mit2 71s09 gfinal sol

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2710

Optics

Final

Exam

Solutions

Spring

lsquo09

1 Considerthefollowingsystem

(a)

If

we

position

an

on-axis

point

source

at

the

center

of

the

object

plane

(front

focal plane of L1) a collimated ray bundle will emerge to the right of L1 and

itsdiameter issetbyS1 thereforeS1istheaperturestop(AS) Similarly S2

limits the lateral extent of an imaged object (consider an off-axis point source)andthusitrsquosourfieldstop(FS)

(b) Theentrancepupilistheimageof theASbytheprecedingopticalcomponentsTofinditslocationweusetheimagingcondition

1 1 1 S of 1

2f 123

+ = S i = =

S o

S i

f 1

rArr

S o

minusf 1

f 1( 23

minus1)

2f 1S o =

S i =

minus2f 1 (virtual)3

rArr

Sotheentrancepupil is locatedat2f 1

totherightof L1 Tofind itsradiuswe

computethelateralmagnification

S iM L

=

minusS o

= 3

rarrrEnP

= 3a1

Fortheexitpupil(ExP)

f 1

( f 3

1 +f 2)f 2

f 22

S o

= +f 2 S i = =f 2

+3 (totherightof L2)

3

f 1

f 1+

f 23 f 2minus

+f 2)minus3f f

1

2( f 3

1

f 2

f 2M L

=( f 1 +f 2)

=

minus3

f 1

(inverted)

rarr rExP

= 3

f 1

a1

3

TheexitwindowisthesameasS2bull

The entrance window is the image of S2 through the preceding optical elebull

ments(ie combinationof L1andL2) Itisf 1

totheleftof L1

1



(c) Solution

Thenumericalapertureistanα

asympα

asympsinα

asympNA

asymp

a1

f 1

2X s

2f 1a2

2a2Thefieldof view(FOV)is FOV=2β = = =

3f 1

3f 1f 2

3f 2

(d)

The

location

of

S1

limits

the

FOV

because

of

the

requirement

for

the

CR

to

gothrough the center of the aperture stop (AS) It canbeseenthatthe least

restrictiveAS location isattheFourierplane(f 1

totherightof L1 f 2

to

lArrrArr

the

left

of

L2)

2

Given

the

following

GRIN

medium

2n(r) =

radic 2

minusr2

0lt r lt1

r=

radic x2 +z

1 r

ge1

(a) TheHamiltonianequationsare(r lt1)

dx =

partH 1 2P x

P x

P x

ds partP x

=

minus2

radic P x

2+P z

2

=

minusn

=

minusradic 2minusx2minusz2

dz =

partH

1 2P z

P z

P z

ds

partP z

=

minus2

radic P x

2+P z

2

=

minusn

=


dP x

partH

1

minus2x x

x

= =

ds =

minus partx =

minus2

n n

radic 2minusx2minusz2

dP z

partH 1minus2z z z= =

ds

=

minuspartz

=

minus2 n n


where

H =n(q )

minus[P 2 +P 2]12 = 0x z

=[2

minusx

2minus

z 2]12minus

[P x

2 +P z

2]12 = 0

n= P 2 +P 2

x z

2



(b) Recallthatwe justfoundn2 =P x

2 +P z

2 917501 2

917501 2

dP x

dP z

1048573x

11141092

1048573z 11141092 x2

+z 2 x2 +z 2

+ = + = =

ds ds n n n2 P 2 +P 2

x z

=

P

2

r

+

2

P

2 =

P

2

2minus

+

n

P

2

2 =

P

2 +

2

P

2minus

1

x z x z x z

partn

(c) Since =0theScreenHamiltonian isnotconserved Thismaybeverifiedby

partz

directsubstitution

h=

minus n2

minusP x

2 =

minus 2

minusx2

minusz 2minus

P x

2

andweseethatparth z

=

= 0

partz 2

minusx2

minusz 2

minusP x

2

3 ConsidertheMichelsoninterferometershownbelow

(a) Webeginbywritingtheanalyticexpression(inphasorform)of asphericalwave

withoriginat(x p z p)usingtheparaxialapproximation

E 0ei

2λπ

(zminuszp)

iπ

(xminusxp)2

E S = e

λ (zminuszp)

iλ(z minusz p)

