1. Diffraction
When monochromatic light from a distance
source (or a laser ) passes through a narrow slit
and then intercepted by a viewing screen ,the
light produces on the screen a diffraction pattern
like that in figure this pattern consists of abroad
and intense (very bright )central maximum and
A number of
Narrower and less
Intense maxima
(called secondary or side maxima) to both side
In between the maxima are minima

2. Diffraction by a Single Slit
1 . Let us consider how
plane waves of light of
wavelength are diffracted
by a single long narrow
slit of width a in an
otherwise opaque
screen B, as shown in
cross section in figure (a).

3. .We can justify the central bright fringe seen in
figure by noting that the waves from all points in
the slit travel about the same distance to reach
the center of the pattern and thus are in phase
there. As for the other bright fringes, we can say
only that they are approximately halfway
between adjacent dark fringes.
3 , to locate the first dark fringe at point p1 , we
first mentally divide the slit into two zones of
equal widths a/2 . Then we extend to p1 a light
ray r1 ,from the top point of top zone and a

4. light ray r 1 from the
top point of the bottom
zone. A central axis is
drawn from the center of
the slit to screen C, and
P 1 is located at an angle
θto that axis. The first
dark fringe can be located at
It means

5. A slit of width a is illuminate by white light ,For
SAMPLE PROBLEM 1
A slit of width a is illuminate by white light ,For
What value of a will the first minimum for red
Light of λ=650 nm be at θ=15 ° ? C
Incident wave
（a）

6. SOLUTION : at the first minimum ,m=1 in Eq.
solving for a ,
wefind
(answer)
For the incident light to flare out that much ( )
The slit has to be very fine indeed ,amounting to
About four times the wavelength .Note that a
fine Human hair may be about 100 µm in
diameter

7. 4 .To find the second dark fringes above
and below the central axis, we divide the slit
into four zones of equal widths , a/4 as
shown in above figure (a). We then extend rays
r1 , r 2, r 3, and r 4 from the top points of the
zones to point P2, the location of the second
dark fringe above the Central axis .the second
dark fringe can then be located at

8. We then extend rays r 1 , r 2, r 3, and r 4
from the top points of the zones to point P2,
the location of the second dark fringe above the
Central axis .the second dark fringe can then be
located at or
5. The dark fringes can be located with the
following general equation :
For m=1,2,3,….

9. Above equation is derived
for the case of .D>>a
However, it also apply if
we place a converging
lens between the slit and
the viewing screen and
then move the screen in
so that it coincides with
the focal plane of the
lens. Intensity in single
-Slit diffraction : we can

10. Prove the expression for the intensity I of the
pattern as ,where
And Im , is the greatest value of the intensity in
the pattern ,and it occurs at the central maximum
Where (θ=0) Figure shows plot of the intensity of
A single-slit diffraction pattern for three different
Slit width .

11. Diffraction from a Circular Aperture
1 . Here we consider diffraction by a circular
aperture, through which
light can pass. Figure
shows the image of a
distant point source of light
formed on photographic
film placed in the focal
plane of a converging lens
(1) The image is not
a point But a circular disk surrounded

12. by several progressively fainter secondary rings.
(2) The first minimum for the diffraction pattern
of a circular aperture of diameter d is given by .

13. 2 .Resolvability: In figure (b) the angular
separation of the two point sources is such that
the central maximum of the diffraction patter
of one source is centered on the firstminimum
of the diffraction pattern of the other, a
condition called Rayleigh’s criterion for
resolvability. Thus two objects that are barely
resolvable by this criterion must have an angular
separation ,

14. SAMPLE PROBLEM 2
A converging lens ,32 mm in diameter and with a
Length f , of 24 cm ,is used to form images of
Objects . Considering the diffraction by the
Lens ,what angular separation must two distant
Point objects have to satisfy Rayleigh’s criterion ?
Assume that the wavelength of the light from the
Distant object is λ=550 nm

15. .
SOLUTION :From Eq we have
Of small angular separation ,it is desirable .these
When we wish to use a lens to resolve objects
These can be done either by increasing the
lens Diameter or by using light of a shorter
Wavelength .

16. Diffraction by a double slit
In double-slit experiment ,we implicitly
Assumed that the slits were narrow compared to
The wavelength of the light illuminating them ;that
Is ,a<<λ. For such narrow slits ,the central
Minimum of the diffraction pattern of either slits
Covers the entire viewing screen ,Moreover ,the
interference of light from the two slits produces
bright fringes that all have approximately the
same intensity.

17. 2. In practice with visible light, however, the
condition a<<λ is often not met. For relatively
wide slits, the interference of light from two slits
produces bright fringes that do not have the same
intensity. In fact, their intensity is modified by
the

18. diffraction of the light through each slit. See
the figures.

19. 3. With diffraction effects taken into account, the
intensity of a double-slit interference pattern is
given by , in which
and

20. Diffraction Grating 1 . One of the most useful tools in the study of
Light and of objects that emit and absorb light is
The diffraction grating .some what like the
double-slit arrangement ,these devise has a
much Greater number N of slits ,often called
rulings, perhaps as many as several thousand
Per millimeter

21. 2 . With monochromatic light on a diffraction
Grating if we gradually increase the number of
Slits From two to a large number N, The maxima
Are now narrow
(and so called lines )
They are
Separated by relatively
wide dark regions ,as
Shown in above figure

22. Look at the figure
An idealized
Diffraction grating
Containing five
Slits ,The scale is
Distorted for
Clarity

23. . The separation between rulings is called the
grating spacing. The maxima-lines are located
at , in which the
integers are then called the order numbers, and
the lines are called the zeroth-order line (the
central line, with m=0), the first-order line, the
second-order line, and so on.
4 . Width of the lines: (1) we measure the half-
width of the central line as the angle ΔθKλ
from the center of the line at θ=0 outward to
Where the line effectively ends and darkness,

24. we measure the half-width of the central line
as the angle ΔθKλ from the center of the line at
θ=0 outward to Where the line effectively ends
and darkness, Effectively begins with the first
minimum .
(2) the Half-width of the central line is given as(3)
(3) The half-width any other line is given as :

25. (4) An application of diffraction grating :Figure
Shows a simple grating spectroscope in which a
Grating to determine the wavelength

26. X-Ray Diffraction 1 .X rays are electromagnetic radiation whose
wavelengths are of the order of 1Ǻ . Figure
shows that x rays are produced when electrons
escaping from a heated filament F are
accelerated
by a potential
difference V and strike
a metal target T.

27. 2 . Bragg’s law: .
Figure shows how the inter
planer spacing d can
be related to the unit
cell dimension :a0

28. Questions
Home work
Exercise and problems :
3E P
16P 39P
43E 66E
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