1. Ray Optics, Mirrors, Lenses, Image and Optical Instruments
Chapter 22, 23
Ray Optics, Mirrors, Lenses, Image and Optical Instruments

2. A Brief History of Light
1000 AD
It was proposed that light consisted of tiny particles
Newton
Used this particle model to explain reflection and refraction
Huygens
1670
Explained many properties of light by proposing light was wave-like

3. A Brief History of Light, cont
Young
1801
Strong support for wave theory by showing interference
Maxwell
1865
Electromagnetic waves travel at the speed of light

4. A Brief History of Light, final
Planck
EM radiation is quantized
Implies particles
Explained light spectrum emitted by hot objects
Einstein
Particle nature of light
Explained the photoelectric effect

5. in the absence of medium.
C = f l
C: speed of light
C = 3 x 108 m/s
in vacuum
EM wave can travel
in the absence of medium.
In other medium, the
speed of light is
smaller than in vacuum.

6. 3 x 108 = f l
nano = 10-9

7. How do we see an object? Detection of light directly emitted by object
Detection of light reflected by object (most common)

8. Ray Optics – Using a Ray Approximation
Light travels in a straight-line path in a homogeneous medium until it encounters a boundary between two different media
The ray approximation is used to represent beams of light
A ray of light is an imaginary line drawn along the direction of travel of the light beams

9. The Index of Refraction
Speed of light c= 3x 108 m/s in vacuum.
Speed of light is different (smaller) in other media
The index of refraction, n, of a medium can be defined

10. Index of Refraction, cont
For a vacuum, n = 1
For other media, n > 1
n is a unitless ratio
Normal air:
Water: 1.33
Flint glass: 1.66
Diamond 2.42

11. Reflection of Light
A ray of light, the incident ray, travels in a medium
When it encounters a boundary with a second medium, part of the incident ray is reflected back into the first medium
This means it is directed backward into the first medium

12. Specular Reflection
Specular reflection is reflection from a smooth surface
The reflected rays are parallel to each other
All reflection in this text is assumed to be specular

13. Diffuse Reflection
Diffuse reflection is reflection from a rough surface
The reflected rays travel in a variety of directions
Diffuse reflection makes the road easy to see at night

14. Law of Reflection The normal is a line perpendicular to the surface
It is at the point where the incident ray strikes the surface
The incident ray makes an angle of θ1 with the normal
The reflected ray makes an angle of θ1’ with the normal

15. Law of Reflection, cont
The angle of reflection is equal to the angle of incidence
θ1= θ1’

16. Reflection and Mirrors
q1 = q1 q1
1
Law of reflection
Specular Reflection
Diffuse Reflection

17. Image forms at the point where the light rays converge.
When we talk about an image, start from an ideal point light source.
Every object can be constructed as a collection of point light sources.
VIRTUAL
IMAGE
|q| p
Image forms at the point where the light rays converge.
When real light rays converge  Real Image
When imaginary extension of L.R. converge  Virtual Image
Only real image can be viewed on screen placed at the spot.

18. For plane mirror: p = |q|
IMAGE
VIRTUAL
For plane mirror: p = |q|
How about left-right? Let’s check?
p: object distance
q: image distance

19. Parallel light rays: your point light source is very far away.
Spherical Mirror
R: radius of curvature
focal Point
f: focal length = R/2
Optical axis
concave
convex
Parallel light rays: your point light source is very far away.
Focal point:
(i) Parallel incident rays converge after reflection
(ii) image of a far away point light source forms
(iii) On the optical axis

20. Spherical Aberration Reflected rays do not converge:
Not well-defined focal point
not clear image
Spherical Aberration
f = R/2 holds strictly for a very
narrow beam.
Parabolic mirror can fix this problem.

21. Case 1: p > R p f q
P > q
Real Image

22. Case 2: p = R
p = q
Real Image

23. Case 3: f < p < R
p < q
Real Image

24. Case 4: p = f
q = infinite

25. Case 5: p < f
q < 0
Virtual Image

26. For a small object, f = R/2 (spherical mirror)
Mirror Equation
1/p + 1/q = 1/f
For a small object, f = R/2 (spherical mirror)
1/p + 1/q = 2/R
Alert!!
Be careful with the sign!!
Negative means that it is inside the mirror!!
p can never be negative (why?)
negative q means the image is formed inside the mirror
VIRTUAL
How about f?

27. Focal point inside the mirror
For a concave mirror: f > 0
Focal point inside the mirror
f < 0
1/p + 1/q = 1/f < 0 : q should be negative.

28. All images formed by a convex mirror are VIRTUAL.
1/p + 1/q = 1/f < 0 : q should be negative.
All images formed by a convex mirror are VIRTUAL.
Magnification, M = -q/p
Negative M means that the image is upside-down.
For real images, q > 0 and M < 0 (upside-down).

29. Example: An object is 25 cm in front of a concave spherical
mirror of radius 80 cm. Determine the position and
characteristics of the image.
1/p + 1/q = 1/f f = R/2 = 40 cm
Object is at the center: p = 25 cm
1/q = 1/40 – 1/25 =
q = cm < 0 (Virtual Image, 66.7 cm behind mirror)
M = -q/p = -(-66.7)/25 = 2.7
Erect, 2.7 times the size of the object

30. Example: What kind of spherical mirror must be used, and
what must be its radius, in order to give an erect image 1/5 as
Large as an object placed 15 cm in front of it?
M = -q/p  -q/p=1/5
So q = -p/5 = -15/5 = -3 cm
1/p + 1/q = 1/f  1/15 - 1/3 = 1/f
1/f = (1-5)/15
f = -15/4 = cm
R = 2 f = -7.5 cm Convex

