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Lecture 22

Wave Optics-3 Chapter22

PHYSICS 270Dennis Papadopoulos

April 2, 2010

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R = A[coswt+ cos(wt+f)+cos(wt+ 2f)+cos(wt+ 3f) + ....]

= 2pdsinq

l

O QS =f

A = 2rsin(f/2)

O QT= nf

AR = 2rsin(nf/2) = Asin(nf/2)

sin(f/2)

I= Iosin2(nf/2)

sin2(f/2)

Io A2

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I= Iosin2(nf/2)

sin2(f/2)

f = 2pdsinq

l

Phase Arrays

Gratings

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I= Iosin

2(nf/2)

sin2(f/2)

= 2pdsinq

l

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

d 0;f 0

Take nf the difference from one end to the other constant,say nf = F

but sendf to zero. sinf f

I = n2Iosin2(F /2)

F 2= Imax

sin2(F /2)

F 2

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w = 2.44l LD

= D

Dc = 2.44l L

When to use ray optics and when wave optics

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Actual double slit interference pattern(a wavelength)

Convolution of ideal double slitand single slit patterns

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Raleigh Criterion

Two objects are resolvableif > min =1.22 /D, namely

the angle of the first darkfringe of the diffractionpattern

Objects not resolvable if

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Resolution limit -- Rayleighs criteriacircular aperture, slightly different:

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The Resolution of OpticalInstruments

The minimum spot size to which a lens canfocus light of wavelength is

where D is the diameter of the circular apertureof the lens, and fis the focal length.In order to resolve two points, their angularseparation must be greater than min , where

is called the angular resolution of the lens.

The same criterion applies to the focusing spot of mirrors if D is the diameterof the mirror

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Raleigh Criterion

Two objects are resolvable if > min =1.22 /D, namely the angle of the

first dark fringe of the diffraction pattern

Objects not resolvable if

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EXAMPLE

The Hubble space telescope has a diameter 2.4 meters. It is used tophotograph objects 30000 light years away ( 1 light year is 9.46x1015 meters). Assume that it uses red light with 650 nm wavelength. Whatis the distance between two stars that can be resolved?

=1.22l /D

s = Rq = R(1.22l /D) =1011

km

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SECTIONS 24.3-24.4-24.5

The Eye-Angular Magnification- Resolution

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VisionThe human eye is roughly spherical, about 2.4

cm in diameter.The transparent cornea and the lens are theeyes refractive elements.

The eye is filled with a clear, jellylike fluidcalled the aqueous humorand the vitreoushumor.

The indices of refraction of the aqueous andvitreous humors are 1.34, only slightlydifferent from water.

The lens has an average index of 1.44.The pupil, a variable-diameter aperture in the

iris, automatically opens and closes tocontrol the light intensity.

The f-number varies from roughly f/3 to f/16,ver similar to a camera.

f-number =f/D

f/3 means that f-number is 3

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Focusing andAccommodationThe eye focuses by changing the focal length

of the lens by using the ciliary muscles tochange the curvature of the lens surface.

Tensing the ciliary muscles causes

accommodation, which decreases the lenssradius of curvature and thus decreases itsfocal length.

The farthest distance at which a relaxed eye

can focus is called the eyes far point (FP).The far point of a normal eye is infinity; that is,the eye can focus on objects extremely faraway.

The closest distance at which an eye canfocus usin maximum accommodation is theUsually 25cm

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h/s

qNP h/25cm

Cannot focus any

closer than the nearpoint of the eye ~25 cm

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sinq q h/s h/f

AngularMagnification

M= q/qNP = 25cm/f

Angular Magnification

Exercise: Compareconcepts and scalingof angular vs. lateralmagnification