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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
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
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
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
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
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
7
Read: Ch5: 1-2Homework: Ch5: 1,5,6Note: In P5.1 the expression should be (so+si-x)2.Due: January 30
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
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?
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.
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
12
Read: Ch5: 2Homework: Ch5: 7,10,11,15Due: February 6
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
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)
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
16
Read: Ch5: 2Homework: Ch5: 20,25,26,32,33Due: February 6
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
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
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
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.
21
Read: Ch5: 3-5Homework: Ch5: 54,60,61,64,65,67,68Due: February 13