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Transcript
Chapter 25:Optical Instruments
Cameras
Homework assignment : Read Chap.25,
Sample exercises
: 4,21,24,41,43
 Principle of a camera
Area of image :
Aimage  s'2
Focal length vs. s’ : f  s '
2
Area of image : Aimage  f
Intensity of light at film:
I  D 2 / Aimage
D / f
2
D
2
Define f-number as :
f - number  f / D
Intensity of light at film:
I  1 /( f - number)
2
s
s’
Eyes
 Structure of the eye
near point (~25 cm, changes):
the closest distance for
which the lens can
accommodate to focus light
on retina.
far point :
the farthest distance for
which the lens of the relaxed
eye can accommodate to
focus light on retina.
Eyes
 Conditions of the eye
farsightedness (hyperopia)
nearsightedness (myopia)
Presbyopia (old age-vision) : due to a reduction in accommodation ability
Antigmatism
: due to asymmetry in the cornea or lens
Power of a lens (unit diopter):
P=1/f P in diopter, f in m, f=+20 m -> P=+5.0 diopter
Angular size
h
d (usually dmin=25 cm)
angular size :
h
s 
d
Magnifying glass
s’
virtual image
1 1 1
 
s' f s
the eye is most relatex
 s'
m


when s  f  s'   but
s
the minimum distance at
which an eye can see image
of an object comfortably and clearly.
hi mh h
h
tan  i   i  
 ( )s f 
s'
s'
s
f
i
h/ f
d
M  m  
 min , d min  25 cm for human eye.
 s h / d min
f
Microscope
L  i  f1 , f 2
f1 , f 2 small
magnifier
Object is placed
near F1 (s1~f1).
Image by lens1
is close to the
focal point of
lens2 at F2.
h1 | m | h0  Lh0 / f1 ( | m | i / s1  L / s1  L / f1 )
1  h1 / f 2  ( Lh0 ) /( f1 f 2 ) ( i  hi / i  h / f for a magnifier )
M  m  1 /  object  [( Lh0 ) /( f1 f 2 )] /[ h0 / d min ]  ( Ld min ) /( f1 f 2 )
Refracting Telescope
Image by lens1 is at its focal point which is
the focal point of lens 2
image distance
after lens1
h1  mh0  ih0 / s  i s  f1 s ( s  )
magnifier
 i  h1 / f 2   s f1 / f 2
m   i /  object   i /  s  f1 / f 2
image height by lens1 at its focal point
angular size of image by lens2; eye
is close to eyepiece
Reflecting Telescope
m  i / object  f1 / f 2
Resolution of Single-Slit and Circular Apertures
 Resolution of single-slit aperture
The ability of an optical system such as the eye, a microscope, or a
telescope to distinguish between closely spaced objects is limited
because of wave nature of light.
- Light from two independent
sources which are not
coherent.
- If the angle subtended by
the sources at the aperture
is large enough, then the
diffraction patterns are
distinguishable (resolvable).
- If the angle is too small, then
the two sources are not
distinguishable (unresolvable).
Resolution of Single-Slit and Circular Apertures
 Rayleigh’s criterion
When the central maximum of one image falls on the first minimum
of another image, the images are said to be just resolved. This limiting
condition of resolution is know as Rayleigh’s condition.
From what we learned in
Chapter 24 about the
diffraction due to a single-slit
the first minimum occurs at:
sin  

a
According to Rayleigh’s
criterion, this gives the
smallest angular separation
for which two images are
resolvable.

 min  sin  min 
a
a
Resolution of Single-Slit and Circular Apertures
 Resolution of circular aperture
D
Central maximum has an angular
width given by:
sin   1.22 ( / D)
Comparable to the slit geometry, where the
central maximum is defined by sin =  /a.
Following the same argument as for the singleslit case, the limiting angle of resolution of the
circular aperture is:
 min  1.22

D
Resolution of Single-Slit and Circular Apertures
 Resolution of circular aperture (cont’d)
Resolution of Single-Slit and Circular Apertures
 Example 25.6 : Resolution of a microscope
Sodium light of wavelength 589 nm is used to view an object under a
microscope. The aperture has a diameter of 0.90 cm.
(a) Find the limiting angle of resolution.
 min

 589 109 m 
  8.0 105 rad
 1.22  1.22
2
D
 0.90 10 m 
(b) Using visible light of any wavelength you desire, find the maximum
limit of resolution (the shortest visible wavelength - violet ).
 min

 4.0 107 m 
5

 1.22  1.22

5
.
4

10
rad
2

D
0
.
90

10
m


(c) What effect does water between the object and the objective lens
have on the resolution with 589-nm light?
 min
a
589 nm
 443 nm
n
1.33
 6.0 10 5 rad
w 

Resolution of Single-Slit and Circular Apertures
 Example 25.7 : Resolving craters on the Moon
The Hubble Space Telescope has an aperture of diameter 2.40 m.
(a) What is its limiting angle resolution at a wavelength of 600 nm?
 min

 6.00 107 m 
  3.05 107 rad
 1.22  1.22
D
 2.40 m 
(b) What is the smallest lunar crater the Hubble Space telescope can
resolve (The Moon is 3.84x108 m from Earth)?
s  r  (3.84 108 m)(3.05 10-7 rad)  117 m
Resolution of Single-Slit and Circular Apertures
 Resolving power of the diffraction grating
• Devices based on both the prism and the diffraction grating can be
used to make accurate wavelength measurements. However, the
diffraction grating device has a higher resolution.
• Resolving power of the diffraction grating:
R

2  1


 Nm (1  2   )

N : the number of grating slits illuminated
m : the order number
Resolution of Single-Slit and Circular Apertures
 Example : Light from sodium atoms
Two bright lines in the spectrum of sodium have wavelengths of 589.00
nm and 589.59 nm, respectively.
(a) What must the resolving power of a grating be to distinguish these
wavelengths?
R

589.00 nm
589 nm


 1.0 103
 589.59 nm  589.00 nm 0.59 nm
(b) To resolve these lines in the 2nd-order spectrum, how many lines of
the grating must be illuminated?
R 1.0 103
N 
 5.0 102 lines
m
2
The Michelson Interferometer
• Light from a source is split into 2
beams, reflected from 2 mirrors,
and recombined.
• Path difference r = 2(d2 – d1)
• Recombined light shows an
interference pattern at the detector
• If one mirror is moved a distance
/2 then the path difference changes
by  and exactly one fringe moves
across the detector.
• Precise distance measurement by
counting fringes!
• Alternatively, if one mirror is moved
a distance of /4, a bright fringe
becomes a dark fringe. This shift of
the fringe makes it possible to
measure the wavelength accurately.
d1
d2