Transcript lecture 21 - Las Positas College

Young’s Double Slit Review
L
d

Path length difference = d sin
Which condition gives 1) dsin   m
destructive interference? 2) d sin   (m  1)
2
where m = 0, or 1, or 2, ...
d

1
Multiple Slits
Diffraction Grating
L
1
d

2
3
d
4
d
Path length difference 1-3 =
Path length difference 1-2 =
Path length difference 1-4 =
Constructive interference for all paths when
2
Intensity
L
1
d
2


3
d
All 3 rays are interfering constructively at the point shown. If the
intensity from ray 1 is I0 , what is the combined intensity of all 3
rays?
1) I0
2) 3 I0
3) 9 I0
3
Multiple Slit Interference
(Diffraction Grating – N slits with spacing d)

sin   m
d
For many slits, maxima are still at
Peak location
depends on
wavelength!
Region between maxima gets suppressed more and
more as no. of slits increases – bright fringes
become narrower and brighter.
2 slits (N=2)
0

2
dsin 
intensity
intensity
10 slits (N=10)
dsin 
0

2
4
X-Ray Diffraction


dsin
Constructive interference:
d  0.5nm
Crystal solid such
as sodium
2dsin  m
in NaCl
For =.017nm
X-ray
d
1st maximum will be at 100
5
Diffraction/ Huygens
Every point on a wave front acts as a source of tiny
wavelets that move forward.
•
•
Light waves originating
at different points
within opening travel
different distances to
wall, and can interfere!
We will see maxima and
minima on the wall.
6
Single Slit Interference
a
y


ysin= path
length
difference
dE(y) = (dy/a)E0cos(kR+kysin so, Enet =
E
cos(kR  kysin  )dy
a
from
a
y
2
to

y
o
a
2
7
Single Slit Interference
so do the integral!
E
net
(R, )  2E
kasin 
)
2
cos(kR)
kasin 
2
sin(
0

2E sin( )cos(kR)
2

0

2
 where
 
 sin
2 
Note: interference only occurs when w > 
8
E 2 max
I
2 o c
I
1
2 o c
Intensity of Single slit
Pattern
2Eo sin
(


I
/2  
2 )2
a
 sin  

2
sin 2 (  /2)
finally : I  I max
/2
a sin   , sin  =

a
a
where /2 =  sin

I max
4Eo2
 4 Io 
2 o c
First diffraction
minimum
9
A simple Single slit Pattern
• Check out This tutorial
(http://micro.magnet.fsu.edu/primer/java/scienceopticsu/diffraction/basicdiffraction/index.html
10
Diffraction through a Circular Aperture
A laser is shone at a screen through a very small hole. If
you make the hole even smaller, the spot on the screen
will get:
(1) Larger
(2) Smaller
Which drawing correctly depicts the pattern of light on
the screen?
(1)
(2)
(3)
(4)
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Diffraction from Circular Aperture
Central
maximum
1st
diffraction
minimum

Diameter D
light
Maxima and minima will be a series of bright and
dark rings on screen
First diffraction minimum is at
12
Intensity from Circular Aperture
I
First diffraction
minimum

1.22
D
sin   
These objects are
just resolved
Two objects are just resolved when the maximum of
one is at the minimum of the other.
13
Resolving Power
To see two objects distinctly, need objects  min
objects
objects is angle between
objects and aperture:
objects  tan (d/y)
min
min is minimum angular separation
that aperture can resolve:
sin min  min = 1.22 /D
D
y
d
Improve resolution by increasing objects or decreasing min 14
Question..
To increase the resolving power of a pinhole camera,
should the hole be made larger or smaller?
15
ACT: Resolving Power
sin min  min

 1.22
D
Q1: What detail do you think a fly sees when he examines
your lunch from a distance?
Q2: Can a “spy Satellite” recognize you from space?
16
The largest Star
• Betelguese is the largest star “nearby” a mere 427 light
years from Earth. Its angular diameter is .04 arc
seconds. Can its disk be resolved by the Hubble space
telescope?
17