chapter 25:optical instruments cameras homework assignment : read chap.25, sample exercises :...
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Chapter 25:Optical InstrumentsCameras
Homework assignment : Read Chap.25,Sample exercises : 4,21,24,41,43
Principle of a camera
s s’
D
Intensity of light at film:
22
2
/
/
fD
ADI image
Area of image : 2'sAimage
Focal length vs. s’ : 'sf Area of image :
2fAimage
Define f-number as :
Dff /number-
Intensity of light at film: 2number)-/(1 fI
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 eyefarsightedness (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
d
hs angular size :
d (usually dmin=25 cm)
h
Magnifying glass
sfs
11
'
1 when 'sfs but
s
sm
'
cmdf
d
dh
fhmM
f
h
s
h
s
mh
s
h
s
i
fsi
ii
25,/
/
)(''
tan
minmin
min
for human eye.
the minimum distance atwhich an eye can see imageof an object comfortably and clearly.
virtual image
s’
the eye is most relatex
Microscope
)/()(]//[)]/()[(/
)magnifier afor //()/()(/
)///||(/||
21minmin02101
210211
1111001
ffLddhffLhmM
fhihffLhfh
fLsLsimfLhhmh
object
ii
21
21
,
,
ff
ffiL
small
Object is placednear F1 (s1~f1).Image by lens1is close to thefocal point of lens2 at F2.
magnifier
Refracting Telescope
21
2121
1001
///
//
)(/
ffm
fffh
sfisihmhh
siobjecti
si
ss
angular size of image by lens2; eyeis close to eyepiece
image height by lens1 at its focal point
Image by lens1 is at its focal point which isthe focal point of lens 2
image distanceafter lens1
magnifier
Reflecting Telescope
21 // ffm objecti
Resolution of Single-Slit and Circular Apertures
Resolution of single-slit apertureThe ability of an optical system such as the eye, a microscope, or atelescope to distinguish between closely spaced objects is limitedbecause 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 criterionWhen the central maximum of one image falls on the first minimumof another image, the images are said to be just resolved. This limitingcondition of resolution is know as Rayleigh’s condition.
a
From what we learned inChapter 24 about thediffraction due to a single-slitthe first minimum occurs at:
a
sin
According to Rayleigh’scriterion, this gives thesmallest angular separationfor which two images areresolvable.
a
minmin sin
Resolution of Single-Slit and Circular Apertures
Resolution of circular aperture
D Central maximum has an angular width given by:
)/(22.1sin D
Comparable to the slit geometry, where thecentral maximum is defined by sin = /a.Following the same argument as for the single-slit case, the limiting angle of resolution of thecircular aperture is:
D
22.1min
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 amicroscope. The aperture has a diameter of 0.90 cm.(a) Find the limiting angle of resolution.
rad 100.8m 1090.0
m 1058922.122.1 5
2
9
min
D
(b) Using visible light of any wavelength you desire, find the maximum limit of resolution (the shortest visible wavelength - violet ).
rad 104.5m 1090.0
m 100.422.122.1 5
2
7
min
D
(c) What effect does water between the object and the objective lens have on the resolution with 589-nm light?
rad 100.6
nm 4431.33
nm 589
5min
na
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?
rad 1005.3m 40.2
m 1000.622.122.1 7
7
min
D
(b) What is the smallest lunar crater the Hubble Space telescope can resolve (The Moon is 3.84x108 m from Earth)?
m 117rad) 10m)(3.05 1084.3( -78 rs
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:
)( 2112
NmR
N : the number of grating slits illuminatedm : the order number
Resolution of Single-Slit and Circular Apertures
Example : Light from sodium atomsTwo bright lines in the spectrum of sodium have wavelengths of 589.00nm and 589.59 nm, respectively.
(a)What must the resolving power of a grating be to distinguish these wavelengths?
3100.1nm 0.59
nm 589
nm 00.589nm 59.589
nm 00.589
R
(b) To resolve these lines in the 2nd-order spectrum, how many lines of the grating must be illuminated?
lines 100.52
100.1 23
m
RN
The Michelson Interferometer
d1
d2
• 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.