goemmetric optics chapter 3 biology

Chapter 3

Geometric Optics

1

Dr. Mohamed Saudy 1st. Year Biology

The Nature of Light

Before the beginning of the nineteenth century, light

was considered to be a stream of particles

The particles were either emitted by the object being

viewed or emanated from the eyes of the viewer

Newton was the chief architect of the particle theory

of light

He believed the particles left the object and stimulated the

sense of sight upon entering the eyes

2

Nature of Light – Alternative View

Huygens argued that light might be some sort of a wave motion

Thomas Young (in 1801) provided the first clear demonstration of the wave nature of light

He showed that light rays interfere with each other

Such behavior could not be explained by particles

3

Dual Nature of Light

In view of these developments, light must be

regarded as having a dual nature

Light exhibits the characteristics of a

wave in some situations and the

characteristics of a particle in other

situations

4

The Ray Approximation in

Geometric Optics

Geometric optics involves the study of the

propagation of light

It uses the assumption that light travels in a

straight-line path in a uniform medium and

changes its direction when it meets the

surface of a different medium or if the optical

properties of the medium are non-uniform

The ray approximation is used to represent

beams of light. 5

Ray Approximation

The rays are straight lines perpendicular to the wave fronts

With the ray approximation, we assume that a wave moving through a medium travels in a straight line in the direction of its rays

6

Ray Approximation, cont.

If a wave meets a barrier, we will assume that λ<<d d is the diameter of the

opening

This approximation is good for the study of mirrors, lenses, prisms, etc.

Other effects occur for openings of other sizes

7

Active Figure 35.4

Adjust the size of the opening

Observe the effects on the waves passing through

PLAY

ACTIVE FIGURE

8

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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

9

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

10

Diffuse Reflection

Diffuse reflection is

reflection from a rough

surface

The reflected rays travel in a

variety of directions

A surface behaves as a

smooth surface as long as

the surface variations are

much smaller than the

wavelength of the light

11

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

12

Law of Reflection, cont.

The angle of reflection is equal to the angle of incidence

θ1’= θ1

This relationship is called the Law of Reflection

The incident ray, the reflected ray and the normal are all in the same plane

Because this situation happens often, an analysis model, wave under reflection, is identified

13

Active Figure 35.6

Use the active figure to vary the angle of incidence

Observe the effect on the angle of reflection

PLAY

ACTIVE FIGURE

14



Refraction of Light

When a ray of light traveling through a

transparent medium encounters a boundary

leading into another transparent medium, part

of the energy is reflected and part enters the

second medium

The ray that enters the second medium is

bent at the boundary

This bending of the ray is called refraction

15

Refraction, 2

The incident ray, the reflected ray, the

refracted ray, and the normal all lie on the

same plane

The angle of refraction depends upon the

material and the angle of incidence

v1 is the speed of the light in the first medium and

v2 is its speed in the second

2 2

1 1

sin

sin

θ v

θ v

16

Refraction of Light, 3

The path of the light

through the refracting

surface is reversible

For example, a ray

that travels from A to

B

If the ray originated

at B, it would follow

the line AB to reach

point A

17

Following the Reflected and

Refracted Rays

Ray is the incident ray

Ray is the reflected ray

Ray is refracted into the lucite

Ray is internally reflected in the lucite

Ray is refracted as it enters the air from the lucite

18

Active Figure 35.10

Use the active figure

to vary the incident

angle

Observe the effect

on the reflected and

refracted rays

PLAY

ACTIVE FIGURE

19


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

20

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

21

Active Figure 35.11

Use the active figure to observe the light passing through three layers of material

Vary the incident angle and the materials

Observe the effect on the refracted ray

PLAY

ACTIVE FIGURE 22


Light in a Medium

The light enters from the left

The light may encounter an

electron

The electron may absorb

the light, oscillate, and

reradiate the light

The absorption and

radiation cause the average

speed of the light moving

through the material to

decrease

23

The Index of Refraction

The speed of light in any material is less than

its speed in vacuum

The index of refraction, n, of a medium can

be defined as

speed of light in a vacuum cn

speed of light in a medium v

24

Index of Refraction, cont.

