Chapter 26
Geometrical Optics
Snell’s Law
Thin Lens Equation
1) Index of Refraction, n
Speed of light is reduced in a medium
Speed of light in a vacuum
c
n

Speed of light in medium
v
Air
Water
Glass
Diamond
1.000293
4/3
1.5
2.4
2) Snell’s Law
a) Reflection and Transmission
light splits at an
interface
Transmitted
ray
Transmitted
ray
(b) Refraction: Transmitted ray is bent at interface
1   2
toward normal if
n increases
1   2
away from normal
if n decreases
c) Derivation of Snell’s Law
1   2
toward normal
if n increases
1
v1 f
sin1 

h
h
v1

hf
2
v2
v2 f

sin2 

hf
h
h


sin 1 sin  2

v1
v2


but v  c /n



 n1 sin 1
 n 2 sin  2
Example: Rear-view mirror
Example: Apparent Depth
x
tan1 
d
x
d’
1

2
d
x
tan2 
d
d tan 1  d tan  2

For small angles, sin  tan
so d sin 1  d sin  2


sin 1
 d   d
sin  2
n1
d  d
n2
3) Total internal reflection
a) The concept
For small values of 1, light splits at an interface
For larger values of 1, 2 > 90º and refraction is not possible
Then all light is reflected internally
2
Note: this is only possible if n1 > n2
b) Critical incident angle
If 1   c , then  2  90º

Snell’s law:
n1 sinc  n2 sin

2
n2
sin  c 
n1
(n1  n 2 )
Some critical angles
Water-air: 49º
Glass - air: 42º
Diamond - water: 33º
Diamond - air: 24º
Why diamonds sparkle
c) Prisms (glass-air critical angle = 45º)
Prisms in binoculars
– Longer light path
– Image erect
d) Fibre optics
Low loss transmission of light, encoded signals.
Fibre optic bundles, coherent bundles
Imaging applications: endoscopy
4) Dispersion
• Index of refraction depends on wavelength
Rainbow
Sun Dogs (parhelia)
5) Image Formation
a) Seeing an object
Diffuse reflection
b) Image formation with a pinhole
Diffuse reflection
Diffuse
reflection
screen
Characteristics of pinhole imaging
– Infinite depth of field (everything in focus)
– Arbitrary magnification
– Low light (increasing size produces blurring)
Diffuse reflection
Diffuse
reflection
screen
c) Ideal lens
Characteristics of the ideal lens
– All rays leaving a point on object meet at one point on image
– Only one perfect object distance for selected image distance
(limited depth of field -- better for smaller lens)
6) Thin lenses
a) Converging - thicker in the middle
(i) Parallel coaxial rays converge at focus
Reversible
(ii) Symmetric - rays leaving focal point
emerge parallel (f’ = f)
(iii) Ray through centre undeviated
Summary of ray tracing rules for converging lens
b) Diverging - thinner in the middle
(i) parallel, coaxial rays diverge as if from focus
Reversible
(ii) symmetric - rays converging toward focus emerge
parallel
(iii) ray through centre undeviated
Summary of ray-tracing rules for diverging lens
c) Real lenses:
- usually spherical surfaces
- approximate ideal lens for small angles (paraxial
approximation)
7) Image Formation with thin
lenses (ray tracing)
(a) Converging lens - real image
Use 2 of 3 rays:
camera
/CCD sensor
(b) Converging lens - virtual image
(c) Diverging lens - virtual image
8) Thin Lens Equation
a) The equation
1 1
1
 
f di d o
b) Sign Convention (left to right)
(i) Focal Length:
f > 0 converging
f < 0 diverging
(ii) Object distance
do > 0 left of lens (real; same side as incident light)
do < 0 right of lens (virtual; opposite incident light)
(iii) Image distance
di > 0 right of lens (real; opposite incident light)
di < 0 left of lens (virtual; same side as incident light)
(iv) Image size
hi > 0 erect
hi < 0 inverted
c) Lateral magnification
Definition:
hi
m
ho
d i
From geometry (and sign convention): m 
do

