aoptics2-lens

Dr.G.Mirjalili Physics Dept.Yazd University

Spherical single thin lensSpherical single , thin lens, thick lens


F d f ti f th t t b hi h th i i li ht i d i t dSurface power

For curved refracting surfaces, the extent by which the incoming light is deviated from its original path is known as the “surface power” (D) and is dependent on:

(I) the difference in refractive index(I) the difference in refractive index

(II) the radius of curvature of the surface, r.

Glass Block

n n'

r

F = n' - n rD


Surface powerSurface Power, D:

p

D = n’ – n r must be measured in metresr

⇒ unit for D is the reciprocal metre , i.e. 1/metres or m-1

The reciprocal metre is better known as the dioptre

Higher values for D result from1- large differences between the refractive indices 2 h h l f i ll (i h i )2- when the value for r is small (i.e. the curvature is steep)

Rays aimed at the centre of curvature of a spherical refractingRays aimed at the centre of curvature of a spherical refracting surface will pass through undeviated. Why ?


Vergence(1)How do we specify the extent to which the incident/emergent rays converge or diverge?

Rays of Light Diverging from a Point Object

The arc shown is called a wavefront, & the curvature of the a efront is called the ergencewavefront is called the vergence


Vergence(2)g ( )Note: with increased distance from source, wavefront gets less curved (i.e. has less vergence)( g )

so inverse relationship exists

Vergence ∝ 1

distance-to-source


Vergence(3)⇒ the vergence of the wavefront is determined by:

g ( )

(i) distance from point of interest to source

(ii) refractive index of medium in which rays are travelling

Unit for Vergence? V=n/d

Reciprocal metre.... ⇒ Dioptre

-ve vergence: ⇒ rays diverging

+ve vergence: ⇒ rays converging


Sample Calculation 1:Light is made to converge by a lens in air to a point that is 75cm from the lens. Find the vergence of the wavefront when the ray l th l d t 50 d 100 f th lleaves the lens, and at 50cm and 100cm from the lens.

Converging 50cm behind lens

+75cm

Diverging 100cm behind lensWhich is the point of interest?

Dr.G.Mirjalili Physics Dept.Yazd UniversityConverging 50cm behind lens

+75cmDiverging 100cm behind lens

On leaving the lens: Vergence = 1 = 1 = +1.33D (“Image Vergence”) l’ +0.75

50cm from lens: Vergence = 1 = 1 = +4Dl’ 0 25l’ +0.25

100cm from lens: Vergence = 1 = 1 = 4D100cm from lens: Vergence = 1 = 1 = -4Dl’ -0.25


VergenceVergence

8

V= 1/(-0.08) = -12.5 m-1 V=1/(+0.08) = 12.5m-1

8 cm8 cm

V=1/∞ = 0


Vergence1→2

112

VnL

LnV ⇒=1

d dLnV

VL

−=2

1

L

)(1

1Vn

Vnn

)(1))(( 1

12

1

11

2

ndV

VV

nVd

Vn

n

dVn

nV−

=⇒−

=−

=


Refraction at spherical interfaces“Some rules”

1 Light travels left to right1. Light travels left to right2. V = origin – measure all distances from here3. R = positive to the right of V, negative to the left4 S iti f l bj t (i t th l ft f V) ti f4. S = positive for real objects (i.e. one to the left of V), negative for

virtual5. S’ = positive for real image (to right of V), negative for virtual

imagesimages6. Heights – y,y’ – positive up, negative down

VR+R-

+V

+- _


Refraction by Spherical SurfacesAt point P we apply the law of refraction to obtain

i iθ θ1 1 2 2sin sinn nθ θ=

Using the small angle approximation we obtainapproximation we obtain

1 1 2 2n nθ θ=

Substituting for the angles θ1 and θ2we obtain

( ) ( )′( ) ( )1 2n nα ϕ α ϕ′− = −

Neglecting the distance QV and iti t t f th l i

n > n

writing tangents for the angles gives

h h h hn n⎛ ⎞⎛ ⎞− = −⎜ ⎟⎜ ⎟n2 > n1 1 2n n

s R s R= ⎜ ⎟⎜ ⎟ ′⎝ ⎠ ⎝ ⎠


Refraction by Spherical Surfaces IIRearranging the equation we obtain

1 2 1 2n n n ns s R

−− =

′

Using the same sign convention as for mirrors we obtainfor mirrors we obtain

1 2 2 1n n n nR−

+ =′s s R′

n > nn2 > n1


Guss`s Formula and surface powerGuss s Formula and surface power

α

γn2, n`n1, n

γ

γR

S,O S`, I

nnnn −=+

''Rss '

S = positive for realS positive for real objects (i.e. one to the left of V), negative for virtual


Other sign conventionOther sign convention

• In some references Gauss` formula is defined as:

nnnn −i

nR

nnon 2121 =+

ioO= negative for real object +

+

R- R+ O- i+

+__


Vergence & surface powerVergence & surface power

in

Rnn

on 2121 =

−+

n1 n2

O iO i

n1/o = inter vergence

(n n )/R = surface power(n2-n1)/R = surface power

n2/i = exit vergence

Inter vergence+ surface power (optics system power) =exit vergence


ExampleExampleExample:Example:o = 5mm, n2 = 1.5,(n1=1)radius of the lens=1 43mmradius of the lens=1.43mmWhat is exit vergence?What is i?

O i

in

Rnn

on 2121 =

−+

2121

in

Rnn

on

=−

+ in

fn

on

111

2

2

21 =+

G ` f l`VV =+ β

1/(5 10 3) (1 5 1)/(1 43 10 3) 1000/1 5

ifo 2

=+ Gauss formula

1/(5x10-3)+(1.5-1)/(1.43x10-3)=1000/1.5

V=1000/1.5 i=1.5/V i=10 mm


n ( i )

Image Formation—Single Curves Surface

object

n (air)

n’ (glass)objectimage

n (glass)

robject distance,

oimage

di tdistance, i

n+ n’=

(n’ - n)Gauss’s Equation for refraction by a single

o +i

=rrefraction by a single spherical surface


Gaussian lens equatioin vergence notation

n’n (n’ - n)in

on + =r

( )

V =no OBJECT VERGENCEo

V` =n’i

IMAGE VERGENCEV i

β(n’ - n) POWER OF REFRACTING SURFACEβ = r( ) POWER OF REFRACTING SURFACE

β + V = V`

Dr.G.Mirjalili Physics Dept.Yazd UniversityFOCAL LENGTH AND LENS CURVATURE

R = 4 mmR 4 mm

R = 8 mm

R = 10 mm

Longer R less curved lens⇒Longer R less curved lens⇒Less lens curvature longer focal length⇒


From Gauss’s Equation to the Vergence Equation

1 1 1+ = 1 11 - = 1 11 + =-s s’ f+ = -s s’f = s s’f

+ =

1P 1U = 1VfP =Lens power

sU =Object vergence

s’V =Image vergence

f s s’ in meters

Lens power Object vergence Image vergence

f, s, s in meters

P U VP + U = V


Optical system & VergenceOptical system & Vergence

V V`Inter Exit

Optical

V VInter vergence

Exit vergence

psystem

O iββ

*,,,int vergenceVexitrsystempoweopticsvergenceerV =+ β


exampleexampleA ll bj i l d 0 1 i f f f G dA small object is located 0.1 m in front of a convex surface. Ground

with a 0.03 m radius on a block of glass of index 1.66. find the position of image;

(a) If the object and glass are in air.(b) If both are immersed in water (n=4/3)S l tiSolution :(a) from Gauss`formula, we find that;n1/o+(n2-n1)/R+n2/i

o i

n1/o+(n2 n1)/R+n2/i1.66/i=1.00/-0.1+1.66-1.00/+0.03→ i=+0.138 m (image is real)(b):1.66/i=4/3/-0.1+(1.66-4/3)/+0.03→i=1.66/-2.444=-0.679 (image is virtual)

Dr.G.Mirjalili Physics Dept.Yazd UniversitySpherical Surface Focusing Example

E lExample:s = 5mm, n2 = 1.5, s' = 10mm (n1=1)

''( 1 )

What is the radius of the lens?

