telescope is:
θ = 1.220 𝜆 /𝐷
…………………………..…………………..(27)
Where: λ: is the wavelength of light.
D: is the diameter of circular aperture which limits the beam forming
the primary image,
Example (5): Calculate the minimum angle of resolution of a telescope if the
diameter of the objective lens is 4cm and its focal length is 30cm,and
the light wavelength is 5.6×10-6cm .
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Solution:
θ =1.22𝜆/𝐷
=1.22 ×5.6×10-6/4
=1.71×10-5 rad
5-6-7 Resolving Power of a Microscope
The resolution R (here measured as a distance):
𝑅=
1.22𝜆
………………….………………….(28)
𝑁 𝐴 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 +𝑁𝐴 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒
Where:
𝑁𝐴 = 𝜂 𝑠𝑖𝑛𝜃
Here NA: is the numerical aperture.
θ : is half the included angle of the lens, which depends on the diameter of the
lens and its focal length.
η : is the refractive index of the medium between the lens and the specimen.
λ : is the wavelength of light .
7-7 The Double Slit.
When we studied interference in Young’s double-slit experiment, we ignored the
Diffraction effect in each slit. We assumed that the slits were so narrow that on the
screen you saw only the interference of light from just two point sources.
The diffraction pattern of two slits of width b that are separated by a distance d is
the interference pattern of two point sources separated by d multiplied by
the diffraction pattern of a slit of width b as shown in figure (15).
In other words, the locations of the interference fringes are given by the equation
dsinθ=mλ
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D
Figure (15): Diffraction pattern by Double Slit
Figure (16) shows the Path differences of parallel rays leaving a double slit.
Figure (15): Path differences between lights in Diffraction pattern by Double Slit
The light intensity in the double-slit interference pattern is:
𝐼 = 4𝐴2𝑜
𝑠𝑖𝑛 2 𝛽
𝑐𝑜𝑠 2 𝛾
𝛽2
…………………………………………(29)
The factor 𝑐𝑜𝑠2𝛾 : is the characteristic of the interference pattern produced by two
beams of equal intensity.
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𝑎𝑏
1
𝑘𝑏 𝑠𝑖𝑛𝜃 = (𝜋𝑏𝑠𝑖𝑛𝜃)/𝜆 and 𝐴 𝑜 =
2
𝑟
Where 𝛽 =
the quantity β is a convenient variable, which signies one-half the phase difference
between the contributions coming from opposite edges of the slit.
𝜋 3𝜋 5𝜋
The intensity will be zero wherever 𝛾 = 2 , 2 , 2 , ….. and also when 𝛽 =
𝜋, 2𝜋, 3𝜋, …. . The first of these two sets are the minima for the interference pattern,
𝜋
and since by definition = ( ) 𝑑𝑠𝑖𝑛𝜃 , they occur at angles θ such that:
𝜆
𝑑𝑠𝑖𝑛𝜃 =
𝜆 3𝜆 5𝜆
2
,
2
,
2
1
, … . . = (𝑚 + ) 𝜆
2
𝑚𝑖𝑛𝑖𝑚𝑎 …………….(30)
m= 0,1,2,3,…..
The second series of minima are those for the diffraction pattern, and these,
𝜋
since 𝛽 = ( ) 𝑑𝑠𝑖𝑛𝜃, occur where
𝜆
𝑚𝑖𝑛𝑖𝑚𝑎 ………………………(31)
𝑏𝑠𝑖𝑛𝜃 = 𝜆, 2𝜆, 3𝜆 , … . . = 𝑃𝜆
the smallest value of p being 1
The positions of the maxima will then be determined solely by the 𝑐𝑜𝑠2 𝜸
factor, which has maxima for = 𝑜, 𝜋, 2𝜋, … .. , that is, for:
𝑚𝑎𝑥𝑖𝑚𝑎 ………………...(32)
𝑑𝑠𝑖𝑛𝜃 = 0, 𝜆, 2𝜆, 3𝜆 , … . . = 𝑚𝜆
Example (6): Suppose that in Young’s experiment, slits of width 0.020 mm are
separated by 0.20 mm. If the slits are illuminated by monochromatic
light of wavelength 500 nm, how many bright fringes are observed in
the central peak of the diffraction pattern?
Solution
The angular position of the first diffraction minimum is:
−7𝑚
𝜃 ≈ 𝑠𝑖𝑛𝜃 = 𝜆 = 5.0×10
𝑏
2.0×10−5𝑚
= 2.5 × 10 −2 𝑟𝑎𝑑
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Using dsinθ=mλ
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for θ=2.5×10−2rad, we find
𝑑𝑠𝑖𝑛𝜃 (0.20 𝑚𝑚)(2.5 × 10−2𝑟𝑎𝑑)
𝑚=
=
= 10
𝜆
5.0 × 10−7𝑚
8-7 Comparison of the single slit and double-slit patterns
Single slit diffraction pattern
Double slit diffraction pattern
`1-The single-slit diffraction pattern
1-The double-slit or multiple-slit
is due to interference between the
interference pattern results from
light passing through one half of
interference between light passing
the slit width vs. light passing
through the separated slits.
through the other half.
2-In the double slit ,the maxima
2- Intensities in single slit pattern are
appear on either side but the
not constant but decreases to zero
intensity is too weak to observe then
on either side of the central maxima
after two or three maxima
3-Intensity pattern of double slit
3-Intensity pattern in single slit is
pattern is four times that of single
quite less
slit pattern
4-the spacing of single slit diffraction 4-spacing of double slit depend on a
pattern depends on the width of the
and b, where b is width of opacity
slit
5-it consists of diffraction pattern of
equally spaced interference maxima
5-it consists of bright maxima and
and minima within the central
minima of gradual lower intensity
maxima
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Example (7): Light of wavelength 589 nm is used to view an object under a
microscope. If the aperture of the objective has a diameter of 0.900
cm, (a) what is the limiting angle of resolution?
Solution:
a)
b) If it were possible to use visible light of any wavelength, what would be
the maximum limit of resolution for this microscope?
To obtain the smallest limiting angle, we have to use the shortest wavelength
available in the visible spectrum. Violet light (400 nm) gives a limiting angle
of resolution of
(c) Suppose that water fills the space between the object and the objective.
What effect does this have on resolving power when 589-nm light is used
Solution:
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Example (8): Two bright lines in the spectrum of sodium have wavelengths of
589.00 nm and 589.59 nm, respectively.
(a) What must the resolving power of a grating be so as to distinguish
these wavelengths? (b) To resolve these lines in the second-order
spectrum, how many lines of the grating must be illuminated?
Solution:
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