1 chapter 5 geometrical optics january 21,23 lenses 5.1 introductory remarks image: if a cone of...

21
1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain point P, then P is called the image of S. Diffraction-limited image: The size of the image for a point source is not zero. The limited size of an optical system causes the blur of the image point due to diffraction effect: Geometrical optics: When <<D, diffraction effects can be neglected, and light propagates on a straight line in homogeneous media. Physical optics: When ~ D or >D, the wave nature of light must be considered. D f d

Upload: quentin-campbell

Post on 01-Jan-2016

213 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

1

Chapter 5 Geometrical opticsJanuary 21,23 Lenses

5.1 Introductory remarksImage: If a cone of rays emitted from a point source S arrives at a certain point P, then P is called the image of S.Diffraction-limited image:The size of the image for a point source is not zero. The limited size of an optical system causes the blur of the image point due to diffraction effect:

Geometrical optics:When <<D, diffraction effects can be neglected, and light propagates on a straight line in homogeneous media.Physical optics:When ~ D or >D, the wave nature of light must be considered.

D

fd

Page 2: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

2

5.2 LensesLens: A refracting device that causes each diverging wavelet from an object to converge or diverge and to form the image of the object.

Lens terminology: •Convex lens, converging lens, positive lens•Concave lens, diverging lens, negative lens

•Focal points •Real image: Rays converge to the image point•Virtual image: Rays diverge from the image point•Real object: Rays diverge from the object point•Virtual object: Rays converge to the object.

S

P

S

P

P

S

Page 3: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

3

5.2.1 Aspherical surfacesDetermining the shape of the surface of a lens: The optical path length (OPL) from the source to the output wavefront should be a constant.Example: Collimating a point source (image at infinity)

02)1(

2

constant)(

constant

constant

2222

22222

22

22

cyxcnxn

cxcnxnyx

cxnyx

xdnyx

ADnFA

ADnFAnOPL

titi

titi

ti

ti

ti

ti

The surface is a hyperboloid when nti>1, and is an ellipsoid when nti<1.

F

A(x,y)D

dx

y

nint

Aspherical lens can form perfect image, but is hard to manufacture.Spherical lens cannot form perfect image (aberration), but is easy to manufacture.

Example: Imaging a point source. The surface is a Cartesian oval.

)(42))(1( 2222 2222222 yxccdndxnyxn tititi

Page 4: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

4

5.2.2 Refraction at spherical surfacesTerminology: vertex, object distance so, image distance si, optical axis.

S

A

P

i

t

siso

lilo

CV

n1n2

R

o

o

i

i

ioi

i

o

o

ti

ti

i

io

o

i

t

i

o

i

o

l

sn

l

sn

Rl

n

l

n

l

Rsn

l

Rsn

nn

l

Rs

l

Rs

lRs

lRs

122121

21

1

)()(

sinsin

sinsin

)sin(sin

sin)sin(

Gaussian (paraxial, first-order) optics: When is small, cos ≈1, sin ≈

R

nn

s

n

s

n

io

1221 Paraxial imaging from a single spherical surface:

iiii

oooo

sRRsRRsl

sRRsRRsl

cos)(2)(

cos)(2)(

22

22

Page 5: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

5

R

nn

s

n

s

n

io

1221 Paraxial imaging from a single spherical surface:

Note:

1) This is the grandfather equation of many other equations in geometrical optics.

2) For a planar surface (fish in water): (A bear needs to know this.)

3) Magnification: (P5.6).