For a vacuum, n = 1

We assume n = 1 for air also

For other media, n > 1

n is a dimensionless number greater than

unity

n is not necessarily an integer

25

Some Indices of Refraction

26

Frequency Between Media

As light travels from one medium to another, its frequency does not change Both the wave speed and

the wavelength do change

The wavefronts do not pile up, nor are created or destroyed at the boundary, so ƒ must stay the same

27

Index of Refraction Extended

The frequency stays the same as the wave

travels from one medium to the other

v = ƒλ

ƒ1 = ƒ2 but v1 v2 so λ1 λ2

The ratio of the indices of refraction of the

two media can be expressed as various

ratios

1 1 21

2 2 12

cλ v nn

cλ v nn

28

More About Index of

Refraction

The previous relationship can be simplified to

compare wavelengths and indices: λ1n1 =

λ2n2

In air, n1 = 1 and the index of refraction of the

material can be defined in terms of the

wavelengths

in vacuum

in a mediumn

λ λn

λ λ

29

Snell’s Law of Refraction

n1 sin θ1 = n2 sin θ2 θ1 is the angle of incidence

θ2 is the angle of refraction

The experimental discovery of this relationship is

usually credited to Willebrord Snell and is therefore

known as Snell’s law of refraction

Refraction is a commonplace occurrence, so identify

an analysis model as a wave under refraction

30

Snell’s Law – Example

Light is refracted into a crown glass slab

θ1 = 30.0o, θ2 = ?

n1 = 1.00 and n2 = 1.52

From Table 35.1

θ2 = sin-1(n1 / n2) sin θ1 = 19.2o

The ray bends toward the normal, as expected

31

Example 1 Green light with a wavelength of 5×10-7 m in a

vacuum enters a glass plate with refractive index 1.5.

A. What is the velocity of light in the glass?

B. What is the wavelength of the light in the glass?

Solution: A) From the definition n=c/v, the velocity in the

glass is

B) In the vacuum, v=c and n=1, thus

32

Example 2 A beam of light of wavelength 550 nm

traveling in air, is incident on a slab of transparent material.

The incident beam makes an angle of 400 with the normal,

and the refracted beam makes an angle of 260 with the

normal. Find the index of refraction of the material.

Solution Using Snell's law of refraction, and taking n1=1

for air

33

Example 3 A light ray of wavelength 589 nm traveling through air is

incident on a smooth, flat slab of crown glass of refractive index 1.52 at an

angle of 300 to the normal, as in figure. Find the angle of refraction.

Solution We rearrange Snell's law of refraction with these data n1=1 for air

and n2=1.52 for crown glass to obtain:

Because this is less than the incident angle of 300,

the refracted rays is bent toward the normal.

Its change in direction is called

the angle of deviation and is given by

12 1

2

1 0

2

sin sin

1sin 30 0.329

1.52

sin (0.329) 19.2

n

n

0 0 0

1 2 30 19.2 10.8

34

Example 4 A light beam passes from medium 1 to medium 2, with the later medium being a thick slab of material whose index of refraction is n2 as shown in figure. Show that the emerging beam is parallel to the incident beam.

Solution Let us apply Snell's law of refraction

to the upper surface:

Applying this law to the lower surface gives:

Therefore, 3=1, and the slab does not alter the direction of the beam. The beam parallel to itself by a distance d.

