Title: Optics Review
1Optics Review GT Timberlake 3/06
Hermann von Helmholtz (1821 1894)
2SNELLS LAW
3CRITICAL ANGLE
(largest refracted angle)
As qi approaches 90o, qr approaches qcritical.
n sin qi n sin qr
Let qi 90o
n sin 90 n sin qr
n n sin qr
4TOTAL INTERNAL REFLECTION
Only when
Rays go from higher to lower index
Angle of incidence greater than critical angle
5PRISMS
apex
Prisms bend rays toward the base.
base
(PD)
Prismatic Power
6D prism diopters
PD 2q
Prism diopters are about equal to 2 times the
deflection angle (up to 45 degrees)
100 tan q
7How much does a 15 PD prism bend light in degrees?
PD
tan q
Exact
100 cm
q 8.5o
OR
PD 2q
Approximate
15 2q
15
q
7.5o
2
8An angle of 45o corresponds to how many prism
diopters?
PD 2q
PD 2 x 45 90
9PRENTICES RULE
10SPHERICAL CONCAVE MIROR
q
q
r
F
optical axis
C
11MIRROR SIGN CONVENTION
CONCAVE
Radius positive
-s
Object negative
CONVEX
Radius negative
Object negative
12MIRROR EQUATIONS
P U V
Magnification
13An object is located 80 cm to the left of a
concave mirror of radius 50 cm. Calculate (a)
the power of the mirror, (b) the position of the
image, and (c) its magnification.
P U V
(a)
(b)
(c)
1
U - 1.25 D
M -0.455
4 D (-1.25 D) V
V 2.75 D
14An object is located 20 cm to the left of a
convex mirror of radius 50 cm. Calculate (a) the
power of the mirror, (b) the position of the
image, and (c) its magnification.
P U V
(a)
(b)
(c)
1
U
-5
-0.2 m
-9
U -5 D
-0.5 m
M 0.555
-4 D (-5 D) V
2
V -9 D
1
s
-0.111 m
-9
15SIGN CONVENTION
1. In diagrams light travels from left to right.
DIRECTION OF LIGHT
NEGATIVE
POSITIVE
16SIGN CONVENTION
2. Distances to the left of a surface (opposite
direction of light propagation) are negative.
Distances to the right are positive.
DIRECTION OF LIGHT
NEGATIVE
POSITIVE
17SIGN CONVENTION
Typically an object is to the left of a lens or
other surface, so its distance is negative. The
image is to the right of the lens, so the
distance is positive.
POSITIVE image distance
object
image
NEGATIVE object distance
18SIGN CONVENTION
19SIGN CONVENTION
Sign of vergence found using radius of curvature
rule.
wavefronts
image
20REFRACTION BY SINGLE SPHERICAL SURFACES
n
n
AND
s
s
21A cornea has an anterior radius of curvature of
7.7 mm and a posterior radius of curvature of 6.8
mm and a center thickness of 0.5 mm. What will
be its dioptric power if it is submerged in
water? Assume
nwater 1.333
ncornea 1.376
naqueous 1.336
22(No Transcript)
23aqueous
For anterior surface
For posterior surface
1.336 1.376
P
0.0068 m
Pp -5.88 D
Pa 5.58 D
24Pa 5.58 D
Pp -5.88 D
Pt -0.30 (-0.0119)
Pt -0.288 D
Thats why you cant see well under water.
25Convex Lens
Focal Points
26Thin Convex Lens Ray Trace
Q
F
F
M
27Virtual Images
Convex Lens
28Concave Lens
Focal Points
29Virtual Images
Concave Lens
30LENS SPHERICAL MIRROR EQUATIONS
Lateral magnification
31COMPLETE THIN LENS EQUATION
P U V
32A lens of index 1.52 has a power of -10 D in air.
What will be its power when placed in a liquid
of index 1.72?
Assume r1 r2
1.00 1.52
1.52 1.00
For lens in air
-10 D
-r
r
1.04
-0.52
0.52
-10 D
-r
r
r
33For lens in liquid
Two simultaneous equations
1.04
1.04
r
-10 D
-10 D
r
P
-10 D
-0.46
1.04
-0.46
-0.46
P
r
r
P
P 4.42
Now have a lens!
341.000
1.000
1.517
1.517
1.517
1.517
1.517
1.773
1.773
35Thin Lens Combinations
361. Where is the -1 D lens intermediate image?
2. Describe intermediate image (e.g., inverted,
real, minified)
3. What is size of intermediate image relative
to object?
4. What is location of final image?
5. Describe the final image.
6. What is size of final image relative to
object?
37Thin Lenses in Contact
Lens 1
Lens 2
F2
F1
L1
P1
L2
P2
Ptotal P1 P2
38Lens Makers Formula
f focal length
n index of refraction of lens
r1, r2 radii of curvature of front and back
surfaces
39PRINCIPAL POINTS AND PLANES
Primary Principal Plane
Primary Principal Point
H
F
40NODAL POINTS PLANES
Direction of incoming ray same as outgoing ray
q
N
N
q
Ray segment before reaching lens parallel to ray
segment after leaving lens
41THICK LENS POWER IN TERMS OF SURFACE POWERS, LENS
THICKNESS AND INDEX
Total lens power, Pt , is
42A 2 cm thick lens of index 1.60 is in air. The
radii of curvature of the first and second
surfaces are 3 cm and 5 cm respectively. What is
the total lens power?
Pt 20D 12D
Pf 20 D
Pt 32D (0.013 x 240)D
Pt 32D 3 D
Pt 29 D
Pb 12 D