For simplicity we take x p

= z p

= 0 At the observation plane the interference

patternisgivenby

I =

|E S 1

+E S 2

|2 =

|E S 1

|2 +

|E S 2

|2 +E S 1

E S lowast

2

+E S lowast

1

E S 2

3



917501

917501

917501

where

e λ π 2 e λ π

2E 0

i2π1

i x E 0

i

2π1

i xE S 1

= e

λ1 E S 2

= e

λ2

4 iλ1

4 iλ2

1

=z 0

+ 2z 1

+z c

2

=z 0

+ 2z 2

+z c

E 02 E 02

= =|E S 1

|2

16(λ1)2

|E S 2

|2

16(λ2)2

So

E 02 E 0

2 E 02

I = + + [e

i∆φ +eminusi∆φ]16(λ1)2

16(λ2)2 16λ212

E 02 1 1 2

=

16λ2 12

+

22

+

12

cos∆φ ∆φ=φ2

minusφ1

2π x2 2π x2

φ2 = +

2

φ1 = +

1λ

22

λ

21

2π 2

1 1rArr∆φ=

λ x

22

minus

21

+2∆z ∆z =z 2

minusz 1

917501

917501

E 02

1 1 2 2π 2

∆

I = + + cos xrArr

16λ2 21

22

12

λ 212

minus∆

∆=2∆z =2

minus1

E 02 1 1 2

I max

= + +

16λ2 21

22

12

E 02 1 1 2

I min

=

16λ2 2

1

+

22

minus

12

(b) If theflatmirrorM2isreplacedbyaconvexsphericalmirrorof radius2(z 0

+z 2)thesphericalwavegetscollimatedandtheinterferencepatternbecomes 2

E 0

E 0

2 E 0

2 E 0

2

I =E S 1

+ 4λ e

iφ

=

16(λ)2

+

16(λ1)2

+8λ21

sin∆φ

Plane

Wave Chirp

Function

4



=z 0

+z 2

2π

φ = 2λ

∆φ =φ1

minusφ

4

Consider

the

4-f

system

shown

below

(a) The pupil mask can be implemented by placing two pinholes (small apertures)onecenteredwithrespecttotheopticalaxisandthesecondoneat1cmoff-axisThe2ndpinholeisphasedelayedbyapieceof glassof thicknesstwhere

2πφ=π= t(15

minus1) t=λ

λ

rArr

(b) Theinputtransparencyis

=gt

middot 1048573 1048573 1048573 1048573 1048573 1048573 1114109

=1

α

infinsinc(αq )e

i2π

qx

gin

gillumination

Λ

q=minusinfin

AttheFourierplaneinfin 1048573

q 1114109Gin

=α sinc(αq )δ u

minus

Λ

u=

x

λf q=minusinfin 1048573 1114109 917501

917501

1

infin q λf λf 05

times10minus4

10

Gin(x) =

2

sinc

2

δ xminusq Λ

Λ

=

5

times10minus4

middot

= 1

q=minusinfin

5



917501

917501

917501

917501

917501

917501

Afterthepupilmaskonlythe0thand+1orderspass The+1ordergetsphase

delayed

byeiπ =

minus1

1 1 1

Gout(x) = δ (x)

minus sinc δ (xminus1)

2 2 2

1 1

δ (xminus

1)=

δ (x)

minus2 π

1 1

gout(u) =

2

minus

πeminusi2πu

u=

x

I out

=

1 1

2

minus

π

eminusi

2πx

λf

λf

2

1 1 1 2π x=

4

+

π2

minus

cos

π λ f

1 4π

Thecontrastisv= 0906=

π = 1 1+

4 +π2

4

π2

5

(a)