Rnn

sn

sn −

=+'

''

Rss1 12 2

sns

const nR

+ = =−

'( )

n n`

15

1510

210

1510

05s s R

R+ = + =

. . .s s,

5 10 10 103510

05R

R=

. . nnnn 2121 −1035 5

RR =. i

nR

nnon 2121 =+

535

143R mm= =.

.


More Spherical Surface Focusing ExamplesFor the same lense, consider s = 6mm. What is s’?

R = 1.43 mm

6 mm8 18 mm

5 mm8.18 mm

10 mm

16

+1.5

′ s =

0.51.43

1.5′ s

= 0.35 − 0.1666 = 0.1833

′ s =1.5

0.18333= 8.18mm

Dr.G.Mirjalili Physics Dept.Yazd UniversityInteresting Case

Let s ∞

from1s

+n2

′ s =

(n2 −1)R

= constant

as “s” increases “s' “ decreases (and vice versa.)

As the point “O” goes to ∞, O' moves in to a certain

distance -- called the focal length f':

n n1 2A

O'

f'

Dr.G.Mirjalili Physics Dept.Yazd UniversityFocal lengths

f' is the focal length in the glassf is the focal length in the glass.

Now let O' go to ∞:

n n1 2B

ff

• If we had a light source at f' (in the glass,) the lens collimates the output Ap

• Similarly for a light source at f in B

Dr.G.Mirjalili Physics Dept.Yazd Universityf and f’

What are f and f'? ns

ns

n nR

1 2 2 1+ =−

'( )

Since s = ∞ f':s s R

n2 (n2 −n1)

and since s' = ∞ f:

2

′ f =

( 2 1)R f f``R

and since s ∞ f:

n1

f=

(n2 −n1 )R

n2n1

f R

so fn

Rn n

fn

'( )2 2 1 1

=−

=( )2 2 1 1

Dr.G.Mirjalili Physics Dept.Yazd UniversityGeneral Case

General case (we do not prove):General case (we do not prove):

For any system of lens (no matter how complicated)

ººº ?n1 n2

f ′ f

f f '

S t f lfn1

=fn2

Sometimes

System of lenses

Sometimes

is written asn1+n2 =

(n2 −n1)

(It is easier to measure f than R and n)

s+

′ s R1

+n2 =

1(It is easier to measure f than R and n)s

+′ s

=f


If l l h i f h• If we want to calculate the image of the system, we can calculate the image of spherical surface one after another.

• The image of first surface is regarded as the object of the second surface and so on.

E i ll tt ti t th i f bj t di t• Especially, pay attention to the signs of object distance, image distance and curvature radius of each spherical surface.surface.


Example : A l b ll h di f 10 dExample : A glass ball has a radius of 10cm andrefractive index of 1.5. Use paraxial rays formula tocalculate the image of an object that is 40cm from the ballcalculate the image of an object that is 40cm from the ball.

n2 (glass)

On1 (air)

n2 (glass)

n1 (air)I1I·OI1I2p1 p2

u u220cmv2u1

v1

u220cm

Solution: for the first surface, we knowSolution: for the first surface, we know,10,40,5.1,1 121 cmrcmunn ====


151511vr

nnvn

un

1015.15.1

401

1

12

1

2

1

1 −=+

−=+

cmvvrvu

601040

1

111

=

This is the first step. I1 is the virtual image for the first lens and it is regarded as the object of the second spherical surface. So for the second surface, we have:

vunn )20(151 ===cmrcm

vunn1040

)20(,1,5.1 1221

−=−=−−===

cmrcm 10,40 ==


S b tit ti th b d t t th i lSubstituting the above data to the paraxial ray formula, we have:

nnnn 1221 −=+

rvu 22

=+

105.111

405.1 −

=+−

cmvv411

1040 2

=−

cmv 4.112 =


Thin lensesThin lenses• A lens is a simple coaxial system and it is an optical system• A lens is a simple coaxial system and it is an optical system including two refracting surfaces.

If th thi k f l i h ll th t• If the thickness of a lens is much smaller than curvature radius, the object and image distances, the thickness of the lens can be negligible in comparison with them. Such a lens is calledcan be negligible in comparison with them. Such a lens is called thin lens.

• Lens can converge and diverge light• Lens can converge and diverge light.

D<<r, S, S`

D


Thin lens:

Thickness small in comparison to distances f ti l ti ( di f tof optical properties (radius of curvature,

focal length, image and object distances)

r

focal point

TT

Dr.G.Mirjalili Physics Dept.Yazd UniversitySome special rays for converging and diverging

lenseslenses

(1). Light rays parallel to the axis of a converging lens are refracted through the focal point on the opposite side of the lens.

First and second focal point f l

Planes through the focal points of a lens. of a lens are called focal planes.


(2). Light rays parallel to the axis of a diverging lens are( )refracted so that their backward continuations pass throughthe focal point on the same side of the lens.


(3) Definition of converging and diverging lensesdiverging lenses

A lens is converging if the glass is thinner aroundthe circumference than at the center anddiverging if the situation is reversed.


(4) Graphical method(4). Graphical methodA light ray through the center of a thin lens continues un-deviated and un-displaced. A light ray parallel to the axis passes through the second focal point of a converging lens

F Real image

Virtual image


The lens equation nThe lens equationn0

n0

According to the gformula of single spherical surface, we

u v1

v

p ,can use it one by one and finally calculate u v1

ythe lens equation.

u1 = u, v1 = -u2, nnnn 1221 −=+

v2 = v, rvu+


For the first surface, n1= n0, n2 = n

00 nnnn −=+

11 rvu=+

For the second surface, n1= n, n2 = n0

00 nnnn −=+−

21 rvv


The abo e t o eq ations can be addedThe above two equations can be added together, then we have

⎞⎜⎛− 0 1111 nn 00 nnnn −

⎟⎠

⎞⎜⎜⎝

⎛−=+

210

0 1111rrn

nnvu 1

0

1

0

rnn

vn

un

=+

⎠⎝ 210

If lens is in the air, n0 = 1, 2

00

1 rnn

vn

vn −

=+−0

⎞⎜⎛ 11)1(11

⎟⎠

⎞⎜⎜⎝

⎛−−=+

21

11)1(11rr

nvu ⎠⎝ 21


first focal distance for thin lensfirst focal distance for thin lens

⎞⎛ 11111

11−

⎤⎡ ⎞⎛⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=

∞+

210

0

1

1111rrn

nnf 210

01

11⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

rrnnnf

⎠⎝ 2101 rrnf 210 ⎦⎣ ⎠⎝ rrn

second focal distance for thin lens:second focal distance for thin lens:

