C.K.Cheung
Diffraction:
The result of superposing ( light ) waves from coherent sources on the same portion
of wavefront after the wavefront has been distorted by some obstacle (e.g. slits, edges
etc.,)
Note:
1/
2/
Effect obvious if obstacles' dimension (size ) :  ~ 100 
frequency of wave not changed.
Types of diffraction
1/
Fresnel diffraction
2/
Fraunhofer diffraction
1
C.K.Cheung
Fraunhofer Diffraction:
adjusable slit
screen
Initially, the slit is opened wide  no diffraction effect. As the slit is narrowed (~100
to ~  ), a pattern is obtained as shown.
Note:
The central maximun is twice in width.
2
C.K.Cheung
Theory:
P'
A
a
2
a/2
a
2
a/2
plane waves
D

P
C
E

B
slit width
=a
The slit is imagined to consist of strips of equal width, parallel to the length of the slit.
For first minimum:
CP' - AP' =

2
and for all pairs of corresponding slits in AC and CB (i.e. A' & C' ) because the same
path difference
 CD =


or
2
sin  =

exists.
2
a

( sin   )
2
2

a
(1st min.)
3
C.K.Cheung
For second minimum:
By similiar 'pairing' process:
cancel in pairs
a
4
cancel in pairs
a
'
Since,

a

sin  ' 
4
2
sin ’ =
2
a
Hence, in general,minimum positions are:
sin  =
n
a
n = 1, 2, 3,
4
etc.,
C.K.Cheung
Note
1/
If a >>  (> 103  ) , sin  =

~ 0 ,  ~ 00,  most of the light emerging from
a
the slit is in the direction of the incident light  little diffraction.

2/
If a = ,  sin  =

= 1,  = 900,  central maximum fills the entire forward
a
hemisphere

5
C.K.Cheung
For first maximum
At P, wavelets from all the imaginary strips in the slit arrive in phase  Central
maximum.
Other maxima locations ( calculation very complicated )
Over-simplified explanation:
Cancel each other
a
3
Light projected on
screen
a

Since,

a

sin  =
3
2
sin  =
3
2a
For secondary maximum:

sin ’ =
5
2a
Hence, the maxima occur half-way between minima
6
C.K.Cheung
Intensity
Sin 
 3
a
 2
a

a

a
0
2
a
3
a
Note:
1/ Width of central bright band twice that of any other bright band.
2/ By calculaton:
I1 = (5%) I0 ;
I2 = (2%) I0
 the successive max. decrease rapidly in intensity, the position of first min. are
significant
7
C.K.Cheung
Diffraction at multiple slits ( diffraction grating )
Monochromatic
source
S
Multiple slits
screen
(grating)
Theory:
Consider any two sucessive slits (i.e. interference of double slits):
d

X
S
L
Y
M
..\..\powerpoint\diffraction1.ppt
8
C.K.Cheung
All points in the slits are secondary centres on the same wavefront  coherent
sources.
In a direction  :
Consider XL and YM
Path difference = S sin 
For constructive interference
 S sin  = n 
n = 1, 2, 3 …… until  sin    1
Other pairs of slits throughout the grating can be treated in the same way and parallel
diffracted rays (i.e. parallel to XL & YM ) are collected by a converging lens to the
same position on the screen.
n = +2
n = +1
n=0
n=-1
n=-2
e.g.
If  = 5.89x10- 7 m
S = 17x10- 7 m
n = 1  sin  =

  = 20.7 0
S
n = 2  sin ’ =
2
 ’ = 450
S
n = 3  sin ’’ =
3
= 1.06 > 1  no 3rd other
S
9
C.K.Cheung
Given that :
If :
IS = intensity due to interference between slits
ID = intensity due to diffraction of 1 slit. ( the central one )
Then:
The resultant intensity , IR, at a point on the screen:
IR = IS x ID
intensity
IS

ID

missing order
missing order

..\..\powerpoint\diffraction2.ppt
Hence, intensity variation is modulated ( controlled ) by diffraction of 1
slit ( the central slit )
10
C.K.Cheung
Note:
If number of slits increases but slit separation remains unchanged  subsidiary
maxima between principal maxima:
N = 2 slits
N = 7 slits
Note:
1/
The position of principal maxima are unchanged  contrast increased  sharper
pattern
2/
Number of subsidiary maxima = (No. of slits –2)
11
C.K.Cheung
e.g.
81’ IIB
3.
line filament
lamp
Student B
yellow
filter
several
metres
metre rule
x
metre rule
Student
A
diffraction
grating
Figure 1
Student A views a line filament lamp covered with a yellow filter through a
diffraction grating with its lines parallel to the filament (Figure 1). The grating is
held at one end of a metre rule which is aimed at the lamp. At the other end of
the metre rule, a second rule is placed at right angles to the first rule.
(a) Student A asks student B to move a pencil held vertically along the second rule
and tells him to stop when it coincides with the yellow band in the first image
of the lamp as seen through the grating. If the distance between the first rule
and the pencil is x = 0.37 m (see Figure 1) and the diffraction grating has 6.0 
105 lines per metre, calculate the wavelength of the yellow light.
X

