Lenses

Plane Surface

Snell’s Law

Point Object (n’>n as shown)

q

AA’

-s’

q’

h

Angles

Small-Angle Approximation

n n’

s

Plane Surface

qAA’

-s’s

q’

x’x

B’ B

Extended Object

Angles

Small-Angle Approximation

Transverse Magnification

Snell’s Law

Spherical Surface

A A’

q’

q

a bg

s

R

C

d

h

Exterior Angles of Triangles

Tangents of Angles

Small-Angle Approximation

Spherical Surface

Back Focal Point Object at Infinity

Front Focus Image at Infinity

Optical Power

Units: m-1 = diopters

F’

f’

f

F

Spherical Surface

A

A’

B

B’

q’q

x

-x’

Angles

s’

s

Snell’s Law &Small Angles

TransverseMagnification

n n’

Object Conventions

so

object is REAL when rays diverge from object:

so > 0

object is VIRTUAL when rays converge to object:

so < 0

usually only with lens combinationsso

Image Conventions

si

image is REAL when rays converge :

si > 0

image is VIRTUAL when rays diverge :

si < 0

rays project back to the imagesi

rays focus on the image

R Conventions

R1

R2

R1

R2

R > 0 when line lands on right R < 0 when line lands on left

R1 > 0

R2 < 0

R1 > 0

R2 < 0

f Conventions

f

lens is CONVERGING when rays converge:

f > 0

lens is DIVERGING when rays diverge:

f < 0

f

f fcheck rays from

The Simple Lens (1)

A1 A1’A2

s1’R1

s1

Two Surfaces: Air-Glass Glass-Air

Find Image from FirstSurface:

1

n1’ = n2n1n2’

d

The Simple Lens (2)

A1’A2

s1’-s2

-s2+d = s1’ s2 = d-s1’

Object Distance for SecondSurface:

n1’ = n2n1

n2’2

3

R2 s’2

Find Image from Second Surface:

A2’

Note VirtualObject

d

The Simple Lens (3)

Summarize

n1’ = n2n1

n2’4

A2’

A1

s1

w’

w s’2s2 = d-s1’

Note, w for workingdistance instead of s.This is important later

Thin Lens (1)

+

‘

Thin Lens (2)

Front Focal LengthBack Focal Length

f f’

Special Case: Thin Lens in Air

Lens Makers Equation with d = 0Lens Equation

f f’

ExerciseWhat must be the radius of curvature of a plano-convex lens if the parallel bundle of rays is to come to a focus 100cm from the vertex? The glass lens (n=1.46) is immersed in ethyl alcohol (n=1.36).

ExerciseSuppose that we have a glass rod (ng = 1.50) surrounded by air with the left end ground to a convex hemispherical of 2cm radius. If a point source is located 6cm to the left of the hemisphere’s vertex, where will its image appear?

If the glass rod is immersed in water (nw = 1.33) determine the new location of the image of the point source.

ExerciseCompute the focal length of the bi-concave thin lens, assuming it to be made of fluorite (nl = 1.43) immersed in carbon disulfide (nm = 1.63). It is given that the radii for the first surface is 10cm and the second surface is 20cm.

Repeat the above problem but using nl = 1.66 and nm = 1.33

Exercise

A compound lens consists of two thin bi-convex lenses L1 and L2 of focal lengths 10cm and 20cm, separated by a distance of 80cm. Describe the image corresponding to a 5cm tall object 15cm from the first lens.

Stops and pupils

Stops in optical system Brightness of the image is determined primarily

by the size of the bundle of rays collected by the system (from each object point)

Stops can be used to reduce aberrations.

Stops in optical system How much of the object we see is determined by

the field of view.

Rays from Q do not pass through system. We can only see object points closer to the axis of

the system. Field of view is limited by the system.

Aperture stop

A stop is an opening (despite its name) in a series of lenses, mirrors, diaphragms, etc.

The stop itself is the boundary of the lens or diaphragm

Aperture stop: an element of the optical system that limits the cone of light from any particular object point on the axis of the system

Aperture stop

Entrance pupil (EnP)

o is defined to be the image of the aperture stop in all the lenses preceding it (i.e. to the left of AS - if light travels left to right)

How big does theaperture stop lookto someone at O

EnP – defines thecone of raysaccepted by thesystem

Exit pupil (ExP) The exit pupil is the image of the aperture stop in

the lenses coming after it (i.e. to the right of the AS)

Location of Aperture Stop (AS)

In a complex system, the AS can be found by considering each element in the system.