⎞⎛ 1111 nn1

11−

⎤⎡ ⎞⎛⇒⎟⎠

⎞⎜⎜⎝

⎛−

−=+

∞ 210

0

2

1111rrn

nnf

210

02

11⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

rrnnnf

⎠⎝ 2102f 210 ⎦⎣ ⎠⎝ rrn


Therefore e ha e f f e can s pposeTherefore we have f1 = f2 . we can suppose that they are all equal to f and for air medium, we have

⎞⎜⎜⎛⎞

⎜⎜⎛− 0 11)1(111 nnn

⎠⎜⎜⎝

−−=⎠

⎜⎜⎝

−=21210

0 )1(rr

nrrnf

Finally we obtain the lens equationFinally we obtain the lens equation,


exampleexample

• b) solving using Vergence:

P1=1/+0.2 =+5, P2= -16, P3=+10 )(1 1

12 dV

VV−

=

nV2 = 5/[1-(5)(0.06)]=7.1429 m-1

On leaving lens 2

V`= V+P =(+7.1429)+(-16)=-8.8571

Translation from lens 2 to 3:

V`=(-8.8571)/{1-(-8.8571)(0.06)= -5.7836m-1

On leaving lens 3 ;

V``=(-5.7836)+(+10)= +4.2164

The inverse of that is image distanceThe inverse of that is image distance

I3=1/+4.2164 =23.72 cm


Dr.G.Mirjalili Physics Dept.Yazd University7.3.3 Systems of two lenses

M ti l i t t i l d bi ti f tMany optical instruments include combinations of two or more lenses. In systems of multiple lenses, the image formed by one lens becomes the object for the next lens (see theby one lens becomes the object for the next lens (see the following figure).

ff1

u1 v1 u2 f2


However same as before sometimes the second lens is placedHowever, same as before, sometimes the second lens is placedbetween the first lens and the image. The original imagedisappears, and the second lens may or may not form a newpp , y yimage, but the image that the first lens would have formedstill serves as the object for the second lens. Because the firsti i i iimage is not actually formed but still functions as the objectfor the second lens, we call it a virtual object. The objectdistance u of a virtual object is negativedistance u of a virtual object is negative.


The case of two connected thin lenses:

For the first lens u u

The case of two connected thin lenses:

111f

=+

For the first lens, u1 = u

11 fvu

For the second lens

u1= uv

v1= -u2

For the second lens, u2 = -v1, v2 = v,

111=+− 1111

+=+21 fvv

21 ffvu+=+


Suppose that the focal distance for the group of thin lenses is f, we have

111+=

group of thin lenses is f, we have

21 fff+=

For the more lens systems, we could use this method one by one to solve them. s e od o e by o e o so ve e .Written as dioptric strength,

L++= 21 DDD


Example 7.3: A converging lens of focal length 12 cm is placed 52 cm fromlength 12 cm is placed 52 cm from another converging lens of focal length 8 cm. Calculate the image position of an object that is 16 cm in front of the firstobject that is 16 cm in front of the first lens.


52cmf =12cm f2=8cm52cmf1=12cm 2 8c

air airair

16cm·· ·

airair

v1u2

u1

u2

v2


Sol tion: the image formed b the first lensSolution: the image formed by the first lens can be obtained by the given data and the thin length equation. Substituting the first object distance of 16 cm and the focal jdistance of first lens into thin lens equation, we havewe have

111=+

111 fvu+

cmvv

48 1211

161

1 =⇒=+v 1216 1


As the distance bet een the t o lenses is 52As the distance between the two lenses is 52 cm, the object distance for the second lens is 52 – 48 = 4 cm. Using the focal distance for the second lens of 8 cm, the final image can , gbe obtained as

cmv 81111112 −=⇒=+⇒=+

vfvu 84 22222

The image is virtual.


MagnificationMagnification

• Transverse Magnification• Axial MagnifiacationAxial Magnifiacation• Angular Magnification

Dr.G.Mirjalili Physics Dept.Yazd UniversityMagnification*

object

X ffX

sdo

ff

s’di

immage

1. Any ray that comes in parallel on one side proceeds toward a particular point called the focus (on the other side of the lens at a distance f frompoint, called the focus (on the other side of the lens, at a distance f from the lens).

2. Any ray that arrives at the lens from the focus on one side comes out y yparallel to the axis and the other side.

Dr.G.Mirjalili Physics Dept.Yazd UniversityMagnification cont’d

G

y'

y

A

BC

F

D I

X f f x'

yE H

Triangles ABC and EDB are similar, so:

s (d0 ) s'(di )

y /

f=

yxf x

Triangles DFG and IHG are also similar, so:

y /

x / =yfx f

Dr.G.Mirjalili Physics Dept.Yazd UniversityMagnification cont’d

From equation 1 y /

=fFrom equation 1,

y=

x

From equation 2, y /

y=

x /

fy f

/ / fOr,

y /

y=

x /

f=

fx

= M , the magnification.

Consider triangle ADC and HDI, which are also similar:/ / dy

do

=y /

di, or

y /

y=

di

do

= M .


Vergence & MagnificationVergence & Magnification

γ`

hh

oi

γγ`

h`

hoh

′

= γ

i

ii

′=

γ

γ

MVonnnio

122 ⇒

′=

γγ

hM

MVinnno 2

1

1

2

1

2

′

=′

=⇒′

=γ

lationshipHelmholtzSmithhnhnh

M

Re,21 −′′=

=

γγ


Axial (longitudinal) MagnificationAxial (longitudinal) Magnification

∆o ∆i

o1 i1o2 i22

iiiiio

ioio

iof−

=−

=22

22

11

11

)()(

ooii

ooii

ioioioio

=−−

−=−

21

21

12

12

11222211 )()(

ttt MMMMoi

ooii

===∆∆

=−−

2112

12α

Dr.G.Mirjalili Physics Dept.Yazd UniversityMagnification and Angular Spread (Angular Magnification)

hαo α i

X f f X'

Consider an optical source with an angular optical output spread of

did0

p g p p p

αo . What is the resulting angular spread αi

after it is focused by the lens?after it is focused by the lens?

Recall that:Recall that:

Dr.G.Mirjalili Physics Dept.Yazd UniversityMagnification and Angular Spread cont’d

So,⎛ ⎞ h⎛ ⎞ ⎛ ⎞ h⎛ ⎞

tan αo

2⎛ ⎝

⎞ ⎠ =

hdo

⎛ ⎝ ⎜ ⎞

⎠ ⎟ tan αi

2⎛ ⎝

⎞ ⎠ =

hdi

⎛ ⎝ ⎜ ⎞

⎠ ⎟

Therefore,

tan αo⎛ ⎝

⎞ ⎠

h⎛ ⎞tan

2⎝ ⎠

tan α i⎛ ⎝

⎞ ⎠

=doh =

hdo

⎛ ⎝ ⎜ ⎞

⎠ ⎟ di

h⎛ ⎝

⎞ ⎠ =

di

do

= M.tan

2⎝ ⎠ di

o o

For small θ, (i.e. < 20 degrees or < 0.35 radians) tan θ = θ so:

( if the angles are small )Mi

=αα

0


CouplingCoupling

•

•

• Coupling Laser & Fiber

Dr.G.Mirjalili Physics Dept.Yazd UniversityMagnification and Angular Spread cont’d

C P bl h i i l f l di d ( LED)Common Problem: the emission angles from a laser diode (or LED) can be 40 to 60 degrees. The acceptance angle for a fiber can be 10 to 30 degrees. The solution is to use a lens to increase the coupling efficiency.

