OpticsWe have developed a formalism which we can now applyto electromagnetic waves – lightElectromagnetic waves are oscillations of the electric (E) and magnetic (B) fields

Wave equation for electromagetic waves

oo

oo

c

wheret

E

x

E

1

2

2

2

2

oo are two constants which describe how wellwaves propagate throughelectric and magnetic mediathe o subscript tells about thepropagation in free space – the vacuum (c=299792458 ms-1).

For materials with values of relative permittivity (r) and relative permeability (r) the velocity of the lightis

n

ccv

rr

where n is the refractive index

There are two guiding principles that we shall employ extensively:

1. Huygen’s principle

Each point on a wavefront serves as the source of spherical secondary wavelets that advance with a speed and frequencyequal to those of the primary wave.

Fermat’s Principle (Pierre de Fermat)

The actual path between two points taken by a beam of light is the one which is traversed in the least time (dt/dl=0).

Stricter definitionthe optical path length must be extremal, which means that it can be either minimal, maximal or a point of inflection (a saddle point). Minima occur most often, for instance the angle of refraction a wave takes when passing into a different medium or the path light has when reflected off of a planar mirror. Maxima occur in gravitational lensing. A point of inflection describes the path light takes when it is reflected off of an elliptical mirrored surface.

Relationship between Huygen’s and Fermat’s Principles

speed of lightin a mediumis less than vacuum.Speed is characterised by index ofrefraction (n)

n=c/v

For water n=1.333 air n=1.0003

http://ephysics.physics.ucla.edu/ntnujava/propagation/ereflection_and_refraction.htm

When light strikes the boundary surface, there is a transmitted andreflected component (just as with waves on a string).

reflected

refracted

n1

n2

n1<n2

i

r

i’

i= i’ (angle of incidence = angle of reflection)

n1sin(i)=n2sin(r) (Snell’s law)

From Fermat to Snellius

n1

n2

n1<n2

1

2

d1

d2

l1

l2

L

)sin()sin(

)sin()sin(

))((

)(

)(

..

0))((

2

)1)((2)(

2

2

))(()(

timeMinimise

22

11

2

2

1

1

22/122

212/122

1

2

2/1222

1

2/1221

2

2/1222

1

2/1221

21

v

c

v

c

vv

vxLd

xL

vxd

x

ei

v

xLdxL

v

xdx

dx

dT

v

xLd

v

xdT

ttT

x

Total internal reflection

n1

n2

n1>n2

1

2

air) tofor water 6.48(/)sin(

..

)sin(

2/

)sin()sin(

'

0121

211

2

2211

nn

ei

nn

when

nn

LawsSnell

n2

n1n1>n2

Dispersionthe refractive index is slightly differentfor different wavelengths

n1

n2

n1>n2

1

2

René Descartes, French philosopher

Mirrors!

image is virtual

s s’

y y’

P P’

ray diagram

Angled Mirrors

Get multiple images

Spherical Mirrors:

image is real –light rays pass through itand image could be projectedon a screen – not true forvirtual images.

Spherical aberrations-non axial rays (paraxial)come to a different focus, and thus image is blurred.The non-paraxial rays are usuallyremoved using an aperture.

C

P

P’

'

112

..'

approx. angle small using

2

..

2

ssr

eis

l

r

l

s

l

ei

A

s rs’

l

when the object distance, s, is large compared with the radiusof curvature, r

equation)(mirror 1

'

11

length focal thecalled is '2

'

s as so

21

Fss

then

Fs

rs

rs

Concave mirror

Convex mirror

- parallel rays strike themirror and are focused at F at a distance r/2-Incoming plane wavesbecome spherical waves converging at F

-the outgoing wavefrontsappear to emanate from F behind the mirror. Rays are perpendicular to the wavefronts and appear to diverge from F

Ray diagrams

Draw 3 rays1. parallel2. focal3. radial

Note. if s < F image isbehind mirror and virtualand need a different construction

Fss

1

'

11Convention for

s is +ve/-ve if object is in front/behind mirrors’ is +ve/-ve if image is in front/behind mirrorF/r +ve/-ve if mirror is concave/convex

Ray diagram for convex mirror

Magnification:for similar triangles:

s

s

y

ym

''

negative magnification hence image is invertedpositive magnification: image is said to be erect

Better do an example or two!

Concave mirror, 40 cm radius of curvature, object 1cm high,placed 100 cm from the mirror – where is the image and what is the magnification

rC

1 cm

100 cm

F 20 cmcmssFs

Fss

25'100

4

100

1

20

111

'

1

1

'

11

25.0100

25'

s

sm

Thus image is 0.25 cm highand inverted

Concave mirror, 40 cm radius of curvature, object 1cm high,placed 10 cm from the mirror – where is the image and what is the magnification

r

C

1 cm

10 cm

F 20 cm

cmssFs

Fss

20'20

1

10

1

20

111

'

1

1

'

11

210

20'

s

sm

Thus image is 2 cm highand erect and virtual

An object is 2 cm high and 10 cm from a convex mirrorwith a radius of curvature 10 cm. (a) locate the image(b) find the image height

10

10

2

cf

333.3'

3333.010

3

10

1

5

111

'

1

1

'

11

ssFs

Fss

333.010

333.3'

s

sm

Lenses!