To

compute

the

OTF

we

first

need

the

ATF

u1

=

x=

1cm

= 02micromminus1 =200mmminus1

λf 05microm

times10cm

est

δx 02cm

δu=

λf asymp

05microm

times10cm


H (u) =H (u)

otimesH (u)

= rectu

minusrect

uminusu1

rect

uminusu

minusrect

uminusu1

minusu

du

δu

δu

δu

δu

(autocorrelation)

6



917501

917501

917501 917501

=

rect

urect

uminusu

duδu δu

rarr

917501 917501

rect

u minusrect

uminusu1

minusu

duminus

δu δu

rarr

917501

917501

rect

uminusu1

rectuminusu

duminus

δu δu

rarr

917501 917501

+ rect

uminusu1

rect

uminusu1

minusu

duδu δu

rarr

SotheresultantOTFandMTFare(afternormalization)

(b) OnlytheDCandthe

plusmn1stharmonicsatu=

plusmn200mmminus1 (period=5microm)ie

1 1 1 2πx I (x) =

2

minus

2

times

π

times2cos

5microm

=

DC

term

minusH (200mmminus2)

times

1st

harmonic

times

2

cos

2πx

5microm

(c) Solution

7



F minusrarr sinc2(δux)sotheiPSFis

1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8



MIT OpenCourseWarehttpocwmitedu

271 2710 OpticsSpring 2009

For information about citing these materials or our Terms of Use visit httpocwmiteduterms



(c) Solution

Thenumericalapertureistanα

asympα

asympsinα

asympNA

asymp

a1

f 1

2X s

2f 1a2

2a2Thefieldof view(FOV)is FOV=2β = = =

3f 1

3f 1f 2

3f 2

(d)

The

location

of

S1

limits

the

FOV

because

of

the

requirement

for

the

CR

to

gothrough the center of the aperture stop (AS) It canbeseenthatthe least

restrictiveAS location isattheFourierplane(f 1

totherightof L1 f 2

to

lArrrArr

the

left

of

L2)

2

Given

the

following

GRIN

medium

2n(r) =

radic 2

minusr2

0lt r lt1

r=

radic x2 +z

1 r

ge1

(a) TheHamiltonianequationsare(r lt1)

dx =

partH 1 2P x

P x

P x

ds partP x

=

minus2

radic P x

2+P z

2

=

minusn

=


dz =

partH

1 2P z

P z

P z

ds

partP z

=

minus2

radic P x

2+P z

2

=

minusn

=


dP x

partH

1

minus2x x

x

= =

ds =

minus partx =

minus2

n n


dP z

partH 1minus2z z z= =

ds

=

minuspartz

=

minus2 n n


where

H =n(q )

minus[P 2 +P 2]12 = 0x z

=[2

minusx

2minus

z 2]12minus

[P x

2 +P z

2]12 = 0

n= P 2 +P 2

x z

2




2 +P z

2 917501 2

917501 2

dP x

dP z

1048573x

11141092

1048573z 11141092 x2

+z 2 x2 +z 2

+ = + = =


x z

=

P

2

r

+

2

P

2 =

P

2

2minus

+

n

P

2

2 =

P

2 +

2

P

2minus

1

x z x z x z

partn


partz

directsubstitution

h=

minus n2

minusP x

2 =

minus 2

minusx2

minusz 2minus

P x

2

andweseethatparth z

=

= 0

partz 2

minusx2

minusz 2

minusP x

2




E 0ei

2λπ

(zminuszp)

iπ

(xminusxp)2

E S = e

λ (zminuszp)

iλ(z minusz p)