Laser Fiberα β

Example 2-4:Lens

Example 2 4:Suppose a source radiates with a 40 degree full cone angle,and it has the dimensions 20 µm x 20 µm (must be an LED). Design a lens system to decrease the beam spread to 10 degrees Alsolens system to decrease the beam spread to 10 degrees. Also determine the image size.

Dr.G.Mirjalili Physics Dept.Yazd UniversityLens System Example Problem

Solution:Solution:

α⎛ ⎞ 40⎛ ⎞ tan αo

2⎛ ⎝

⎞ ⎠

α i⎛ ⎞ = M =

tan 402

⎛ ⎝

⎞ ⎠

10⎛ ⎞ =

0.36390 08748

= 4.16tan i

2⎛ ⎝

⎞ ⎠ tan

2⎛ ⎝

⎞ ⎠

0.08748

Note: M ≈

αo

2α =

20= 4Note: α i

25

This means that the image size is 83.2 µm 20x4 16=83 220x4.16 83.2

Dr.G.Mirjalili Physics Dept.Yazd UniversityLens Systems

Starting from:1

+1 1

⇒do +

do do ⇒ 1 +1 do

( Multiply both sides by d0 )

do

+di

=f

⇒ o

do

+ o

di

= o

f⇒ 1 +

M= o

f

So if M = 4.16, then: d0 di

f

1+1

4 16=

do

f= 1.24

f

4.16 f

d0=1.24f

Dr.G.Mirjalili Physics Dept.Yazd UniversityLens Systems cont’d

If the lens has a focal length of 10 cm, then:d = 1.24 f = 12.4cmdo 1.24 f 12.4cm

And since

di

d= M ⇒ di = 4.16do ⇒ di = 51.6cm

20µm x 20µm

80µm x 80µm

d0

40 deg 10 degLaser Fiber

f f

d

f f

d0 i

Dr.G.Mirjalili Physics Dept.Yazd UniversitySystems Lens cont’d

Note: We could have chosen f = 1 mm, then our

d 1 24 d 5 16do = 1.24mm,di = 5.16mm

Thi l ti i bl if li tThis solution is reasonable if we are coupling to a large diameter fiber,i.e. a fiber diameter of greater than 80 µmthan 80 µm.

However single mode fibers have core diameters onHowever, single mode fibers have core diameters on the order of only 4 to 12 µm. Fortunately, laser diodes have small apertures, i.e. only 1 to 2 µm in p , y µheight and 2 to 5 µm in width.

Dr.G.Mirjalili Physics Dept.Yazd Universityfor Ray Rules Tracing

F R l f R T i1. Rays travelling through the center of the lens are

d i t d ( i th thi l i ti ll l

Four Rules for Ray Tracing:

undeviated,(using the thin lens approximation, parallel surfaces).

2. Incident rays travelling parallel to the lens axis pass through the focal point after emerging from the lens.

3. An incident ray travelling parallel to a central ray intersects that ray in the focal plane after transmission through the lensthat ray in the focal plane after transmission through the lens.

4. An incident ray passing through the focal point travels parallel to the lens axis after it emerges from the lens.

Dr.G.Mirjalili Physics Dept.Yazd UniversityRay Tracing Rules Illustration

4

f

32

f f

1

Figure 2-10: The numbers refer to the rules.


Magnification vs. do/f

From our basic equations:1s

+1s / =

1f

⇒1do

+1di

=1f

, andf o i f

1+1M

=do

f⇒

1M

=do

f−1

f M =1

M f M f

Therefore, M = do

f−1

5

4

and the text considers values of3M

2 1 < < 2do

fFig 2-18

11.2 1.4 1.6 1.8 2.01

df

0

Fig. 2-18

Dr.G.Mirjalili Physics Dept.Yazd UniversityOther Possibilities:

dCase 1:

do

f= ∞

• The object is at infinity• Only the parallel rays make it to the lens (this is why the sun'srays focus to a point)rays focus to a point).

Dr.G.Mirjalili Physics Dept.Yazd UniversityOther Possibilities cont’d

Case 2: 2 <do < ∞Case 2:

M =1

=1

2 <f

< ∞

Example: d /f=3 M =3 −1

=2

The image is real, inverted, reduced (demagnified), and located

Example: d0/f=3,

The image is real, inverted, reduced (demagnified), and locatedbetween f and 2f.

A

P

AB'

P'P

B fA'

Pf2f

B fObject Image

Figure 2-11: The image formed by a converging lens of an object that is located at a finite distance beyond 2f'.that is located at a finite distance beyond 2f .


C 3do 2Case 3: o

f= 2

1 1M =1

2 − 1=

11

= 1

The image is real, inverted, the same size as the object, andis located at 2f on the other side of the lens.

A B'

P P'

ff'2f 2f

A'B

Figure 2-12: The image formed by a converging lens is real, inverted, d th i th t l bj t it lfand the same size as the actual object itself.


Case 4: 1 <do ≤ 2Case 4: 1 < o

f≤ 2

M 1 1 2/f M =1.5 −1

=0.5

= 2Example: d0/f=1.5,

The image is real, inverted, enlarged (magnified), and locatedbeyond 2f on the opposite side of the lens.

AB'

f 2f P'P2f'

A

f' f2fB

A'


Case 5: do 1Case 5: of

= 1

1 1M =1

1 −1=

10

= ∞

No image is formed.

A

f 2ff'P P'

B

Figure 2-13: No image is formed when the object is located at the principle focus.


C 6 0 ≤do < 1Case 6: 0 ≤f

< 1

M =1

=1

= −2Example: d /f=0 5 M =0.5 −1

=−0.5

= 2Example: d0/f=0.5,

The image is virtual, enlarged, and located on the same side of the lens as the object!

A'

A

f' f

A

BB

B'The image formed of an object less than a focal length from the lens.

Dr.G.Mirjalili Physics Dept.Yazd UniversityMagnification

Magnification as a function of do/f

45

tion 1

2

3

Mag

nific

at

2

-1

0

-4

-3

-2

do/f-5

0 1 3 4 5 6 7 8 9 102


Using a GRIN to collimate light

Lens

Fiber(a)

Rod Lens

Fiber(b)

GRIN rod lenses can also be used to collimate light from

l di da laser diode.

Dr.G.Mirjalili Physics Dept.Yazd UniversityOther Types of Lenses - Cylindrical lens

Focal LineFig. 2-11

Lens Axis

• Focusing, collimation l i th ti l

f

only in the vertical direction

• 1D version of the• 1D version of the spherical lens

Fig. 2-12f

Dr.G.Mirjalili Physics Dept.Yazd UniversityCylindrical lens

SourceCly. lens

Point Source

Source

SideView

f

Point Source

TopView

Semiconductor lasers (and some LEDs) have asymmetric ray divergences (or beam spread) because the emitting aperture isdivergences (or beam spread) because the emitting aperture is asymmetric.