Believed light rays enter theeye – theory ofperspective

Apparent depth

Images formed by refraction at a single surface

Consider a spherical surface separating twomedia of different refractive indices

PP’

n2n1

s s’

Snell’s law

2211

2211

approx.) (paraxialapprox angle small use if

sinsin

nn

nn

2

1

C

l r

A

r

lnn

s

ln

s

ln

s

l

r

l

s

l

nnnn

nnnn

n

n

forsub

PACFrom

n

nACPFrom

)('

',,

approx. angle small Using

)(

)(

'

1221

1221

2112

2

1

1

1

12

12

r

nn

s

n

s

n )(

'1221

PP’

n2n1

s s’

2

1

C

l r 2

1

C

2

1

C

1

1

C

l r

A

Magnification

s s’

1

2

y

y’

sn

snm

ei

nn

s

s

s

s

y

ym

2

1

2211

1

2

1

2

'

..

sinsin

sin

sin'

tan

tan''

Sign conventionsincident rayn1

refracted rayn2

s + (real object)

s’ - (virtual image)

r,F - if radius of curvature on incident side

s - (virtual object)

s’ + (real image)

r,F + if radius of curvature on refracted side

2

1

Object is at the focus of 2Image is at the focus of 1

Thin lenses

Image is formed byrefraction at each surface separately. Consider a lens of refractive index nl with the refractive index of the medium nm.

v1 v2

A

B

C2 C1PP1

’

C1 and C2 are the centres of curvature of the surfaces Av1B andAv2B respectively.Applying usual equation for a surface….to the first surface

ss1’

)(' 11

Ar

nn

s

n

s

n mllm

In this case the image distance s1’ is negative (virtual image to the left)So rays at second surface behave as if they came from P1’, i.e. object atP1’ (image of first surface becomes object of the second).If the lens has thickness d: s2 =s1’+d

At second surface, medium on incident wave side has refractiveindex nl and refracted side of nm.

22 ' r

nn

s

n

s

n lmml

s2 is to the left and is hence positive, but s1’ is -ve (s2=-s1’-d)

)('' 21

Br

nn

s

n

sd

n lmml

Add A+B

''

11)(

' 1121 sd

n

s

n

rrnn

s

n

s

n llml

mm

For a thin lens d0 (i.e. last two terms vanish)

21

11)(

' rrnn

s

n

s

nml

mm

21

11)(

' rrnn

s

n

s

nml

mm

When s s’ f (the focal length)

21

21

11)(1

11)(

rrn

nn

f

rrnn

f

n

m

ml

mlm

This is called the lensmakers equation!

fss

1

'

11

The thin lens equation

Sign convention

Incident light Refracted light

Object (s): +veImage (s’): -ver: -ve if C is on incidentside

Object (s): -veImage (s’): +ver: +ve if C is on refractedside

Incident: r2<0 Refracted: r1>0

vefeirr

nf

..0

11)1(

1

21

n=nl/nm

Converging lens (positive lens)

Diverging lens(negative lens)

Maurits Cornelis Escher (Dutch, 1898-1972):Convex and Concave

Diverging lens

Incident: r1<0 Refracted: r2>0

vefeirr

nf

..0

11)1(

1

21

First focal point

Second focal point

P=1/f is called the powerof the lens measured in units of dioptres (1/m)

Ray tracing

F F’

Parallel rayCentral Ray

Focal Ray

y

y’

Note: central ray is undeflected as faces of lens are parallel – just like looking though a window (get slight displacement)

FF’

Parallel ray

Central ray

Focal rayy

y’

ExamplePlane – convex lens of refractive index of 1.5 and convexradius of curvature of 15 cm. What is the focal length

r2<0r1>>0

dioptresP

cmf

f

rrn

f

3.3

30

15

5.0

15

11)15.1(

1

11)1(

1

21

ExampleAn object 1.2 cm high is placed 4 cm from a double convex lens (radii of curvature 10 and 15 cm refractive index 1.5).What is the focal length.Locate the image, and perform the ray tracing. Is the imagereal or virtual, and what is its height?

r2=-15r1=10

dioptresP

cmf

f

rrn

f

3.8

12

6

5.0

15

1

10

1)15.1(

1

11)1(

1

21

C1

10 cm15 cm

C2

F F’

Focal

Parallel

Central

cmss

fss

6'6

1

4

1

12

1

'

1

1

'

11

cmhs

sm

8.12.15.1'

5.14

6'

So image is to left of lens and thusvirtual

Image is magnified

Fresnel Lenses

Systems with more than one lens

F F’

Focal

Parallel

Central

Imagine a second lens (f=+6 cm) is placed 12 cm to the right ofthe previous lens

F2 F2’

cms

s

fss

9'

9

1

18

1

6

1

'

1

1

'

11

2

222

Changed image from virtual to real

cmhorcmh

mmms

smor

x

sm tottot

9.08.15.0'9.02.175.0'

)(18

9'

12

9'21

2

22

2

When two lenses are added together (no spacing) the focal length is

21

111

fff

The eye

Far sighted

Near sighted

Apparent size of an object

The near point is the closest distance for which the eye can forma sharp image (usually about 25 cm) – this changes with age!

Optical correctionA persons near point is 75 cm – what kind of lens is required to bring it to 25 cm?

diopterp

cmf

f

fff

67.2

75.0

2

75

21

1

75

1

25

1

111

2

2

2

21

Hubble – celestial eyeEskimo nebula

The Telescope

8 inch refractor at Chabot Space and Science Centre in Oakland, California.

Job of telescope is to make objectswhich are far away appear close- and magnify them!

fs

fs

fss

'

11

'

1

1

'

11

'

11

'

1:

1

'

11

'

1

'

11

e

eeeeee

oo

ooo

s

ffsfss

fs

fss

Magnifying power

e

o

o

e

ee

oo

f

fM

f

y

f

y

''

Telescopes need to collect light-large lenses suffer from gravitational saggingSolution: reflector (Newtonian telescope)– using lighter mirrorsTelescopes classified by F-number(ratio of focal length of objective/diameter)