= z p


patternisgivenby

I =

|E S 1

+E S 2

|2 =

|E S 1

|2 +

|E S 2

|2 +E S 1

E S lowast

2

+E S lowast

1

E S 2

3



917501

917501

917501

where

e λ π 2 e λ π

2E 0

i2π1

i x E 0

i

2π1

i xE S 1

= e

λ1 E S 2

= e

λ2

4 iλ1

4 iλ2

1

=z 0

+ 2z 1

+z c

2

=z 0

+ 2z 2

+z c

E 02 E 02

= =|E S 1

|2

16(λ1)2

|E S 2

|2

16(λ2)2

So

E 02 E 0

2 E 02

I = + + [e


16(λ2)2 16λ212

E 02 1 1 2

=

16λ2 12

+

22

+

12

cos∆φ ∆φ=φ2

minusφ1

2π x2 2π x2

φ2 = +

2

φ1 = +

1λ

22

λ

21

2π 2

1 1rArr∆φ=

λ x

22

minus

21

+2∆z ∆z =z 2

minusz 1

917501

917501

E 02

1 1 2 2π 2

∆

I = + + cos xrArr

16λ2 21

22

12

λ 212

minus∆

∆=2∆z =2

minus1

E 02 1 1 2

I max

= + +

16λ2 21

22

12

E 02 1 1 2

I min

=

16λ2 2

1

+

22

minus

12



E 0

E 0

2 E 0

2 E 0

2

I =E S 1

+ 4λ e

iφ

=

16(λ)2

+

16(λ1)2

+8λ21

sin∆φ

Plane

Wave Chirp

Function

4



=z 0

+z 2

2π

φ = 2λ

∆φ =φ1

minusφ

4

Consider

the

4-f

system

shown

below


2πφ=π= t(15

minus1) t=λ

λ

rArr


=gt

middot 1048573 1048573 1048573 1048573 1048573 1048573 1114109

=1

α

infinsinc(αq )e

i2π

qx

gin

gillumination

Λ

q=minusinfin


q 1114109Gin

=α sinc(αq )δ u

minus

Λ

u=

x

λf q=minusinfin 1048573 1114109 917501

917501

1

infin q λf λf 05

times10minus4

10

Gin(x) =

2

sinc

2

δ xminusq Λ

Λ

=

5

times10minus4

middot

= 1

q=minusinfin

5



917501

917501

917501

917501

917501

917501


delayed

byeiπ =

minus1

1 1 1

Gout(x) = δ (x)


2 2 2

1 1

δ (xminus

1)=

δ (x)

minus2 π

1 1

gout(u) =

2

minus

πeminusi2πu

u=

x

I out

=

1 1

2

minus

π

eminusi

2πx

λf

λf

2

1 1 1 2π x=

4

+

π2

minus

cos

π λ f

1 4π


π = 1 1+

4 +π2

4

π2

5

(a)

To

compute

the

OTF

we

first

need

the

ATF

u1

=

x=

1cm


λf 05microm

times10cm

est

δx 02cm

δu=

λf asymp

05microm

times10cm


H (u) =H (u)

otimesH (u)

= rectu

minusrect

uminusu1

rect

uminusu

minusrect

uminusu1

minusu

du

δu

δu

δu

δu

(autocorrelation)

6



917501

917501

917501 917501

=

rect

urect

uminusu

duδu δu

rarr

917501 917501

rect

u minusrect

uminusu1

minusu

duminus

δu δu

rarr

917501

917501

rect

uminusu1

rectuminusu

duminus

δu δu

rarr

917501 917501

+ rect

uminusu1

rect

uminusu1

minusu

duδu δu

rarr


(b) OnlytheDCandthe



1 1 1 2πx I (x) =

2

minus

2

times

π

times2cos

5microm

=

DC

term


times

1st

harmonic

times

2

cos

2πx

5microm

(c) Solution

7




1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8









2 +P z

2 917501 2

917501 2

dP x

dP z

1048573x

11141092

1048573z 11141092 x2

+z 2 x2 +z 2

+ = + = =


x z

=

P

2

r

+

2

P

2 =

P

2

2minus

+

n

P

2

2 =

P

2 +

2

P

2minus

1

x z x z x z

partn


partz

directsubstitution

h=

minus n2

minusP x

2 =

minus 2

minusx2

minusz 2minus

P x

2

andweseethatparth z

=

= 0

partz 2

minusx2

minusz 2

minusP x

2




E 0ei

2λπ

(zminuszp)

iπ

(xminusxp)2

E S = e

λ (zminuszp)

iλ(z minusz p)


