Transcript Lecture 16 Diffraction Chp. 37

Lecture 16 Diffraction Chp. 37
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Topics
– Young’s double slit interference experiment
– Diffraction and the wave theory
– Single slit diffraction
– Intensity of single slit diffraction
– Circular aperture and double slit diffraction
– Diffraction grating
– Dispersion and resolving power
– Warm-up problem
Demos
– Diffraction grating and slits
– Inverted mirage
– Measuring diameter of a strand Debra’s hair
Young’s Double Slit
Interference Experiment
m=2
q
y
m=1
m=0
m=1
m=2
D
d sin q  m m = 0,1,2, 3... Maximum
tanq 

ym 

ym
D
ym  Dtanq  Dsinq
mD
d
Maxima

or
Minima
m ym +/-
m ym +/-
0
1
2
3
0
1
2
3
0
D/d
2D/d
3D/d
D/2d
3D/2d
5D/2d
7D/2d
1
dsinq  (m  ) m= 0,1,2,3... Minimum
2
(m 1/2)D
ym 
d

 dsinq

What about the intensity of light along the screen?
I  4I0 cos2 12 

2d

sin q
13E Suppose that Young’s experiment is performed with blue-green
light of 500 nm. The slits are 1.2mm apart, and the viewing screen is
5.4 m from the slits. How far apart the bright fringes?
From the table on the previous slide we see that the separation
between bright fringes is D /d
D /d  (5.4m)(500109 m) /0.0012m
 m
 0.00225m  2.25m
Diffraction of a single slit
Find minimum
ym





path length difference between1 and
r r2 
a

sinq  for the first minimum
2
2
asinq  
asin q  m m = 1,2, 3.. minima
ym
tanq 
or ym  Dtanq  Dsinq
D
D
y m  Dtanq  m
a
Find maximum
First maximum lies on the axis at q= 0
or ym=0. Other maxima lie in between
the minima. To find them we need to
find the intensity along the screen.
Intensity of single slit
diffraction
Maxima/Minima conditions
phasedifference
 
2

2

path length difference
x sinq
sin  2
I(q )  I0   2 
 2 

2

asinq

1
Maxima when  (m  ) m= 0,1,2,3..
2
2
Minima when

2

 m
1 
sinq  (m  )
2 a
sinq  m
m=1,2,3..


a

Exact solution for maxima
let  

2
sin 2  
I(q )  I0  2  I0 (sin2  )( 2 )
  
To find maxima of a function, take derivative and set equal to 0
dI
0
d
I0 2sin  cos  2  I0 2sin 2   3  0
cos  sin  /
  tan
Transcendental equation. Solve graphically


sinq  m  m  m 1
a



 m
sinq  m  m
  0.2,0.4,0.6,0.8,1.0
a
5 5

sinq  m

a
m

m

10 10
8. A 0.10-mm-wide slit is illuminated by light of wavelength 589nm.
Consider a point P on a viewing screen on which the diffraction patters
of the slit is viewed; the point is at 30o from the central axis of the slit.
What is the phase difference between the Huygens wavelets arriving at
point P from the top and midpoint of the slit? (Hint: see Eq. 37-4.)
We note that nm = 10-9 m = 10-6 mm.
From Eq. 37-4,
 2 
   x sin q 
  
2

 0.10m m
o

sin
30



6
2 
 58910 m m
 266 .7 rad
This is equivalent to 266.7 - 84 = 2.8
rad = 160o
6. Sound waves with frequency 3000 Hz and speed 343 m/s diffract
through a rectangular opening of a speaker cabinet and into a large
auditorium. The opening, which has a horizontal width of 30.0 cm,
faces a wall 100 m away. Where along that wall will a listener be at
the first diffraction minimum and thus have difficult hearing the
sound? (Neglect reflections).
Let the first minimum be a distance y
from the central axis which is
perpendicular to the speaker. Then
sin q 
y
D y
2
2

m 
 (for m  1).
a
a
q
Therefore,
y
D
a  2  1


D
af
vs   1
2
100m
0.300m  3000Hz 343m s 
2
1
 41.2m
y
Diffraction and Interference by a double slit
I = I (double slit interference) x I(diffraction)
I(q )  Im cos 2 ( ) 
sin 2 
2
dsinq


dsinq = m2
m2 = 0,1,2.. maxima
asinq

asinq = m
m=1,2,3.. minima

Sample problem 37-4
(a) How many bright interference fringes fall within the central
peak of the diffraction envelope?
The idea here is to find
  405nm
Given d  19.44nnm
the angle where the first
minimum occurs of the diffraction
a  4.050nm
envelope.
asin q  
d sin q  m2 

m2  d /a  19.44/4.050 4.8

We have m=0 and
m=1,2,3 and 4 on both
sides of central peak.
Ans is 9

Diffraction by a circular aperature
The first minimum for the diffraction of light from a circular
aperature is given by:
sinq 1.22 /d where d is the diameter of the aperature.
Our ability to resolve two distant point like objects is determined
when the first minimum of one objects diffraction pattern overlaps
the central maximum of another. This is called Raleigh’s criterion.
Raleigh' s Criterion
q R  1.22 /d
Example
Example 15E. The two headlights of an approaching automobile are 1.4 m
apart. Assume the pupil diameter is 5.0 mm and the wavelength of light is
550 nm. (a) At what angular distance will the eye resolve them and (b) at
what distance?
(a)
qR  1.22  /d
qR 1.22(550106 mm) /5.0mm
qR 134.2 106 rad
(b)
qR
qR  s /D
D  s /q R
D
6
D  1.4m /134.2 10 rad
D  0.010106 m  10km


qR  1.22  /d
s
Diffraction Grating
Double slit -- N slits or rulings.
w
d sin q  m m = 0,1,2, 3..
d = w/N
where w is the entire width of the grating
Can be used to measure wavelength of light
Measure angles of diffracted lines with a spectroscope using
formula below. Then relate to wavelength
tan q  (y m  y 0 ) /D
 = dsinq /m
Resolving power of grating.
Measure of the narrowness of lines
Highest R
R   /  Nm
Dispersion of a grating. Measure
of how well lines are separated
m
D  q / 
d cosq
Highest D
Show diffraction gratings with increasing N
and single slit diffraction with varying slit
width a.
Babinets Complementarity Principle
In the diffraction region the intensity is the same whether you
have an aperature or opaque disk. You can also replace a slit
with a wire or hair strand. A compact disk is an example of a
diffraction grating in reflection instead of transmission.
Experiment: Measure diameter of a strand of hair from Debra.
asinq  
for the first diffraction minim um
a   /sin q
sinq  tanq  y1 /D
a  D / y1
a  560 nm (D/ y1 )
Mirage
eye
sky
1,09
1.09
1.08
1.08
1.07
1.07
1.06
Hot road causes gradient in the index of refraction that increases
as you increase the distance from the road
In the demo before you the gradient is in the opposite direction
Warm-up
HRW6 37.CQ.02. [73994] You are conducting a single-slit diffraction experiment with light of
wavelength .
(a) What appears, on a distant viewing screen, at a point at which the top and bottom rays through
the slit have a path length difference equal to 5.?
the m = 5 maximum
the m = 5 minimum
the m = 4 minimum
the m = 4 maximum
(b) What appears, on a distant viewing screen, at a point at which the top and bottom rays through
the slit have a path length difference equal to 4.5.?
the minimum between the m = 4 and m = 5 minima
the minimum between the m = 4 and m = 5 maxima
the maximum between the m = 4 and m = 5 maxima
the maximum between the m = 4 and m = 5 minima