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

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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

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Young
• 1801
• Strong support for wave theory by showing interference
Maxwell
• 1865
• Electromagnetic waves travel at the speed of light

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Planck
• EM radiation is quantized
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Implies particles
• Explained light spectrum emitted by hot objects
Einstein
• Particle nature of light
• Explained the photoelectric effect

C = f l
C: speed of light
C = 3 x 10 8 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.

3 x 10 8 = f l nano = 10 -9 http://laxmi.nuc.ucla.edu:8248/M248_99/iphysics/spectrum.gif

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Detection of light directly emitted by object
Detection of light reflected by object
(most common)

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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

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Speed of light c= 3x 10 8 m/s in vacuum.
Speed of light is different (smaller) in other media
The index of refraction, n, of a medium can be defined
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

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For a vacuum, n = 1
For other media, n > 1 n is a unitless ratio
Normal air: 1.0003
Water: 1.33
Flint glass: 1.66
Diamond 2.42

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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

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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

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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

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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 θ with the normal
1
The reflected ray makes an angle of θ with the normal
1
’

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The angle of reflection is equal to the angle of incidence
θ
1
= θ
1
’

Reflection and Mirrors q
1 q
1 q
1
= q
1
Law of reflection
Specular Reflection Diffuse Reflection

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 p |q|
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.

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

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

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.

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

Case 2: p = R p = q
Real Image

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

Case 4: p = f q = infinite

Case 5: p < f q < 0
Virtual Image

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?

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

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).

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
= -0.015
q = -66.7 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

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 = -3.75 cm
R = 2 f = -7.5 cm Convex

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

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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

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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

Snell’s Law
All three beams (incident, reflected, and refracted) are in one plane.
q
1 q
1 n > 1 n
1 sin q
1
= n
2 sin q
2 q
2

water q
1 q
1 q
1
> q
2 q
2

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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

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A particular angle of incidence will result in an angle of refraction of 90°
• This angle of incidence is called the critical angle sin q  n
2 n
1 for n
1
 n
2

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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

sin q
Examples of critical angles (relative to vacuum) crit
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1 n
Substance
Vacuum
Air
Ice
Water
Ethyl Alcohol
Glycerine
Crown glass
Sodium chloride
Quartz
Heavy flint glass
Tooth enamel
Sapphire
Heaviest flint glass
Diamond
N
1
1.00029
1.31
1.333
1.36
1.473
1.52
1.54
1.544
1.65
1.655
1.77
1.89
2.42
Critical angle
90.0
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88.6
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49.8
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48.6
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47.3
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42.8
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41.1
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40.5
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40.4
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37.3
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37.2
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34.4
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31.9
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24.4


How could fish survive from spear fishing?
Fish vision q f
= 2 q c q c
= sin -1 (1/1.33)
= 49 

> n core n clad

Lenses
Converging lens
Diverging lens

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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

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These are examples of
converging lenses
They have positive focal lengths
They are thickest in the middle

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These are examples of
diverging lenses
They have negative focal lengths
They are thickest at the edges

Glass lens (n
G
= 1.52)

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)

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 .

two focal points: f
1 and f
2
Parallel rays: image at infinite!!

Virtual image
Magnifying glass
Virtual image

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

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

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

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

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

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Single converging lens
• Simple analysis: put eye right behind lens
• Put object at focal point and image at infinity
• Angular size of object is q , bigger!
Outgoing rays f q q f y
Rays seen coming from here
Image at
Infinity
1
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1
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1 q f p

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One can show m q
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25
(standard) f m q

25
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1 (maximum) f
 f must be in cm

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Find angular magnification of lens with f = 5 cm m q m q
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25
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5
5 Standard

25
1 6 Maximum
5

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View Distant Objects
(Angular) Magnification M=f obj
/f eye
Increased Light Collection
Large Telescopes use Mirrors