9) Compound Lenses
Image of first lens is object for the second lens.
Apply thin lens equation in sequentially.
Overall magnification is the product:
hi2
hi2 ho2
hi2 h i1
m


 m1m 2
ho1
ho2 ho1
ho2 h o1



m  m1m2

Example: Problem 26.66
f1  9.0 cm
f 2  6.0 cm
d  18.0 cm
Find final image and magnification.
10) Vision and corrective lenses
a) Anatomy of the eye
120 x 106 rods - detect intensity: slow, mono, sensitive
6 x 106 cones - detect frequency: R - 610 nm, G - 560 nm, B - 430 nm
b) Optics
- Accomodation: focal length changes with object distance
- near point: nearest point that can be accomodated
- normally < 25 cm
- far point: furthest point that can be accomodated
- normally ∞
c) Myopia
- far point < ∞
- near-sighted (far-blind)
- correction: object at ∞ --> image at far point
Correction: object at ∞ --> image at far point
1
1
1


f do d i

1 1
1
 
f  FP
f  FP

(ignoring the eye-lens distance)


d) Refractive Power
Refractive power in diopters
1

f (in meters)
For a far point of 50 cm, f = -50 cm,
Lens prescription: -2
e) hyperopia (hypermetropia)
- near point > 25 cm
- far-sighted (near-blind)
- correction: object at 25 cm --> image at near point
Correction: object at 25 cm --> image at near point
1
1
1


f do d i


1
1
1


f 25 cm NP
(25cm)NP
f 
0
NP  25cm
(ignoring the eye-lens distance)
For near point of 40 cm, f = 66 cm
Power = + 1.5 (reading glasses)
Examples:
Problem 26.73
Age 40: f = 65.0 cm --> NP’ = 25.0cm
Age 45: NP’ --> 29.0 cm
(a) How much has NP (without glasses) changed?
(b) What new f is needed?
Problem 26.75
FP = 6.0 m corrected by contact lenses. (Find f)
An object (h = 2.0 m) is d = 18.0 m away.
(a)
Find image distance with lenses.
(b)
Find image height with lenses.
11) Angular Magnification
a) Angular size
h

d
b) Angular magnification

Angular size with optical device
M 
 Angular size without optical device

12) Magnifier
With magnifier:
ho
 
do

1
1
1
where


f do d i
1 1 
so   ho   
 f di 
(Magnifier allows object to be close to the eye)
Without magnifier:
ho

N

1 1 

so M =  N   

 f di 
1 1 
We had   ho   
 f di 
Highest magnification (di = -N):

Lowest magnification (di = -∞):

N
M  1
f
N
M 
f
(tense eye)
(relaxed eye)
(Magnification quoted with N = 25 cm, for relaxed eye)

Example:
Problem 26.82
Farsighted person has corrective lenses with f = 45.4 cm.
Maximum magnification of a magnifier is 7.50 (normal vision).
What is the maximum magnification of the magnifier for the farsighted
person without lenses?
13) Compound Microscope
• Simple magnifier: M = N/f
– to increase M, decrease f
– practical limits to decreasing f (and therefore size):
• small lens difficult to manufacture and use
• increases aberrations
• Microscope introduces an additional lens to
form a larger intermediate image, which
can be viewed with a magnifier
Magnification:
 hi1 di2

M
ho N

For image at ∞, di2 = fe

hi1 f e hi1 N 
M
  h N  h  f 
 o  e 
o
 m oM e
do1  f o


L
di1 N 
M 
 
 do1  f e 

For max M, do1  fo
For di2 = ∞, di1 + fe = L

N L  f e 
M   

 f e  f o 
Example:
Problem 26.88
Microscope with fo = 3.50 cm, fe = 6.50 cm, and L = 26.0 cm.
(a) Find M for N = 35.0 cm.
(b) Find do1 (if first image at Fe)
(c) Find lateral magnification of the objective.
14) The Astronomical Telescope
Qu i c k T i m e ™ a n d a
T I F F (Un c o m p re s s e d ) d e c o m p re s s o r
a re n e e d e d to s e e th i s p i c t u re .
• Magnifier requires do < f, but do
-> ∞ for stars
• Introduce objective to form nearby
image, then use magnifier on the
image
For do  , di  f o
Magnification:

hi f e
 hi f e


M
ho do hi f o


 fo
M
fe


Long telescope, small eyepiece
Example:
Problem 26.94
Yerkes Observatory: fo = 19.4 m, fe = 10.0 cm.
(a) Find angular magnification.
(b) If ho = 1500 m (crater), find hi, given do = 3.77 x 108 m
(c) How close does the crater appear to be.
Galilean Telescope (Opera glasses)
Reflecting Telescope