Dr.G.Mirjalili Physics Dept.Yazd UniversityPlanar Source Emitting Apertures

emittting aperatureemittting aperature

θ

laserθ//

θ⊥Side View

Planar Source Emitting A t

θAperature

Top View

Dr.G.Mirjalili Physics Dept.Yazd UniversityThe Graded Index Rod Lens (GRIN)

r r

r n2

r

a

2an

1n

2n(r)z0

Fig. 5-4g

Fig. 5-6

Dr.G.Mirjalili Physics Dept.Yazd UniversityRay paths inside a GRIN Rod

2a

Figure 5-5 Ray paths along a GRIN fiber.

P

(c)

GRIN rod. (c) A typical ray path.

Dr.G.Mirjalili Physics Dept.Yazd UniversityGRIN cont’d

P/4 P/4

(a) (b)(a)

Graded Index Rod. (a) A quarter-pitch lens collimates light emerging from ( ) q p g g ga point.(b) A quarter-pitch lens focuses a collimated light beam.


Using a GRIN to collimate light

Lens

Fiber(a)

Rod Lens

Fiber(b)

GRIN rod lenses can also be used to collimate light from

l di da laser diode.

Dr.G.Mirjalili Physics Dept.Yazd UniversityThe Numerical Aperture

Numerical Aperture (NA or N A )Numerical Aperture (NA or N.A.)Originally defined for microscope objectives.

NA = n sinθNA n sinθ

Dr.G.Mirjalili Physics Dept.Yazd UniversityThe Numerical Aperture

β

αθ

β

op t ica l sys temn1

Where θ is the angle of the outermost ray that enters (and is useful to) the system.to) the system.

Consider an "optical receiver system" consisting of aConsider an optical receiver system consisting of a lens and a photodetector:


Determining the Numerical Aperture

f

(a)

θf

(b )

β

(c)

d θ

θ

d / 2

β

For this system the maximum angle θ is given by

d / 2Photo Detector

For this system, the maximum angle θ is given by d2

⎛ ⎝

⎞ ⎠ dtan(θ ) = 2⎝ ⎠

f=

d2 f

Example: this receiver system has a focal length of 10cm and the photodetector has a diameter of 1cm. Find NA.p

Dr.G.Mirjalili Physics Dept.Yazd UniversityDetermining the Numerical Aperture

θ d 1 0 05

−1

tan .θ = = =df2

120

0 05

(2.862º)θ = tan1(0.05) = 0.04995rad

NA= n1 sin(θ ) = 1sin(2.862º )1 ( ) ( )NA= 0.04993

θ β Fib e r

α NA n≡ 1 sinθNA = sinθ , n=1


Acceptance Angle

0.5

0.6

ure

0.4

Ap

ertu

0.2

0.3

eric

al

0.1Nu

m Sin g le Mod e Fib er

30201000.0

Accep ta n ce An gle

Figure 2-23: Numerical aperture and g pacceptance angle. NA = sin θ


Thick lens*Thick lens**Thi k l i f i l*Thick lenses contain two systems of coaxial spherical surfaces.

*Thickness of the thick lenses cannot be negligible while the thickness of thin lenses can be ignored.while the thickness of thin lenses can be ignored.

*such a system can be solved by spherical surface, b t it t i l t f t i i l d t il i ll fbut it contain a lot of trivial details especially for coaxial optical system of more spherical surfaces.


Thick Lenses and Cardinal PointsThick Lenses and Cardinal Points

Aims & Objectivesl b t t f f ti t f t d b k f f-learn about concept of refraction at front and back surfaces of a

thick lens

-introduce the step-along and step-back equations

- learn how to determine image location with a thick lens

-discover the properties of principal planes

-find the cardinal points of a thick lens


Thick Lenses and Cardinal Points

Image size is determined by:g yshape

thicknessthickness

refractive index

Thick Lenses versus Thin Lenses?

Thin Lenses:D = D1+D2 (shape & refractive index)D D1+D2 (shape & refractive index)

Thick Lenses:DE = D1 + D2 - (t/ng)D1D2 (shape, thickness & refractive index)

Dr.G.Mirjalili Physics Dept.Yazd UniversityThick Lens Terminology

D1 D2n1 n1

' or n g or n 2n2

'1 2

F F 'A1 A2

D1 and D2 are the first (front) and second (back) surface powers

t

111

nnD −′= 22 nnD −′

=1

1 rD =

22 r

D =

Dr.G.Mirjalili Physics Dept.Yazd UniversityD1 and D2 are the first

(front) and second (back)n n ' or n or n n2'

D1 D2

(front) and second (back) surface powers

n1 n1 or n g or n 2n2

F F 'A1 A2

′

1

111 r

nnD −′=

′

t 2

222 r

nnD −′=

A1 and A2 are the front and back vertices.

Distance from A1 to A2 is the axial thickness of the lens.

The line joining the centre of curvatures of the first and second j gsurfaces (C1 and C2) is called the principal axis, or optical axis.

f and f’ are the first and second focal points for the lens as a wholef and f are the first and second focal points for the lens as a whole.


Ray 1 H H 'Ray 1 H H '

P P 'F F '

Ray 2Ray 2

G G '


Ray 1 H H '

For Ray 1: the two refractions which take placeRay 1 H

P

H '

P 'F F '

refractions, which take place at the front and back surfaces are equivalent to

Ray 2

surfaces, are equivalent to one refraction at H’

G G '

For Ray 2: the two refractions, which take place at the front andFor Ray 2: the two refractions, which take place at the front and back surfaces, are equivalent to one refraction at G

xR di t d t d P lA1 A2

P P '

O

y

Rays directed towards P leave the lens unchanged in direction as though they had come from P’as though they had come from P

Dr.G.Mirjalili Physics Dept.Yazd UniversityBack Vertex Focal Length: (fv’)

Distance from back vertex of the lens to the second focal point of the lens

H '

P 'F F 'A2

fv '

fE 'e'

e' = f ' - fE 'e = fv - fE

Dr.G.Mirjalili Physics Dept.Yazd UniversityH '

P 'F F 'A2

fv '

1Back Vertex Power: fE '

e'

' f ' f ' vv f

D′

=′ 1

“the vergence of the light leaving the back vertex when rays parallel to the principal axis are incident on the front surface”

vf

parallel to the principal axis are incident on the front surface

( )( )

2121 DDntDDD g−+

=′ ( ) 11 DntD

gv −

=


Attention: In ‘Ophthalmic Lenses and Dispensing’ you will derive another (equivalent) expression for back vertex power:

( )[ ] 21 DDDv +=′ ( )[ ] 2

11 Dnt gv −

Dr.G.Mirjalili Physics Dept.Yazd UniversityFront Vertex Focal Length: (fv)

Distance from front vertex of the lens to the first focal point of the lens

PF A1

Gfv

e

fE

e

e = fv - fE


Front Vertex Power:

PF A1 v fD 1

−=vf

Gfv

e

“ the vergence of the incident wavefront at the front vertex of the

fE e = fv - fE

“ the vergence of the incident wavefront at the front vertex of the lens with the special condition that the emergent vergence is parallel”parallel

( )( )

21211 DDntDDD g

v

−+=−= ( ) 21 Dntf gv

v −


DE is the equivalent power

i.e. the power of the thin lens which could be used to replace the hi k lthick lens

DE = D1 + D2 - ((t/ng)D1D2) E 1 2 (( g) 1 2)

( )1 DtDD E

v =⇒(front vertex power)( ) 21 Dnt g−

DE

(back vertex power)( ) 11 DntDD

g

Ev −

=′

D 11−==

vvE ff

D =′

=


Draw in fE, fE’ fv and fv’

fE’: “ second equivalent focal length”f : “ first equivalent focal length”fE: first equivalent focal length

N t Ray 1 H H 'Note: (1) where the first and last

refractive indices are

Ray 1 H

P

H '

P 'F F 'fv f ‘vrefractive indices are the same, the values for fE and fE’ are Ray 2

P P F v v

equal in size but opposite in sign.