31. Example: Where should an object be placed with reference to
a concave spherical mirror of radius 180 cm in order to form a
Real image having half its size?
M = -q/p  -q/p=-1/2
So q = p/2
f = R/2 = 90 cm
1/p + 1/q = 1/f  1/p + 2/p = 1/f
3/p=1/90
p = 270 cm

32. Refraction Details, 1
Light may refract into a material where its speed is lower
The angle of refraction is less than the angle of incidence
The ray bends toward the normal

33. Refraction Details, 2
Light may refract into a material where its speed is higher
The angle of refraction is greater than the angle of incidence
The ray bends away from the normal

34. All three beams (incident, reflected, and refracted) are in one plane.
Snell’s Law
All three beams (incident, reflected, and refracted) are in one plane. q1
q1 q2
n > 1
n1sinq1 = n2sinq2

35. q1
q1
q1> q2 q2
water

37. Total Internal Reflection
Total internal reflection can occur when light attempts to move from a medium with a high index of refraction to one with a lower index of refraction
Ray 5 shows internal reflection

38. Critical Angle
A particular angle of incidence will result in an angle of refraction of 90°
This angle of incidence is called the critical angle

39. Critical Angle, cont
For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary
This ray obeys the Law of Reflection at the boundary
Total internal reflection occurs only when light attempts to move from a medium of higher index of refraction to a medium of lower index of refraction

40. Examples of critical angles (relative to vacuum)
Substance N
Critical angle
Vacuum 1
90.0
Air
88.6
Ice
1.31
49.8
Water
1.333
48.6
Ethyl Alcohol
1.36
47.3
Glycerine
1.473
42.8
Crown glass
1.52
41.1
Sodium chloride
1.54
40.5
Quartz
1.544
40.4
Heavy flint glass
1.65
37.3
Tooth enamel
1.655
37.2
Sapphire
1.77
34.4
Heaviest flint glass
1.89
31.9
Diamond
2.42
24.4

41. qf = 2qc Fish vision qc = sin-1(1/1.33) = 49
How could fish survive from spear fishing?
Fish vision
qf = 2qc
qc = sin-1(1/1.33)
= 49

42. ncore nclad
>

44. Lenses
Converging lens
Diverging lens

45. Thin Lenses
A thin lens consists of a piece of glass or plastic, ground so that each of its two refracting surfaces is a segment of either a sphere or a plane
Lenses are commonly used to form images by refraction in optical instruments

46. Thin Lens Shapes These are examples of converging lenses
They have positive focal lengths
They are thickest in the middle

47. More Thin Lens Shapes These are examples of diverging lenses
They have negative focal lengths
They are thickest at the edges

48. Glass lens (nG = 1.52)

49. Same shape lenses: the higher n, the shorter f
The focal length of a lens is determined by the shape
and material of the lens.
Same shape lenses: the higher n, the shorter f
Lenses with same n: the shorter radius of curvature,
the shorter f
Typical glass, n = 1.52
Polycarbonate, n = 1.59 (high index lens)
Higher density plastic, n ≈ 1.7 (ultra-high index lens)

50. off the optical axis as a point light source.
Rules for Images
Trace principle rays considering one end of an object
off the optical axis as a point light source.
A ray passing through the focal point runs parallel to
the optical axis after a lens.
A ray coming through a lens in parallel to the optical
axis passes through the focal point.
A ray running on the optical axis remains on the optical
axis.
A ray that pass through the geometrical center of
a lens will not be bent.
Find a point where the principle rays or their imaginary
extensions converge. That’s where the image of the point source.

51. two focal points: f1 and f2 Parallel rays: image at infinite!!

52. Virtual image
Magnifying glass
Virtual image

53. positive negative p q f M 1/p + 1/q = 1/f M = -q/p
Lens equation and magnification
1/p + 1/q = 1/f
M = -q/p
This eq. is exactly the same as the mirror eq.
Now let’s think about the sign.
positive
negative p
real object
virtual object
(multiple lenses) q
real image
(opposite side of object)
virtual image
(same side of object) f
for converging lens
for diverging lens M
erect image
inverted image

54. two focal points: f1 and f2 Parallel beams: image at infinite!!
1/p + 1/q = 1/f
1/2f + 1/q = 1/f
1/q = 1/2f
M = -q/p = -1
two focal points: f1 and f2
1/p + 1/q = 1/f
1/f + 1/q = 1/f
1/q = 0  q = infinite
Parallel beams: image at infinite!!

55. Virtual image
Magnifying glass
1/p + 1/q = 1/f
2/f + 1/q = 1/f
1/q = -1/f
M = -(-f)/(f/2) = 2
Virtual image

56. real image (opposite side)
positive f
Example: A thin converging lens has a focal length of 20 cm.
An object is placed 30 cm from the lens. Find the image
Distance, the character of image, and magnification.
f = 20, p = 30
1/q = 1/f – 1/p
= 1/20 – 1/30
= 1/60
q = 60
real image (opposite side)
M = -q/p
= -60/30
= -2
< 0 inverted

57. Magnifier Consider small object held in front of eye Height y
Makes an angle  at given distance from the eye
Goal is to make object “appear bigger”: ' >  y 

58. Rays seen coming from here
Magnifier
Single converging lens
Simple analysis: put eye right behind lens
Put object at focal point and image at infinity
Angular size of object is , bigger!
Outgoing rays
Rays seen coming from here y   f f
Image at Infinity

59. (angular) Magnification
One can show
f must be in cm

60. Example
Find angular magnification of lens with f = 5 cm

61. Combination of Thin Lenses, example

62. Telescope View Distant Objects (Angular) Magnification M=fobj/feye
Increased Light Collection
Large Telescopes use Mirrors