optics we have developed a formalism which we can now apply to electromagnetic waves – light...
TRANSCRIPT
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)