= z p


patternisgivenby

I =

|E S 1

+E S 2

|2 =

|E S 1

|2 +

|E S 2

|2 +E S 1

E S lowast

2

+E S lowast

1

E S 2

3



917501

917501

917501

where

e λ π 2 e λ π

2E 0

i2π1

i x E 0

i

2π1

i xE S 1

= e

λ1 E S 2

= e

λ2

4 iλ1

4 iλ2

1

=z 0

+ 2z 1

+z c

2

=z 0

+ 2z 2

+z c

E 02 E 02

= =|E S 1

|2

16(λ1)2

|E S 2

|2

16(λ2)2

So

E 02 E 0

2 E 02

I = + + [e


16(λ2)2 16λ212

E 02 1 1 2

=

16λ2 12

+

22

+

12

cos∆φ ∆φ=φ2

minusφ1

2π x2 2π x2

φ2 = +

2

φ1 = +

1λ

22

λ

21

2π 2

1 1rArr∆φ=

λ x

22

minus

21

+2∆z ∆z =z 2

minusz 1

917501

917501

E 02

1 1 2 2π 2

∆

I = + + cos xrArr

16λ2 21

22

12

λ 212

minus∆

∆=2∆z =2

minus1

E 02 1 1 2

I max

= + +

16λ2 21

22

12

E 02 1 1 2

I min

=

16λ2 2

1

+

22

minus

12



E 0

E 0

2 E 0

2 E 0

2

I =E S 1

+ 4λ e

iφ

=

16(λ)2

+

16(λ1)2

+8λ21

sin∆φ

Plane

Wave Chirp

Function

4



=z 0

+z 2

2π

φ = 2λ

∆φ =φ1

minusφ

4

Consider

the

4-f

system

shown

below


2πφ=π= t(15

minus1) t=λ

λ

rArr


=gt

middot 1048573 1048573 1048573 1048573 1048573 1048573 1114109

=1

α

infinsinc(αq )e

i2π

qx

gin

gillumination

Λ

q=minusinfin


q 1114109Gin

=α sinc(αq )δ u

minus

Λ

u=

x

λf q=minusinfin 1048573 1114109 917501

917501

1

infin q λf λf 05

times10minus4

10

Gin(x) =

2

sinc

2

δ xminusq Λ

Λ

=

5

times10minus4

middot

= 1

q=minusinfin

5



917501

917501

917501

917501

917501

917501


delayed

byeiπ =

minus1

1 1 1

Gout(x) = δ (x)


2 2 2

1 1

δ (xminus

1)=

δ (x)

minus2 π

1 1

gout(u) =

2

minus

πeminusi2πu

u=

x

I out

=

1 1

2

minus

π

eminusi

2πx

λf

λf

2

1 1 1 2π x=

4

+

π2

minus

cos

π λ f

1 4π


π = 1 1+

4 +π2

4

π2

5

(a)

To

compute

the

OTF

we

first

need

the

ATF

u1

=

x=

1cm


λf 05microm

times10cm

est

δx 02cm

δu=

λf asymp

05microm

times10cm


H (u) =H (u)

otimesH (u)

= rectu

minusrect

uminusu1

rect

uminusu

minusrect

uminusu1

minusu

du

δu

δu

δu

δu

(autocorrelation)