.1

2oi s

n

ns

.2

1

o

i

o

iT sn

sn

y

yM

Page 6: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

6

Object (first) focal length: when si = ,

.12

1 Rnn

nsf oo

Image (second) focal length: when so = ,

.12

2 Rnn

nsf ii

fo

Fo

Fi

fi

EVERYTHING HAS A SIGN!Sign convention for lenses(light comes from the left):

• so, fo + left of vertex• si, fi + right of vertex• xo + left of Fo • xi + right of Fi

• R + curved toward left• yo, yi + above axis

Virtual image (si< 0) and virtual object (so< 0):

si

VFi C

so

VC Fo

Page 7: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

7

Read: Ch5: 1-2Homework: Ch5: 1,5,6Note: In P5.1 the expression should be (so+si-x)2.Due: January 30

Page 8: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

8

January 26, 28 Thin lenses

5.2.3 Thin lensesThin lens: The lens thickness is negligible compared to object distance and image distance. Thin lens equations:

112121

112

222

111

)(

11)(

||

:surface 2nd For the

:surface1st For the

ii

lml

i

m

o

m

iio

lm

i

m

o

l

ml

i

l

o

m

sds

dn

RRnn

s

n

s

n

dsdss

R

nn

s

n

s

nR

nn

s

n

s

n

Forming an image with two spherical surfaces:

S P' P

1st surface 2nd surface

(R1, nm, nl) (R2, nl, nm)

C1

V1 V2P' P

si1

so2

nm

dsi2

R2R1

C2 nl

S

so1

Page 9: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

9

If the lens is thin enough, d → 0.Assuming nm=1, we have the thin lens equation:

21

11)1(

11

RRn

ss lio

112121 )(

11)(

ii

lml

i

m

o

m

sds

dn

RRnn

s

n

s

n

fss os

is io

limlim

21

11)1(

1

RRn

f l

fss io

111

Remember them together with the sign convention.

Gaussian lens formula:Lens maker’s equation:

Question: what if the lens is in water?

Page 10: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

10

Optical center:All rays whose emerging directions are parallel to their incident directions pass through one special common point inside the lens. This point is called the optical center of the lens.Proof:

R2

R1

C1C2

A

BO

i

t

'i

't

. and on dependnot does

//'

'sinarcsin'

sinarcsin

''emergent incident//

1

2

1

2

1

221

BAO

R

R

AC

BC

OC

OCBCACθθ

n

θθ

n

θθ

θθθθ

it

tt

ii

itti

Conversely, rays passing through O refract parallelly.

emergent incident//' :Proofsine of rule

1

1

2

2

1

2

1

2

1

2 ti θθAC

OC

BC

OC

AC

BC

R

R

OC

OC

For a thin lens, rays passing through the optical center are straight rays.Corollary: For a thin lens, with respect to the optical center, the angle subtended by

the image equals the angle subtended by the object.

Page 11: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

11

Focal plane: A plane that contains the focal point and is perpendicular to the optical axis. In paraxial optics, a lens focuses any bundle of parallel rays entering in a narrow cone onto a point on the focal plane.

Proof: 1st surface, 2nd surface

C C’

Focalplane

Focalplane

Fi

Image plane:In paraxial optics, the image formed by a lens of a small planar object normal to the optical axis will also be a small plane normal to that axis.

CS P

Imageplane

Page 12: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

12

Read: Ch5: 2Homework: Ch5: 7,10,11,15Due: February 6

Page 13: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

13

January 30 Ray diagrams

Finding an image using ray diagrams:Three key rays in locating an image point:1) Ray through the optical center: a straight

line.2) Ray parallel to the optical axis: emerging

passing through the focal point.3) Ray passing through the focal point:

emerging parallel to the optical axis.

f

x

x

f

y

y

fsx

fsx

o

ii

o

ii

oo

||

Newtonian lens equation:

2fxx io

Transverse magnification:o

i

o

iT s

s

y

yM

yo1

2

3

Fo Fi

S

P

S'

P'

O

sosi

f f xi

yi

A

B

xo

Longitudinal magnification:

22

2

Too

iL M

x

f

dx

dxM

Meanings of the signs:

+ –• so Real object Virtual object• si Real image Virtual image• f Converging Diverging lens • yo Erect object Inverted object• yi Erect image Inverted image• MT Erect image Inverted image

Page 14: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

14

Thin lens combinations

I. Locating the final image of L1+L2 using ray diagrams:

1) Constructing the image formed by L1 as if there was no L2.

