Transcript Lecture 21 Wave Optics

Lecture 22
Wave Optics-3 Chapter 22
PHYSICS 270
Dennis Papadopoulos
April 2, 2010
R  A[cost  cos(t   )  cos(t  2 )  cos(t  3 )  ....]
d sin
  2

OQˆ S  
A  2r sin( /2)
OQˆ T  n

AR  2r sin(n /2)  A
sin2 (n /2)
I  Io
sin2 ( /2)
Io  A 2

sin(n /2)
sin( /2)
sin 2 (n /2)
I  Io
sin 2 ( /2)
d sin 
  2
Gratings

Phase Arrays


What happens when slit is too large to be considered a point source ?
Huygens principle replace wave-front by a continuous series of point
sources
sin 2 (n /2)
I  Io
sin 2 ( /2)
d sin 
  2

d  0;  0
Take n the difference from one end to the other const
say
ant,
n = 
but send  to zero. sin  
sin 2 ( /2)
sin 2 ( /2)
I = n Io
 Imax
2
2
2
When to use ray optics and when wave optics
2.44L
D
D
Dc  2.44L
w

Actual double slit interference pattern (a<d
and a> wavelength) –
Convolution of ideal double slit and single
slit patterns
Raleigh Criterion
• Two objects are resolvable if
a>min=1.22/D, namely the angle
of the first dark fringe of the
diffraction pattern
• Objects not resolvable if a<min
• Objects marginally resolvable if
amin
Resolution limit -- Rayleigh’s criteria
For circular aperture, slightly different:
The Resolution of Optical Instruments
The minimum spot size to which a lens can focus light of
wavelength λ is
where D is the diameter of the circular aperture of the
lens, and f is the focal length.
In order to resolve two points, their angular separation
must be greater than θmin, where
is called the angular resolution of the lens.
The same criterion applies to the focusing spot of mirrors if D is the diameter of the mirror
Raleigh Criterion
• Two objects are resolvable if a>min=1.22/D, namely
the angle of the first dark fringe of the diffraction pattern
• Objects not resolvable if a<min
• Objects marginally resolvable if amin
EXAMPLE
The Hubble space telescope has a diameter 2.4 meters. It is used to photograph objects
30000 light years away ( 1 light year is 9.46x1015 meters). Assume that it uses red light
with 650 nm wavelength. What is the distance between two stars that can be resolved?
  1.22 /D
s  R  R(1.22 /D)  1011 km

The Eye-Angular Magnification- Resolution
SECTIONS 24.3-24.4-24.5
Vision
• The human eye is roughly spherical, about 2.4 cm in
diameter.
• The transparent cornea and the lens are the eye’s
refractive elements.
• The eye is filled with a clear, jellylike fluid called the
aqueous humor and the vitreous humor.
• The indices of refraction of the aqueous and vitreous
humors are 1.34, only slightly different from water.
• The lens has an average index of 1.44.
• The pupil, a variable-diameter aperture in the iris,
automatically opens and closes to control the light
intensity.
• The f-number varies from roughly f/3 to f/16, very similar
to a camera.
f-number = f/D
f/3 means that f-number is 3
Focusing and Accommodation
• The eye focuses by changing the focal length of the lens
by using the ciliary muscles to change the curvature of the
lens surface.
• Tensing the ciliary muscles causes accommodation, which
decreases the lens’s radius of curvature and thus
decreases its focal length.
• The farthest distance at which a relaxed eye can focus is
called the eye’s far point (FP). The far point of a normal
eye is infinity; that is, the eye can focus on objects
extremely far away.
• The closest distance at which an eye can focus, using
maximum accommodation, is the eye’s near point (NP).
Usually 25 cm
  h /s
 NP  h /25cm

Cannot focus any closer
than the near point of
the eye ~ 25 cm
Angular Magnification
sin     h /s  h / f
AngularMagnification
M   / NP  25cm / f
Exercise: Compare concepts
and scaling of angular vs.
lateral magnification