(2) Th i l t

G G 'fE f ’E

(2) The equivalent power DE is the power of the thin lens that could bethin lens that could be used to replace the thick lens.

Dr.G.Mirjalili Physics Dept.Yazd UniversityImage Position with a Thick Lens

Note: l2 = l1’ - t2 1

L1

A2

B1 '

B1

B1

l1

tt

l1'

Dr.G.Mirjalili Physics Dept.Yazd UniversityImage Position with a Thick Lens

Note: l2 = l1’ - t2 1

L1 L1'L2 L2'

B2 'A2

B1 ' or B2

B1

B1 or B2

l1

l2 '

t

l2

t

l1'


L1 L1'L2 L2'

Refraction at 1st surface:L ’ = L + D

B2 'A2

B1 ' or B2

B1

2 L1 = L1+ D1Object distance is l1

l2 '

l1 Image distance l1‘

l2

t

l1'


L1 L1'L2 L2'


B2 'A2

B1 ' or B2

B1


l2 '


Refraction at 2nd surface:l2

tRefraction at 2nd surface:

L2’ = L2+ D2Object distance is l

l1'Object distance is l2

Image distance is l2’Image distance is l2


L1 L1'L2 L2'


B2 'A2

B1 ' or B2

B1


l2 '







since l2 = l1’ – t L2 = n2 = n22 1 2 2 2l2 (l1’ - t)


L1 L1'L2 L2'


B2 'A2

B1 ' or B2

B1


l2 '







since l2 = l1’ – t L2 = n2 = n22 1 2 2 2l2 (l1’ - t)

(divide above and below line by n )(divide above and below line by n2)


L1 L1'L2 L2' note: n = n ’ = n

B2 'A2

B1 ' or B2

B1

2 note: n2 = n1 = ng

l2 '

l1

l2

t

l1'

(divide above and below line by n2)

L2 = n2 = 1 = 1(l ’ - t) l ’ - t l ’ - t(l1 - t) l1 - t l1 - t

n2 n2 n1’ ng


L1 L1'L2 L2' note: n = n ’ = n

B2 'A2

B1 ' or B2

B1

2 note: n2 = n1 = ng

l2 '

l1

l2

t

l1'

(divide above and below line by n2)

L2 = n2 = 1 = 1(l ’ - t) l ’ - t l ’ - t(l1 - t) l1 - t l1 - t

n2 n2 n1’ ng

(M lti l t & b tt b L ’)(Multiply top & bottom by L1’)


12

1LL

tlL′′

⋅′=

⇒

1

1

1 Lnt

nl

g

−′

′⇒

( ) 1

12 1 Lnt

LL′−

′=

This is the step along equation

( ) 11 Lnt g

The step back equation:

L( ) 2

21 1 Lnt

LLg+

=′ ( ) 2g


Usage of vergenceUsage of vergenceD t i i th P iti f th C di l• Determining the Position of the Cardinal Points (Thick lens):f `• f `E

• fE• fV• f `VV• P• P`P

Dr.G.Mirjalili Physics Dept.Yazd UniversityThick lens: a numerical calculation

Determining the Position of the Cardinal Points

1) Calculate the surface powers

2) Find the position of the second focal point (F’) and the back vertex focal length, fv’

3) Find the position of the first focal point (F) and the front vertex focal length, fvg , v

4)Calculate the equivalent power DE and first and second equivalent focal lengths fE and fE’

5) Find the position of the first principal point (P) and second principal5) Find the position of the first principal point (P) and second principal point (P’)


n1=1

F A1

r1 = +8cm

t= 3.046cm


D

n1=1 n2' = 1.33

D1 D2

F F 'A1 A2

r1 = +8cmr2 = -12cm

t= 3.046cm


n1' or ng or n2 = 1.523 1 g 2

D D

n1=1 n2' = 1.33

D1 D2

F F 'A1 A2

r1 = +8cmr2 = -12cm

t= 3.046cm

t/ng = 3.046/1.523cm =0.02m


n1' or ng or n2 = 1.523

n 1

g

n2'= 1 33

F1 F2D1 D2

n1=1 n2 1.33

F F 'A1 A2

r1 = +8cmr2 = -12cm

1) Calculate the surface powers

t= 3.046cm D1 = n1’ - n1 = 1.523 - 1r +0 08t/ng = 3.046/1.523cm =0.02m r1 +0.08

= +6.5375D

D2= n2’ - n2 = 1.33 - 1.523r 0 12r2 -0.12

= +1.6083D

Dr.G.Mirjalili Physics Dept.Yazd University2) Find the position of the second focal point (F’) and the back

vertex focal length, fv’v

To find the second focal point we need the light incident upon the lens to be parallel, i.e. L1 = 0 and l2’ = fv’

H 'L1 L1` L2`

L2

P 'F F 'A2

fv '

'fE '

e'

e' = fv ' - fE '

Dr.G.Mirjalili Physics Dept.Yazd UniversityFor the first surface:

L1’ = L1 +D1 H '1 1 1

L1 = 0

H '

P 'F F 'A2D1= +6.5375DL1’ = 0 +6.5375 = +6.5375D

P 'F F 'A2

Now we need to use the step along equation

fv '

along equation

t/n = 0 03046/1 523 = 0 02fE '

e'

t/ng 0.03046/1.523 0.02

L2 = L1’ = +6.5375L2 L1 6.53751 - (t/ng)L1’ 1- ((0.02) . 6.5375)

L2 = +7.5209D

Dr.G.Mirjalili Physics Dept.Yazd UniversityFor the second surface:

L2’ = L2 +D2L2 = +7.5209D D2= +1.6083D

⇒ L2’ = +9.1292D

L2’ = n2’/l2’ so l2’ = n2’/ L2’

n2’ = 1.33 L2’ = +9.1292D

Thus l2’ = +0.14569m

B t l ’ f ’ h th li ht i id t th fi t f i ll lBut l2’ = fv’ when the light incident on the first surface is parallel (i.e. L1 = 0)

Thus fv’= +0.14569m

Dr.G.Mirjalili Physics Dept.Yazd University3) Find the position of the first focal point (F) and the front

vertex focal length, fvv

To find the first focal point the light which emerges from the lens must be parallel, i.e. L2’ = 0 and thus l1 = fv