2) Using the image by L1 as an object (may be virtual), locating the final image. The ray through O2 (Ray 4, may be backward) is needed.

11

1122

12

222

111

111111

111

fssf

dfs

dss

fss

fss

o

oi

io

io

io

II. Analytical calculation of the image position:

Fi2

Fo1

si1

so2d

Fi1

Fo2

d<f1, d<f2

O2O1

4

Total transverse magnification:

1111

21

2

2

1

121 )( fsfsd

sf

s

s

s

sMMM

oo

i

o

i

o

iTTT

si2 is a function of (so1, f1, f2, d)

Page 15: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

15

Back focal length (b.f.l.): Distance from the last surface to the 2nd focal point of the system.Front focal length (f.f.l.): Distance from the first surface to the 1st focal point of the system.

2121111

1212222

1111111

f.f.l.

1

1111111

b.f.l.

1

222

111

fdfsdfsfs

fdfsdfsfs

iii

ooo

sosiso

sisosi

Special cases:

1) d = f1+f2: Both f.f.l. and b.f.l. are infinity. Plane wave in, plane wave out (telescope).

2) d → 0: effective focal length f:

3) N lenses in contact:

b.f.l.

1

f.f.l.

1

21

111

fff

.1111

21 Nffff

Page 16: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

16

Read: Ch5: 2Homework: Ch5: 20,25,26,32,33Due: February 6

Page 17: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

17

5.4 Mirrors

5.4.1 Planar mirrors1) |so|=|si|.2) Sign convention for mirrors: so and si are positive when they lie to the left of the

vertex.3) Image inversion (left hand right hand).

5.4.3 Spherical mirrorsThe paraxial region (y<<R):

42

82

)(

2

3

42

22222

fxRxy

R

y

R

y

yRRxRyxR

x

y

February 2 Mirrors and prisms

Page 18: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

18

The mirror formula:

S CP

F V

if

A

f

si

so

R

(paraxial)

i

i

o

o

s

Rs

s

Rs

PA

CP

SA

SC 211

Rss io

111

,2 fss

Rff

ioio

Four key rays in finding an image point:1) Ray through the center of curvature.2) Ray parallel to the optical axis.3) Ray through the focal point.4) Ray pointing to the vertex.

Transverse magnification:

o

i

o

iT s

s

y

yM

S

P

VC

F1

2

34

Ray diagrams of mirrors:

.2

121

1

Rs

R

sRs ooo

Page 19: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

19

5.5 Prisms

Functions of prisms:1)Dispersion devices.2)Changing the direction of a light beam. 3)Changing the orientation of an image.

5.5.1 Dispersion prismsApex angle, angular deviation

]sincossinarcsin[sin

]sincossinarcsin[sin

)]sincoscos(sinarcsin[

)sin(arcsin)sinarcsin(

)()(

1122

1

1122

11

122

2121

2211

iii

ii

tt

tit

tiit

itti

θθnθ

θθn

θθn

θnθnθ

θθθθ

-θθ-θθ

i1 t1i2

t2

Page 20: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

20

Minimum deviation:

21

21

1

2

1

01it

ti

i

t

i d

d

d

d

The minimum deviation ray traverses the prism symmetrically.

2sin

2sin

sin

sin

2

2

1

1

121

21

121

21

m

t

i

mi

ti

tim

tit

it

θ

θn

θθθ

θθ

θθθ

θθ

At minimum deviation,

21 ti θθ

This is an accurate method for measuring the refractive indexes of substances.

Page 21: 1 Chapter 5 Geometrical optics January 21,23 Lenses 5.1 Introductory remarks Image: If a cone of rays emitted from a point source S arrives at a certain

21

Read: Ch5: 3-5Homework: Ch5: 54,60,61,64,65,67,68Due: February 13