12
C.K.Cheung
Tan  =
x
 0.37
1
 sin  = 0.347
 d sin  = n 
1
(
) sin  = ( 1) 
6 x10 5
  = 5.78 x 10-7 m
(b) If student B keeps moving the pencil along the second metre rule in the same
direction, how many more yellow bands will be encountered?
n=2  sin 2 =
2
 0.69
d
n=3  sin 3 =
3
 1.04 > 1  impossible
d
 one more yellow band will be observed.
13
Explain.
C.K.Cheung
(c) If the filter is removed, sketch the full pattern seen by student A on both sides
of the filament, in the space below. Label significant features.
position of zero order
(8 marks)
Red light
n=1
n=2
n=3
sin1=

d
Blue light
 7.5 x10 7 (6 x10 5 )  0.45 sin1=

d
 4.5 x10 7 (6 x10 5 )  0.27
sin2=
2
 0 .9
d
sin2=
2
 0.54
d
sin3=
3
 1.35 >1
d
sin3=
3
 0.81
d
Hence:
1/ distance between 2nd & 1st order < distance between 1st & zero order,
2/ dispersion greater in higher orders,
3/ zero order image is white in color,
4/ blue fringes are closer to central max in same order
5/ 3rd order blue overlap with 2nd order red. ( optional)
14
C.K.Cheung
85 MC
19. The image of a distant star produced by an astronomical telescope is a diffraction
pattern. If the effective diameter of the objective lens is reduced by one-half by
covering its outer parts with a stop, the area of the central maximum of the
diffraction pattern is
A.
B.
C.
D.
decreased by a factor of 4.
halved.
unchanged.
doubled.
E.
increased by a factor of 4.

Sin  = 1.22

~
d
for  small
If d is halved   is doubled  radius of max. doubled  area of
central max. increases by a factor of 4.
15
C.K.Cheung
84’ MC
37.
d
screen
p
R
A beam of electrons, travelling with uniform momentum p, is incident on a narrow
slit of width d. A fluorescent screen is placed at a distance R from the opening.
What is the width of the central maximum of the diffraction pattern observed on
the screen?
(Given that  =
A.
B.
C.
d/2
2d² / R
2h / p
D.
E.
2hd / Rp
2hR /dp
h
and h = Planck’s constant)
p
d
screen

p
R
Sin  =

d

ans. = 2x =
“
h
x
h
x
 tan   

pd
R
pd R
2 Rh
pd
16
x
C.K.Cheung
91’
24. When a diffraction grating is replaced by another with more lines per mm, which
of the following quantities is/are increased?
(1) the angle of diffraction for every spectral line
(2) the angle separation of red and blue lines in the first order spectrum
(3) the number of orders which can be observed
A.
B.
(1), (2) and (3)
(1) and (2) only
C.
D.
E.
(2) and (3) only
(1) only
(3) only
Statement 1/ for spectral lines  young’s pattern
 s sin  = n 
( s = slit separation )
 sin  =
n
, if s decreases   increases.
s
 correct
Statement 2/

= (red) - (blue) = sin 1
( red )
s
 sin 1
(blue)
s
,
s decreases   increases.
 correct
Ststement 3/
for observed fringes  young’s pattern
 s sin  = n 
( s = slit separation )
 sin  =
n
s
for more slits per mm  s decreases  n will be smaller for
sin 
1
 incorrect.
17
C.K.Cheung
89’
22. Light of wavelength  is incident normally on a diffraction grating with p lines
per millimetre. the second-order diffraction maximum is at an angle  from the
central position. For a second grating with 3 p lines per millimetre illuminated
normally by light of wavelength 5/4, the angle between the first-order diffraction
maximum and the central position is . Which of the following relations is
correct ?
A.
B.
C.
sin  = (5 sin )/12
sin  = sin (5 /12)
sin  = sin (15 /4)
D.
E.
sin  = (15 sin )/8
sin  = sin (15 /8)
All observed fringes in grating diffraction are Young’s fringes.
For 1st grating:
 s sin  = n  
1
sin   2  sin  = 2 p 
p
for 2nd grating:
15 p
1
5

sin   (1)( )  sin  =
4
3p
4
18
15(
sin 
)
2  15 sin 
4
8
C.K.Cheung
96’
15
grating


White
light
20
S
R
Q
P
second-order
spectrum
First-order
spectrum
0
Zeroth order
A beam of white light is shone normally on a diffraction grating. The diagram shows
the spectra of the first two orders, which may not be drawn to scale. The first- order
spectrum starts at an angle of 200 from the zeroth order. The respective angular
separations between the two ends (red and violet) of a spectrum are  and  for the
first- and second- order spectra. Which of the following statements is/are correct?
(1) In the first-order spectrum, P is the violet end.
(2)  is greater than .
(3) There is no third-order spectrum.
Statement 1:
s sin  = (1) 
 correct
if  smaller   smaller
Statement 2:
for n = 1,
sinR - sinV =
1
( r   v )
s
for n = 2,
sinR - sinV =
2
( r   v )
s
 correct
statement 3:
for n = 3,
s sin 3 = (3)V
= (3)(s sin 200)
= 1.02 > 1  impossible
 correct
19
>
C.K.Cheung
97’
19/ A plane transmission grating is placed at the centre of a circular 00-3600
protractor. A beam of monochromatic light is incident normally on the grating.
The zeroth-order maximum occurs at a scale reading of 900 and a first order
maximum occurs at a scale rading of 650. At what scale reading would a secondorder maximum be observed?
..\..\powerpoint\diffraction.ppt
3600
00
650
250
900
For young’s pattern:
n=1
S sin 1 = (1)
n =2 S sin 2 = (2)
 sin2 =
2
 2(sin 25 0 ) = 0.845
S
 2 = 580
 (900  580) on protractor scale reading
 320 or 1480
20