The element which gives the entrance pupil subtending the smallest angle at the object point O is the AS.

Chief Ray Chief ray is defined as the light ray which

passes through the centre of the aperture stop

after refraction, the chief ray will also pass through the centre of the exit and entrance pupils

ABCD Matrix Concepts

Ray Description Position Angle

Basic Operations Translation Refraction

Two-Dimensions Extensible to

Three

Ray Vector

Matrix Operation

System Matrix

Ray Definition

x1

a1

Translation Matrix

Slope Constant Height Changes

1

112

1

1

2

2

10

1xx

n

zx

x1

a1=a2

z

x2

Where 1= n1 and 2= n2 are the optical direction cosine

Refraction Matrix

Height Constant Slope Changes

111

'11'

1

xR

nn

1

112

1

1

1

'11'

1

'1

1

01

xx

R

nnx

x1

a2

a1

(q w.r.t. normal)

n1 n1’

Refraction Matrix

1

112

1

1

1'1

'1

1

01

xx

P

x

1

1

1

'11'

1

'1

1

01

x

R

nnx

1

0112 power

Previous Result

Recall Optical Power

Cascading Matrices

'1

'

12

2

2 1

xx

2

22'

2

'2

xx

1

11'

1

'1

xx

1

11122'

2

'2

xx

Generic Matrix:

Determinant (You can show thatthis is true for cascaded matrices)

V1 R1 T12 R2 V’2

Light Travels Left to Right, butBuild Matrix from Right to Left

1

112212

The Simple Lens (Matrix Way)

z12

Front Vertex,V Back Vertex, V’

Index = n Index = nL Index = n’

Building The Simple Lens Matrix

1

01

10

1

1

01

11

12

2 Pn

z

P l

1

01

10

1

1

01

11

12

2 Pn

z

P lSimple Lens Matrix

z12

V V’

n nL n’

M’ = 2121

The Thin Lens

1

01

10

1

1

01

11

12

2 Pn

z

P l

1

01'

tPM

Simple Lens Matrix

Pt = P1 + P2

Thin Lens in Air

Thick Lens Compared to Thin

1

01

10

1

1

01

11

12

2 Pn

z

P l

221

1121

1

01

PPP

P

n

z

P lt

z12

V V’

n nl n’

The Thick Lens in Air

Thick Lens Power is given by thin lens with correction

Lensmaker’s Equation

z12

V V’

n nL n’

Matrices for Optical Components

Free space

Refraction at a planar boundary

Refraction at a spherical boundary

Transmission through thin lens

Reflection from a planar mirror

Reflection from a spherical mirror

10

1n

d

10

01

1

0112

R

nn

1

101

f

10

01

1

201

R

Properties of a system from its matrix

If D = 0, all rays entering the input plane from the same point emerge at the output plane making the same angle with the axis. The input plane must be the focal plane of the system.

If B = 0, all rays leaving the point O at the input will pass through the same point I at the output. This is the condition for object-image

relationship to occur. In addition, A or 1/D will give the

transverse magnification produced by the system.

Properties of a system from its matrix If C = 0, all rays which enter the system parallel

to one another will emerge parallel to one another in a new direction. This is a telescopic system.

If A = 0, all rays entering the system at the same angle will pass through the same point in the output plane. The system brings a bundle of parallel rays to a focus at the output plane.

Exercise

A glass rod 2.8 cm long and of index 1.6 has both ends ground to spherical surfaces of radius 2.4 cm, and convex to air. An object 2 cm tall is located on the axis, in the air, 8 cm to the left of the left-hand vertex. Using matrix method, find the position and size of the final image.

Exercise

A parallel beam of light enters a clear plastic spherical bead whose diameter is 2cm and refractive index 1.4. At what point beyond the bead is the light brought to a focus?

Exercise

A positive (converging) lens of focal length +9 cm is mounted 7 cm to the left of a negative lens of focal length –11 cm. If an object 3cm tall is located on the axis 24cm to the left of the positive lens, find the position and size of the image.