6



917501

917501

917501 917501

=

rect

urect

uminusu

duδu δu

rarr

917501 917501

rect

u minusrect

uminusu1

minusu

duminus

δu δu

rarr

917501

917501

rect

uminusu1

rectuminusu

duminus

δu δu

rarr

917501 917501

+ rect

uminusu1

rect

uminusu1

minusu

duδu δu

rarr


(b) OnlytheDCandthe



1 1 1 2πx I (x) =

2

minus

2

times

π

times2cos

5microm

=

DC

term


times

1st

harmonic

times

2

cos

2πx

5microm

(c) Solution

7




1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8








917501

917501

917501

where

e λ π 2 e λ π

2E 0

i2π1

i x E 0

i

2π1

i xE S 1

= e

λ1 E S 2

= e

λ2

4 iλ1

4 iλ2

1

=z 0

+ 2z 1

+z c

2

=z 0

+ 2z 2

+z c

E 02 E 02

= =|E S 1

|2

16(λ1)2

|E S 2

|2

16(λ2)2

So

E 02 E 0

2 E 02

I = + + [e


16(λ2)2 16λ212

E 02 1 1 2

=

16λ2 12

+

22

+

12

cos∆φ ∆φ=φ2

minusφ1

2π x2 2π x2

φ2 = +

2

φ1 = +

1λ

22

λ

21

2π 2

1 1rArr∆φ=

λ x

22

minus

21

+2∆z ∆z =z 2

minusz 1

917501

917501

E 02

1 1 2 2π 2

∆

I = + + cos xrArr

16λ2 21

22

12

λ 212

minus∆

∆=2∆z =2

minus1

E 02 1 1 2

I max

= + +

16λ2 21

22

12

E 02 1 1 2

I min

=

16λ2 2

1

+

22

minus

12



E 0

E 0

2 E 0

2 E 0

2

I =E S 1

+ 4λ e

iφ

=

16(λ)2

+

16(λ1)2

+8λ21

sin∆φ

Plane

Wave Chirp

Function

4



=z 0

+z 2

2π

φ = 2λ

∆φ =φ1

minusφ

4

Consider

the

4-f

system

shown

below


2πφ=π= t(15

minus1) t=λ

λ

rArr


=gt

middot 1048573 1048573 1048573 1048573 1048573 1048573 1114109

=1

α

infinsinc(αq )e

i2π

qx

gin

gillumination

Λ

q=minusinfin


q 1114109Gin

=α sinc(αq )δ u

minus

Λ

u=

x

λf q=minusinfin 1048573 1114109 917501

917501

1

infin q λf λf 05

times10minus4

10

Gin(x) =

2

sinc

2

δ xminusq Λ

Λ

=

5

times10minus4

middot

= 1

q=minusinfin

5



917501

917501

917501

917501

917501

917501


delayed

byeiπ =

minus1

1 1 1

Gout(x) = δ (x)


2 2 2

1 1

δ (xminus

1)=

δ (x)

minus2 π

1 1

gout(u) =

2

minus

πeminusi2πu

u=

x

I out

=

1 1

2

minus

π

eminusi

2πx

λf

λf

2

1 1 1 2π x=

4

+

π2

minus

cos

π λ f

1 4π


π = 1 1+

4 +π2

4

π2

5

(a)

To

compute

the

OTF

we

first

need

the

ATF

u1

=

x=

1cm


λf 05microm

times10cm

est

δx 02cm

δu=

λf asymp

05microm

times10cm


H (u) =H (u)

otimesH (u)

= rectu

minusrect

uminusu1

rect

uminusu

minusrect

uminusu1

minusu

du

δu

δu

δu

δu

(autocorrelation)

6



917501

917501

917501 917501

=

rect

urect

uminusu

duδu δu

rarr

917501 917501

rect

u minusrect

uminusu1

minusu

duminus

δu δu

rarr

917501

917501

rect

uminusu1

rectuminusu

duminus

δu δu

rarr

917501 917501

+ rect

uminusu1

rect

uminusu1

minusu

duδu δu

rarr


(b) OnlytheDCandthe



1 1 1 2πx I (x) =

2

minus

2

times

π

times2cos

5microm

=

DC

term


times

1st

harmonic

times

2

cos

2πx

5microm

(c) Solution

7




1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8








=z 0

+z 2

2π

φ = 2λ

∆φ =φ1

minusφ

4

Consider

the

4-f

system

shown

below


2πφ=π= t(15

minus1) t=λ

λ

rArr


=gt

middot 1048573 1048573 1048573 1048573 1048573 1048573 1114109

=1

α

infinsinc(αq )e

i2π

qx

gin

gillumination

Λ

q=minusinfin


q 1114109Gin

=α sinc(αq )δ u

minus

Λ

u=

x

λf q=minusinfin 1048573 1114109 917501

917501

1

infin q λf λf 05

times10minus4

10

Gin(x) =

2

sinc

2

δ xminusq Λ

Λ

=

5

times10minus4

middot

= 1

q=minusinfin

5



917501

917501

917501

917501

917501

917501


delayed

byeiπ =

minus1

1 1 1

Gout(x) = δ (x)