PF A1

Gfv

e

fE e = fv - fE


For the second surface:For the second surface:L2’ = L2 +D2⇒ L2 = L2’ - D2

PF A12 2 2

L2’ = 0 Gf2

D2= +1.6083D⇒ L2 = 0 - 1.6083 = -1.6083D

Gfv

fE

e

e = fv - fE

Now we need to use the step back equation

E e fv fE

L1’= L2 = - 1.60831 +(t/n )L 1+ ((0 02) 1 6083)1 +(t/ng)L2 1+ ((0.02). -1.6083)

L ‘ 1 6618DL1‘ = -1.6618D

Dr.G.Mirjalili Physics Dept.Yazd UniversityFor the first surface:

PF A1

L1’ = L1 +D1 ⇒ L1 = L1’ - D1

Gf

L1‘ = -1.6618DD1= +6.5375D

L 1 6618 6 5375 Gfv

fE

e

e = fv - fE

⇒ L1 = -1.6618 - 6.5375 = -8.1993D

E e fv fE

L1 = n1/l1 so l1 = n1/ L1

n1 = 1.0 L1 = -8.1993D ⇒ l1 = -0.12196m

B t l f h th li ht f th l ll lBut l1 = fv when the light emerges from the lens parallel (i.e. L2’ = 0)

Thus fv= -0.12196m


4)Calculate the equivalent power FE and the first and second i l f l l h f d f ’equivalent focal lengths fE and fE’

DE = D1 + D2 - (t/ng)D1D2 P β β β β (d/ )E 1 2 ( g) 1 2

In our example:P = β1+β2-β1β2(d/n)

D1 = +6.5375D D2 = +1.6083D

t = 0.03046m (t always +ve) n = 1.523 t/ n = 0.02t 0.03046m (t always ve) ng 1.523 t/ ng 0.02

Inputting values into equation

DE = D1 + D2 - (t/ng)D1D2

gives D = +7 9355Dgives DE= +7.9355D

Dr.G.Mirjalili Physics Dept.Yazd UniversityIn our example:

D1 = +6.5375D D2 = +1.6083D

t = 0.03046m (t always +ve) ng = 1.523 t/ ng = 0.02g g

Inputting values into equation

D D + D (t/ )D DDE = D1 + D2 - (t/ng)D1D2

gives DE= +7.9355D

DE = n2’/fE’ and DE = -n1/fE ⇒ fE’ = n2’/ DE and fE = -n1/ DE

so the second equivalent focal length (f ’)so the second equivalent focal length (fE )

fE’ = 1.33/+7.9355= 0.16760m = fE’

and the first equivalent focal length (fE)

fE = -1/+7.9355= -0.12602m = fEfE 1/+7.9355 0.12602m fE

Dr.G.Mirjalili Physics Dept.Yazd University5) Find the position of the first principal point (P) and the

second principal point (P’)

position of the second principal point is given by:position of the second principal point is given by:

e’ = fv’ - fE’v E

using e’ = fv’ - fE’

fv’ = +14.569cm fE’ = +16.760cmv E

thus e’ = -2.191cm

The -ve value indicates that the second principal point is located 2.191cm to the left of the back vertex.


thus e’ = -2.191cmthus e 2.191cm

The -ve value indicates that the second principal point is located p p p2.191cm to the left of the back vertex.

To determine the position of the fist principal point we can use:

f fe = fv - fE

f = 12 196cm f = 12 602cmfv = -12.196cm fE = -12.602cm

thus e =+0 406cmthus e +0.406cm

The +ve value indicates that the first principal point is located p p p0.406cm to the right of the front vertex


Ray 1 H H 'H H

f f ‘e e’P P 'F F 'fv f ve e’

Ray 2

G G 'fE f ’E

Dr.G.Mirjalili Physics Dept.Yazd UniversityPrincipal planes for thick lens

HH H’H’EquiEqui--convex or equiconvex or equi--concave and moderately thickconcave and moderately thick

HH H’H’ HH HH’’

PlanoPlano convex or planoconvex or plano concave lens with Rconcave lens with R22 ∞∞HH H’H’ HH H’H’

PlanoPlano--convex or planoconvex or plano--concave lens with Rconcave lens with R2 2 = = ∞∞

Dr.G.Mirjalili Physics Dept.Yazd UniversityPrincipal planes for thick lens

For meniscus lenses, the principal planes move outside For meniscus lenses, the principal planes move outside th lth lthe lensthe lensRR2 2 = = 33RR1 1 (H’ reaches the first surface)(H’ reaches the first surface)

HH H’H’ HH H’H’HH H’H’HH H’H’

Dr.G.Mirjalili Physics Dept.Yazd UniversityComparison between Thin lens

and Thick lens

• The significance (meaning) of• The significance (meaning) of principal plane 1 & 2p p p


The significance of principal plane 1Dr.G.Mirjalili Physics Dept.Yazd

UniversityThe significance of principal plane 2


Usage of vergenceUsage of vergenceD t i i th P iti f th C di l• Determining the Position of the Cardinal Points (Thin lens):f `• f `E

• fE• fV• f `VV• P• P`P

Dr.G.Mirjalili Physics Dept.Yazd UniversityThin Lens Systems & Cardinal Points

Aims & Objectives

-learn about the concept of equivalent power

- learn how to determine image location with a thin lens systemsystem

- discover the properties of principal planes of thin lensdiscover the properties of principal planes of thin lens systems

-find the cardinal points of a thin lens system

Dr.G.Mirjalili Physics Dept.Yazd UniversityCardinal planes of simple systems

1. Thin lens 1n − 1 n−

11

1r

nD =2

21

rnD =

DDD 21 +=nn 11

21

−+

−=

H, H’H, H’ Lens maker’s equationLens maker’s equation( )

rr

111

21

⎥⎤

⎢⎡ Lens maker s equation Lens maker s equation

is obeyed.is obeyed.( )rr

n

1

121⎥⎦

⎢⎣

−−=

Principal planes, nodal planes, Principal planes, nodal planes,

coincide at centercoincide at centerf1

=

Dr.G.Mirjalili Physics Dept.Yazd UniversityCardinal Points of a Thin Lens System

•Two or more thin lenses form a “thin lens system”•The principles & equations which described the thick lens alsoThe principles & equations which described the thick lens also describe thin lens systems•The equation for the equivalent power of a thin lens pair is:•The equation for the equivalent power of a thin lens pair is:

DE = D1 + D2 – t D1D2

D and D are the powers of the thin lenses and t is the distanceD1 and D2 are the powers of the thin lenses and t is the distance between them; d must be measured in metres.Th iti f th i i l i t f d i th•The positions of the principal points are found in the same way as

for the thick lens, i.e. by using the vertex powers to locate the focal points: F t t B k tpoints: Front vertex

power:D = D

Back vertex power:D ’ = DDv = DE

1- t D2

Dv = DE1- t D1

Dr.G.Mirjalili Physics Dept.Yazd UniversitySample Calculation:

A lens system consists of a +5D lens and a +2D lens separated byA lens system consists of a +5D lens and a +2D lens separated by 8cm. Find the equivalent focal length and the positions of the principal points An object 3cm tall is situated 50cm in front of theprincipal points. An object 3cm tall is situated 50cm in front of the first lens. What will be the position and size of the image produced by the combination?y

Equivalent power: DE = D1 + D2 – t D1D2

Given:D1 = +5D D2 = +2D t = +0.08m (always +ve)D1 5D D2 2D t 0.08m (always ve)

