Transcript Lecture_21_March_31 - Space and Plasma Physics

Lecture 21
Wave Optics-2 Chapter 22
PHYSICS 270
Dennis Papadopoulos
March 31, 2010
White light passes through a
diffraction grating and forms rainbow
patterns on a screen behind the
grating. For each rainbow,
A. the red side is farthest from the center of the
screen, the violet side is closest to the center.
B. the red side is closest to the center of the screen,
the violet side is farthest from the center.
C. the red side is on the left, the violet side on
the right.
D. the red side is on the right, the violet side on
the left.
y ~ L / d
Two slit and five slit diffraction
653 nm with 150 slits
Single Slit Diffraction
a<
Pair wavelets with extra distance traveled /2
a
sin  1   / 2
2
Same for 3 - 4 and 5 - 6 destructive interference
 r12 
a sin  p  p  , p  1,2,..

 p  p(  / a)
Angles of dark
fringes
Pair wavelets with extra distance traveled /4

etc
Angles of dark fringes
a sin  p  p  , p  1,2,..
 p  p(  / a)
Positions on the screen
y p  L tan  p  L  p 

 p
L
a
w 2
L
a
The figure shows two single-slit
diffraction patterns. The distance between
the slit and the viewing screen is the same
in both cases. Which of the following
could be true?
A.
B.
C.
D.
The wavelengths are the same for both; a1 > a2.
The wavelengths are the same for both; a2 > a1.
The slits and the wavelengths are the same for both; p1 > p2.
The slits and the wavelengths are the same for both; p2 > p1.
The figure shows two single-slit
diffraction patterns. The distance between
the slit and the viewing screen is the same
in both cases. Which of the following
could be true?
A.
B.
C.
D.
The wavelengths are the same for both; a1 > a2.
The wavelengths are the same for both; a2 > a1.
The slits and the wavelengths are the same for both; p1 > p2.
The slits and the wavelengths are the same for both; p2 > p1.
Light decides to take the
path of “least time”
which turns out to be
given by the reflection
law.
But how does it find this
path?
Does it check all other
paths? Does it smell the
nearby paths and checks
them against each
other?
The answer is in a way
yes. The “smelling
instrument” is the
wavelength and the
decision process is
interference.
In fact light follows all possible paths that first reflect on the mirror and reach point P.
However we must keep not only the amplitude but also the phase and then add the
result as vectors
Light goes from S through the mirror to P with the same amplitude
through all paths but with different phase since the lengths of the paths
are different
S
P
What is the minimum size
of a mirror to give reflection?
  2
y
 
S
L
minimum mirror size
y
P
  = 
L
2y
L

The ends of he mirror not
important
y
L
  / 2 
y d
L 
 y  ( L / y )
y ->Minimum size of the
mirror
Notice that ≈.3-.6 microns
The Diffraction Grating
Suppose we were to replace the double slit with an opaque
screen that has N closely spaced slits. When illuminated
from one side, each of these slits becomes the source
of a light wave that diffracts, or spreads out, behind the
slit. Such a multi-slit device is called a diffraction grating.
Bright fringes will occur at angles θm, such that
The y-positions of these fringes will occur at
L
D
Dc 
2 .4  L

rays
Single slit diffraction
When laser light shines on a screen after
passing through two closely spaced slits,
you see
A.
B.
C.
D.
a diffraction pattern.
interference fringes.
two dim, closely spaced points of light.
constructive interference.
When laser light shines on a screen
after passing through two closely
spaced slits, you see
A. a diffraction pattern.
B. interference fringes.
C. two dim, closely spaced points of light.
D. constructive interference.
Circular-Aperture Diffraction
Light of wavelength λ passes through a circular
aperture of diameter D, and is then incident on a
viewing screen a distance L behind the aperture,
L>>D. The diffraction pattern has a circular central
maximum, surrounded by a series of secondary bright
fringes shaped like rings.
The angle of the first minimum in the intensity is
The width of the central maximum on the screen is
Tactics: Choosing a model of light

nm
L≈1 m
w  2 .44  L / D
2 .44  L / D c  D c
Dc 
2 .44  L