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Chapter 6 More on geometrical opticsFebruary 4 Thick lenses

Review:

Paraxial imaging from a single refracting spherical surface:

Thin lens equation:

Lens maker’s formula:

Gaussian lens formula:

Assumptions: 1) Paraxial rays. 2) Thin lenses.Additional assumption: 3) Monochromatic light.Question: What happens if these assumptions are not valid?Solution: Study the principles of thick lens and aberrations.

21

11)1(

1

RRn

f l

fss io

111

Note the sign convention: everything has a sign.

R

nn

s

n

s

n

io

1221

21

11)1(

11

RRn

ss lio

2

6.1 Thick lenses and lens systemsThick lenses: When d is not small, the lens maker’s formula and the Gaussian lens formula are not valid.Why thick lens?The image of a distant point source is not a point, but a diffraction pattern because of the limited size of the lenses. Larger D produces clearer images.D/22.1

Some jargons in photography:Field of view (angle of view): The angle in the object space over which objects are recorded on the sensor of the camera. It depends on the focal length of the lens and the size of the sensor.

Depth of field: The region in the object space over which the objects appear sharp on the sensor.

f-number (f/#): The ratio of the focal length to the diameter of the entrance pupil:The f/# affects 1) the brightness of the image, 2) the sharpness of the image, and 3) the depth of field.

Dff //#

~ f

f.o.v.

D

d.o.f.

~ f

f

D

3

Question: What are the formulas for f and si for a thick lens?

Terminology of thick lenses:Principal plane: The plane composed by the crossing points between the incident rays parallel to the optical axis and their emerged rays.Principal points: the intersects between the principal planes and the optical axis. H1 and H2.

Note:1) The principal planes are actually curved, while its paraxial regions forms a plane.2) If one surface of the lens is planar, the tangent of the other surface should be a principal plane. prove now, and soon again. 3) Generally

(e.g., plane-convex lenses) to be proved soon.

|).| and 1.5(for )3/1( 212121 -RRdnVVHH l

Fi

b.f.l.

V1 H1V2H2

Secondaryprincipal plane

Fo

f.f.l.

V2H2V1 H1

Primaryprincipal plane

Note that we may later consider a three-segment ray as virtually two segments.

4

N2

N1 O

Nodal points: The crossing points between the optical axis and the rays passing through the optical center.Coincide with the principal points when both sides of the lens are in the same medium. to be proved soon.

Cardinal points of a lens:Two focal points + two principal points + two nodal points.When both sides of the lens are in the same medium:Fi, Fo, H1, H2 are the cardinal points.

Optical center: All rays whose emerging directions are parallel to their incident directions pass through one common point. This point is called the optical center of the lens. proved in chapter 5.

R2R1

C1C2

A

BO

Points to remember for a lens:V1, V2, C1, C2, O, Fi, Fo, H1, H2 , N1, N2. They are fixed tothe lens on its optical axis.Plus S, P for object and image.

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Read: Ch6: 1No homework

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February 6 Thick lens equations

Locations of principal planes: Note that the principal planes may be external in some cases.

However, people have developed a much simpler method which results in a set of thick lens equations that give exactly the same answer as the above method.

Virtual rays:Fact: Under paraxial optics, it is proved that if we extend a ray’s two segments located in the air they then cross at the two principal planes at the same height.Solution: Virtual rays between lens surfaces and/or principal planes can be used to simplify the problem. The far-most rays are still real.

Goal:

?

?,,, 21

i

i

o

o

ll

y

s

y

s

dRRn

Possible solution: Under paraxial optics we may use the formula for the imaging from a single refracting surface twice to locate the final image.

H1H2

7

Fi

b.f.l.

V1

H1

V2

H2Fo

f.f.l.

f fdl

sosi

h2h1

yo

yi

xixo

Thick lens equations:When f, so and si are measured from the principal planes, we have

fss

Rn

fdnh

Rn

fdnh

RRn

dn

RRn

f

io

l

ll

l

ll

l

lll

111

)1(

)1(

)1(11)1(

1

12

21

2121

h is positive when H is to the right of V.

All are to be proved soon. I hate to believe anything that is not proved by my pencil.

Newtonian form: 2fxx io

Magnification:

o

i

o

i

o

iT s

s

x

f

f

x

y

yM

When Light directed toward the first principal plane will emerge from the second principal plane at the same height.

.1,,,0 have we,0 Mfxfxss ioio

Eqs. 6.1-6.4

io

oill

FHHFf

FFHHhhfdRRn

21

212121 ,,,,,,,

Locating the four cardinal points.

Note the new three key rays. Here the rays inside the lens are virtual. The rays in the air are actual.

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Read: Ch6: 1Homework: Ch6:4,6,8Due: February 13

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February 9 Combination of thick lenses

fss

Rn

fdnh

Rn

fdnh

RRn

dn

RRn

f

io

l

ll

l

ll

l

lll

111

)1(

)1(

)1(11)1(

1

12

21

2121

Thick lens equations:

The procedure of locating an image from a thick lens:

). from distance (image '

) from distance(object '

.,,, points cardinal,,,,,

2

2

1

1

212121

o

iT

i

o

io

o

oill

s

sM

Vs

sh

s

f

sVs

h

FFHHhhfdRRn

FiV1

H1

V2

H2Fo

f fdl

sosi

h2h1

yo

yi

10

Example (P6.12):R1=4 cm, R2= -15 cm, dl =4 cm, nl =1.5, object =100 cm before lens.

minified. is image The1

inverted. is image The0

real. is image The0

.072.028.7

6.100

.02.5surface)back thefrom distance (image'

26.2

28.7

79.6

6.100100)surfacefront thefrom distance(object '

60.0

.26.2,60.0,79.6,,,

2

1

2121

T

T

i

o

iT

i

o

i

io

o

ll

M

M

s

s

sM

cms

cms

cms

cmh

cms

cmf

cmscms

cmh

cmhcmhcmfdRRn

Everything has a sign!

