Transcript LHC-Beauty

Lecture 4: Interference and diffraction of light (I)
Young’s two slit experiment (Y&F 35.2)
 To observe interference of light one
needs to set-up two independent, coherent,
monochromatic (same frequency) light
sources.
 Coherent light sources are difficult to
make since light is emitted by the random
agitation of atoms (we will see later on that
lasers can be sources of coherent light).
S1
S0
S2
 To achieve two coherent sources of light,
we can use a common source and split it
such that the light emerges from two
secondary sources, so the relative phase
between them is always the same.
P1X: Optics, Waves and Lasers Lectures, 2005-06.
1
Young’s two slit experiment (Cont.)
 Young (1800) set up an
experiment to demonstrate
the wave nature of light.
 A monochromatic beam of
light is incident on a single slit
S0. This acts as a source of
wavefronts onto two slits S1
and S2.
 The coherent waves
interfere with each other
forming a pattern of light and
dark bands on a screen
some distance from the two
slits.
P1X: Optics, Waves and Lasers Lectures, 2005-06.
2
Geometry of Young’s two slit experiment
 The triangle S1S2Q means that: S1Q = d sin q
If the screen is at a distance R which is much larger than the separation d
of the two slits (R could be a few metres and d a few millimetres) then r1 and
r2 are nearly parallel to each other and q is small:
R  d  r2  r1  d sin q
If
 y  R tanq  R sin q if q is sm all
d sin q = r2- r1
S1
Q
d
Intensity
q
S2
q
q
r2
y
r1
P
R
P1X: Optics, Waves and Lasers Lectures, 2005-06.
3
 Constructive interference: bright fringes
r2  r1  d sinqm  m 

ym  R sin q m  R  m , m  0,1,2,
d
Destructive interference: dark fringes
1

r2  r1  d sin q   m   
2

1

ym  R sin q m  R m   , m  0,1,2,
2 d

P1X: Optics, Waves and Lasers Lectures, 2005-06.
4
 Example 35-1 (Y&F):
In a two slit interference experiment, the slits are 0.20 mm apart, and the
screen at a distance of 1.0 m. The third bright fringe (not counting the central
bright fringe) is displaced by 7.5 mm. Find the wavelength of light used.
Third fringe: m=3, R=1.0 m, d=0.2 mm, R>>d

ym  d
ym  R  m   

d
Rm
3
3
7.5  10  0.2  10
 500 109 m  500nm
1.0  3
Another possibility: m= -3
 ym= -7.5 mm
 7.5  103  0.2  103
9

 500 10 m  500nm
1.0  ( 3)
P1X: Optics, Waves and Lasers Lectures, 2005-06.
5
 Example L61 (Example book):
Light from a source containing 2 wavelengths of 567 nm and 486 nm
illuminates a double slit arrangement with d=1.0 mm and R=1.50 m. At
what distance from the central fringe on the screen will a bright fringe from
one interference pattern coincide with the bright fringe from the other
interference pattern.

 R  m  ym (  486nm)  ym1 (  567nm)
d
486 109
567 109
 Rm
 R  (m  1)
d
d
 m(486109 )  (m  1)(567109 )
9
567 10
m
7
9
(567  486)  10
9
7  1.5  486 10
3
 ym  R  m

5
.
103

10
m  5.103m m
3
1.0  10
P1X: Optics, Waves and Lasers Lectures, 2005-06.
Bright fringes: y m
6
Intensity in interference patterns (Y&F, 35.3)
 The waves associated with light are called electromagnetic waves. Unlike
the other waves we have encountered, they don’t need a medium to
propagate. They can propagate in vacuum.
 In the two slit experiment we have an electric field coming from S1:
E1 (t )  E sin(t  kr1 )
and another coming from S2:
E2 (t )  E sin(t  kr2 )
(frequency and amplitude are the same since the two sources are coherent).
Principle of superposition:
EP  E1 (t )  E2 (t )  E sin(t  kr1 )  E sin(t  kr2 )
Since:
a b a b
sin a  sin b  2 sin
 cos

 2   2 
P1X: Optics, Waves and Lasers Lectures, 2005-06.
P.1463 Y&F
Trigonometric
functions
7
 Since:
a  b t  kr1  t  kr2
 r1  r2 

 t  k 

2
2
 2 
a  b t  kr1  (t  kr2 )
 r2  r1 

 k

2
2
 2 
Then:

 r1  r2     r2  r1  
EP  2 E sint  k 
  cos k 

 2    2 

 Phase difference between two waves:
  (t  kr1 )  (t  kr2 )  k ( r2  r1 ) 
2

( r2  r1 )
r2-r1 is called the path difference.
 New wave:
r1  r2 
   
EP (t )  2 E cos
 sin t  k

2 
 2  
P1X: Optics, Waves and Lasers Lectures, 2005-06.
8
 New wave has angular frequency  and wave number k but the new
amplitude is:
  
EP (t )  2 E cos

 2 
with:
  k ( r2  r1 ) 
2

( r2  r1 )
If
  0  EP  2 E
If
    EP  0
EP
2E

2
3
4
5

-2E
P1X: Optics, Waves and Lasers Lectures, 2005-06.
9
 Intensity of light is the average power per unit area (energy per unit area
per unit time):
Pav
I
A
W   J 
 2
2 
 m   sm 
The intensity of light is proportional to the square of the amplitude of the
electric field:
  
2   
I  EP (t )  4 E cos 
  I 0 cos 

 2 
 2 
2
2
2
I
I0
0

2
3
4
5
P1X: Optics, Waves and Lasers Lectures, 2005-06.

10
Lloyd’s mirror
 Lloyd’s mirror is just like Young’s two slit experiment but with only one
source of light and a mirror used to create a second virtual source of light.
r
d
S1
a
S2
r1
r2
y
y
I
r2
MIRROR
R
 The source interferes with its own image in the mirror.
P1X: Optics, Waves and Lasers Lectures, 2005-06.
11
 There is a phase change of  radians due to the reflection of the mirror
(equivalent to a path difference of /2).
EP
EP


2

BEFORE MIRROR
2

AFTER MIRROR
 The constructive and destructive conditions are reversed with respect to
Young’s experiment and there is a dark fringe next to the mirror.
• Constructive interference: bright fringes
1
1


r2  r1  d sin q m   m    ym  R sin q m  R m   , m  0,1,2,
2
2 d


• Destructive interference: dark fringes
r2  r1  d sinqm  m 
ym  R sin q m  R  m

d
, m  0,1,2,
P1X: Optics, Waves and Lasers Lectures, 2005-06.
12