⇒ DE = +5 +2 - (0.08 . 5 . 2) = +6.2DE ( )

Second equivalent focal length (fE’):

fE’ = 1/DE ⇒ fE’ = 1/+6.2 = +0.1613m

Dr.G.Mirjalili Physics Dept.Yazd Universityfirst equivalent focal length (fE):

f 1/D f 1/+6 2 0 1613fE = -1/DE ⇒ fE = -1/+6.2 = -0.1613m

In the case of thin lens systems A1 and A2 represent the locations of the first and second thin lenses

Front vertex power of combination:Dv = DE = +6.2

1- t D2 1 - (0.8. 2)

= +7.3810D

fv (front vertex focal length) = -1/ Dv

⇒ fv = -1/+7.3810 = -0.1355m

Dr.G.Mirjalili Physics Dept.Yazd UniversityBack vertex power of combination:

D ’ D +6 2Dv’ = DE = +6.21- t D1 1 - (0.08 . 5)

= +10.3333D

fv’ (back vertex focal length) = 1/ Dv’

⇒ fv’ = 1/+10.3333 = +0.0968m

i f fi i i i (A )Location of first principal point (A1P):

e = f f = 0 1355m ( 0 1613m)e = fv - fE= -0.1355m - (-0.1613m)

= +2.58cm (i.e. to the right of the first thin lens)

Dr.G.Mirjalili Physics Dept.Yazd UniversityLocation of second principal point (A2P’):

e’ = fv’-fE’= +0.0968m - (+0.1613m)

= -6.45cm (i.e. to the left of the second thin lens)( )

What will be the position and size of the image produced by the combination?combination?Given:D1 = +5D D2 = +2D t = +0.08m

l 0 50 h 0 03l1 = -0.50m h1 = 0.03m

l = 0 50m thus L = 2Dl1 = -0.50m, thus L1 = -2D

L1’ = L1 + D1 ⇒ L1’ = -2 + 5 = +3DL1 L1 + D1 ⇒ L1 2 + 5 +3D

Dr.G.Mirjalili Physics Dept.Yazd UniversityWhat will be the position and size of the image produced by the

combination?Given:D1 = +5D D2 = +2D t = +0.08mD1 5D D2 2D t 0.08m

l1 = -0.50m h1 = 0.03m1 1

l1 = -0.50m, thus L1 = -2D

L1’ = L1 + F1 ⇒ L1’ = -2 + 5 = +3D

(Step-along equation): L2 = L1’1 d L ’1-d L1’

L2 = +3 = +3 9474DL2 +3 +3.9474D1-(0.08 . 3)

Dr.G.Mirjalili Physics Dept.Yazd UniversityL2’ = L2 + D2 ⇒ L2’ = +3.9474 + 2 = +5.9474D

l2’ = 1/L2’ ⇒ l2’ = 1/+5.9474

l2’ = +0.1681m = +16.81cm (+ve:to the right of the second lens)

Magnification:h2 ‘ = h1 . L1 . L2h2 h1 . L1 . L2

L1’ L2’

h2 ‘ = 0.03 . -2 . +3.9474+3 +5 9474+3 +5.9474

= -0.013274

= -1.3274cm (-ve sign means image is inverted)


A compound lenspTwo thin lens in contact DE = D1 + D2

T thi l t d bTwo thin lens separated by a distance t DE = D1 + D2 – t D1D2


Lens system : compound lens formed by two thin lenses

Lens 1: f = -30 cmLens 1: f1 -30 cm

Lens 2: f2 = 20 cm

t 10 cmt = 10 cm

D1=1/f1 = -3 33D

t

D1 1/f1 3.33D

D2=1/f2 = +5.0D

D = D + D t D D 1

66.65.01

33.31 2

=−

=−

= Ev D

tDDD

AA1 A2

DE = D1 + D2 – t D1D2

= 1.67 +1.67

3 33D 5.233.3

length) focal(Front cm 151

===′

−=−=

E

vv

DDD

Df

= 3.33D

Therefore fE’ = 1/DE = 30 cmcm15cm30cm15:AFor

length) focal(Back cm 40

5.2333.011

11

1

+=+−=−=+=′

+−

E

v

v

ffePf

DtD

D

fE = -1/DE = -30cm cm 10 cm 30 - cm 40 :AFor cm 15 cm 30 cm 15:AFor

22

11

+==′−′=++

Ev

Ev

ffePffeP

Dr.G.Mirjalili Physics Dept.Yazd UniversityLens system: numberical anaylsis

15 cmfE = 30 cm

A1

A2

P2

P1

F F’

tb f l 40

10 cm fE’ = 30 cmf.f.l. = 15 cm

b.f.l. = 40 cm

The two lenses can be considered as combined to form a single thick lens whose principal points and focal length are calculated. It, in turn, is combined with the third lens, and so on with each successive element.

Th d b t d d t t ith 3 4 5 lThe same procedures can be extended to systems with 3, 4, 5, …. or more lenses.

Dr.G.Mirjalili Physics Dept.Yazd UniversityMore on Diopters

L ti f ILocation of ImageTo locate an image, we need to track a minimum of two rays coming from the same point.It will be convenient to take one rayi) along the optic axis for mirrors and ii) going through the center of the lens.

U i lThe two angles are the same

Use simple geometry to show that

θi +θ0 = constant

The sum θi+θo is a constant. What does this constant represent? Geometrically, we interpret it as double the angle made by the dashed radius line. Optically, it is a measure of the strength of the mirror, i.e., how strongly the mirror focuses light, andmeasure of the strength of the mirror, i.e., how strongly the mirror focuses light, and so we call it the focal angle, θf,θi+θo = θf


θ

θ

ϕ

θc

0

[ ] )2...(....................90

)1..(..............................900

0

=++

=+

c

i

θθϕ

θϕ

[ ])4....(..........)2()1(

)3.(....................902 00

+=⇒=++

ciand θθθθθϕ

constant2)5()4()5.(..........)3()2(

0

0

===+⇒++−=⇒

fci

candθθθθ

θθθ


t1 θθ1

u

v ii

tan1

tan

00 ≈=

≈=

θθ

θθ

vu

c = 2ffc

u

fcc 212tan1

==≈= θθθc 2f

fuvfi111

0 =+⇒=+ θθθ

Example: A searchlightSuppose we need to create a parallel beam of light, as in a searchlight. Where should we place the lightbulb? A parallel beam has zero angle between its rays, so θi = 0. To place the lightbulb correctly, however we need to know a distance not an angle: the distance u between the bulb and thehowever, we need to know a distance, not an angle: the distance u between the bulb and the mirror. Since 1/v = θi = 0, it implies θ0=θf, i.e. u = f. The bulb has to be placed at a distance from the mirror equal to its focal point.

Example: DioptersAn equation like v =1/θi really doesn’t make sense in terms of units. Angles are unitless, since radians aren’t really units, so the right-hand side is unitless. We can’t have a left-hand side with units of distance if the right-hand side of the same equation is unitless In real life optometristsunits of distance if the right hand side of the same equation is unitless. In real life, optometrists define the thing θi=1/v as the “dioptric strength” of a lens or mirror, and measure it in units of inverse meters (m –1), also known as diopters (1 D = 1 m –1).

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