2 2 2

1 1

δ (xminus

1)=

δ (x)

minus2 π

1 1

gout(u) =

2

minus

πeminusi2πu

u=

x

I out

=

1 1

2

minus

π

eminusi

2πx

λf

λf

2

1 1 1 2π x=

4

+

π2

minus

cos

π λ f

1 4π


π = 1 1+

4 +π2

4

π2

5

(a)

To

compute

the

OTF

we

first

need

the

ATF

u1

=

x=

1cm


λf 05microm

times10cm

est

δx 02cm

δu=

λf asymp

05microm

times10cm


H (u) =H (u)

otimesH (u)

= rectu

minusrect

uminusu1

rect

uminusu

minusrect

uminusu1

minusu

du

δu

δu

δu

δu

(autocorrelation)

6



917501

917501

917501 917501

=

rect

urect

uminusu

duδu δu

rarr

917501 917501

rect

u minusrect

uminusu1

minusu

duminus

δu δu

rarr

917501

917501

rect

uminusu1

rectuminusu

duminus

δu δu

rarr

917501 917501

+ rect

uminusu1

rect

uminusu1

minusu

duδu δu

rarr


(b) OnlytheDCandthe



1 1 1 2πx I (x) =

2

minus

2

times

π

times2cos

5microm

=

DC

term


times

1st

harmonic

times

2

cos

2πx

5microm

(c) Solution

7




1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8








917501

917501

917501

917501

917501

917501


delayed

byeiπ =

minus1

1 1 1

Gout(x) = δ (x)


2 2 2

1 1

δ (xminus

1)=

δ (x)

minus2 π

1 1

gout(u) =

2

minus

πeminusi2πu

u=

x

I out

=

1 1

2

minus

π

eminusi

2πx

λf

λf

2

1 1 1 2π x=

4

+

π2

minus

cos

π λ f

1 4π


π = 1 1+

4 +π2

4

π2

5

(a)

To

compute

the

OTF

we

first

need

the

ATF

u1

=

x=

1cm


λf 05microm

times10cm

est

δx 02cm

δu=

λf asymp

05microm

times10cm


H (u) =H (u)

otimesH (u)

= rectu

minusrect

uminusu1

rect

uminusu

minusrect

uminusu1

minusu

du

δu

δu

δu

δu

(autocorrelation)

6



917501

917501

917501 917501

=

rect

urect

uminusu

duδu δu

rarr

917501 917501

rect

u minusrect

uminusu1

minusu

duminus

δu δu

rarr

917501

917501

rect

uminusu1

rectuminusu

duminus

δu δu

rarr

917501 917501

+ rect

uminusu1

rect

uminusu1

minusu

duδu δu

rarr


(b) OnlytheDCandthe



1 1 1 2πx I (x) =

2

minus

2

times

π

times2cos

5microm

=

DC

term


times

1st

harmonic

times

2

cos

2πx

5microm

(c) Solution

7




1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8








917501

917501

917501 917501

=

rect

urect

uminusu

duδu δu

rarr

917501 917501

rect

u minusrect

uminusu1

minusu

duminus

δu δu

rarr

917501

917501

rect

uminusu1

rectuminusu

duminus

δu δu

rarr

917501 917501

+ rect

uminusu1

rect

uminusu1

minusu

duδu δu

rarr


(b) OnlytheDCandthe



1 1 1 2πx I (x) =

2

minus

2

times

π

times2cos

5microm

=

DC

term


times

1st

harmonic

times

2

cos

2πx

5microm

(c) Solution

7




1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8









1048573 1114109

1048573 1114109

x x =sinc2

40microm

times 1

minuscos 2π

5microm

8






mit2 71s09 gfinal sol

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