11

Combination of thick lenses: locating the overall cardinal points

2111

12222

2121

1

21

12111

2

2

1

1

1

12

1

222

111

, , ,

1111

f

fdHH

f

fdfsHH

ff

d

ffff

fd

sf

fsfdsfsss

s

s

s

s

s

sM

fdfsfs

i

i

ioioo

o

i

o

i

o

iT

oi

H1 Fo Fo1 H11 H12Fi1 Fo2 H21 H22 Fi2 Fi H2

df1

f1 f2 f2f f

21221121 ,,,,,, HHfHHdff

Note the sign convention for . and 222111 HHHH

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Read: Ch6: 1Homework: Ch6:12,13,14Due: February 20

yR

nnnn

R

yn

R

yn

nnnnnn

itttii

ttii

ttiittiittii

)()(sinsin

13

February 11 Ray matrices

6. 2.1 Matrix methodRay tracing: Mathematically following the trace of a ray.Example: Ray tracing of a paraxial, meridional ray traversing a spherical lens.Meridional ray: A ray in a plane that contains the optical axis and the object point.(Opp: skew ray).

I. Refraction (at P):

Power of arefracting surface R

nnD it

i

ii

t

tt

it

iitt

y

nD

y

n

yyy

Dynn 10

1

i

ii

t

tt

y

nD

y

n 10

1

Refraction matrix Incident ray vectorRefracted ray vectorit Rrr

P

Cy

nt

i

t

R

ni

14

II. Transfer (from P1 to P2):

1

1

2

2

112

12

1/

01

y

n

ndy

n

ydy

nn

1

1

2

2

1/

01

y

n

ndy

n

12 Trr Transfer matrix

P1

y1

y2

n

1

2

d

P2

i

C PV

ir

R

r

yi

III. Reflection (Mirrors):

i

i

r

r

ir

iir

rii

iii

y

nR

n

y

n

yy

R

yR

y

10

21

2

2

10

21

R

n

Mirror matrix

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Note:1) Different definitions for the ray vectors and matrixes may exist.2) Merit of the current matrices: |R|=|T|=1, and their combinations.

Examples of other definitions of ray vectors and matrices:

i

i

titt

t

i

ii

t

tt y

nnnD

y

y

nD

y

n

//

01

10

1

System matrix A of a lens:Transforming an incident ray before the first surface to the emerging ray after the second surface: A→B→C→D

2221

1211

1

2121

2

12121221

1

1

10

1

1/

01

10

1

aa

aa

n

dD

n

dn

dDDDD

n

dD

D

nd

D

l

l

l

l

l

l

l

l

ll

RTRA

Another popular form

A Cy2

nl

1

2

dl

y1

BD

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Read: Ch6: 2Homework: Ch6: 16,20,22,23,24Due: February 20

211

12

11

12

22111

122

12

2221222

2121

212112

12

1222

112

12221

1211

2

)1()1(

1

'

'1 Similarly,

)1()1(

1

)1(11)1(

11

0

Rn

fdnaf

a

a

a

aHVh

Rn

fdnaf

a

ayyHVh

RRn

dn

RRn

n

dDDDDa

yFHf

yay

ya

yaa

aa

y

l

ll

l

ll

l

lll

l

l

i

Eqs. 6.3-4

Eq. 6.2

17

Application I: Where are the cardinal points (Fi, Fo, H1, H2 )?Let the incident ray be parallel to the optical axis:

February 13 Matrix analysis of lenses

l

l

l

l

l

l

l

l

n

dD

n

dn

dDDDD

n

dD

aa

aa

1

2121

2

2221

1211

1

1A

.,,,,,,,, 212121 HHFFhhfdRRn oill

P1

Fi

V1 H1V2H2

P2

y1 y2

System matrix for a reversed lens

fss

fa

ss

ssa

aaaa

sa

aa

saa

a

as

a

asaa

a

as

sa

a

asaa

aa

a

as

a

ashsy

a

ashsy

yaa

aa

y

ioio

io

o

o

i

oi

o

oi

oo

ii

111

/1

1

1

1

1

11

)1(

11

1)(

1))((

12

12

12212211

12

12

112221

12

22

12

112221

12

22

12

12

112221

1211

12

22

12

1111

12

2222

12221

1211

2

y2

18

Application II: Lens equation

Eq. 6.1

If =, then si = so= 0 Nodal points = Principal points

y1 P

V1 H1V2H2

P1P2

S

so si

19

Application III: Thin lens combination: where are the overall cardinal points?

Thin lens system matrix

10

/1121

fA

112

2222

212

1111

212112

1

21212

12

1

1

111

1

111

10

/11

1

01

10

/11

f

fd

a

aHO

f

fd

a

aHO

ff

d

ffa

f

f

dd

ff

d

fff

d

f

d

f

Fi2

Fo1

Fi1

Fo2

O2O1

d

f1

f2

20

Homework:Starting from Eq. 6.31, please prove Eqs. 6.1-4. Please include detailed drawings showing all the parameters.

Due: February 20

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