chapter 25:optical instruments cameras homework assignment : read chap.25, sample exercises :...

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).


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 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 circular aperture (cont’d)


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


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


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


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.

chapter 25:optical instruments cameras homework assignment : read chap.25, sample exercises :...

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