12 1

2

sin sinn

n

23 2

1

2 13 1 1

1 2

sin sin

sin ( sin ) sin

n

n

n n

n n

35

Prism

A ray of single-

wavelength light

incident on the prism

will emerge at angle

from its original

direction of travel

is called the angle of

deviation

F is the apex angle

36

Dispersion

For a given material, the index of refraction

varies with the wavelength of the light

passing through the material

This dependence of n on λ is called

dispersion

Snell’s law indicates light of different

wavelengths is bent at different angles when

incident on a refracting material

37

Variation of Index of

Refraction with Wavelength

The index of refraction

for a material generally

decreases with

increasing wavelength

Violet light bends more

than red light when

passing into a refracting

material

38

Refraction in a Prism

Since all the colors have different angles of deviation, white light will spread out into a spectrum Violet deviates the most

Red deviates the least

The remaining colors are in between

39

The Rainbow

A ray of light strikes a drop of water in the

atmosphere

It undergoes both reflection and refraction

First refraction at the front of the drop

Violet light will deviate the most

Red light will deviate the least

40

The Rainbow, 2

At the back surface the light is reflected

It is refracted again as it returns to the front surface and moves into the air

The rays leave the drop at various angles

The angle between the white light and the most intense violet ray is 40°

The angle between the white light and the most intense red ray is 42°

41

Active Figure 35.23

Use the active figure to vary the point at which the sunlight enters the raindrop

Observe the angles and verify the maximum angles

PLAY

ACTIVE FIGURE

42



Observing the Rainbow

If a raindrop high in the sky is observed, the red ray is seen

A drop lower in the sky would direct violet light to the observer

The other colors of the spectra lie in between the red and the violet

43

Total Internal Reflection

A phenomenon called total internal

reflection can occur when light is directed

from a medium having a given index of

refraction toward one having a lower index of

refraction

44

Possible Beam Directions

Possible directions of

the beam are indicated

by rays numbered 1

through 5

The refracted rays are

bent away from the

normal since n1 > n2

45

Critical Angle

There is a particular

angle of incidence that

will result in an angle of

refraction of 90°

This angle of incidence is

called the critical angle,

θC

2

1 2

1

sin (for )C

nθ n n

n

46

Active Figure 35.25

Use the active figure to vary the incident angle

Observe the effect on the refracted ray

PLAY

ACTIVE FIGURE

47


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 is directed from a medium of a given index of refraction toward a medium of lower index of refraction

48

Example 5 What is the critical angle for light going from glass with refractive index 1.5 into air?

Solution: Using n1=1.5 and n2=1

Thus, light incident from the glass on a glass-air boundary will be completely reflected if the angle of incidence exceeds 420.

Example 6 Find the critical angle for an air–water boundary. (The index of refraction of water is 1.33)

Solution: with the air above the water having index of refraction n2 and the water having index of refraction n1.

49

Fiber Optics

An application of

internal reflection

Plastic or glass rods

are used to “pipe” light

from one place to

another

Applications include

Medical examination of

internal organs

Telecommunications

50

Fiber Optics, cont.

A flexible light pipe is

called an optical fiber

A bundle of parallel

fibers (shown) can be

used to construct an

optical transmission line

51

Construction of an Optical

Fiber

The transparent core is

surrounded by cladding The cladding has a lower n

than the core

This allows the light in the

core to experience total

internal reflection

The combination is

surrounded by the

jacket

52

Problems

1. The wavelength of red helium–neon laser light in air is 632.8 nm.

(a) What is its frequency? (b) What is its wavelength in glass that

has an index of refraction of 1.5? (c) What is its speed in the

glass?

2. A ray of light is incident on a flat surface of a block of crown glass

that is surrounded by water. The angle of refraction is 19.6°. Find

the angle of reflection.

3. A laser in a compact disc player generates light that has a

wavelength of 780 nm in air.

I. Find the speed of this light once it enters the plastic of a compact

disc (n=1.55).

II. What is the wavelength of this light in the plastic?

Answer (v=1.94×108 m) and (n=503 nm)

53

Notation for Mirrors and

Lenses

The object distance is the distance from the object

to the mirror or lens

Denoted by p

The image distance is the distance from the image

to the mirror or lens

Denoted by q

The lateral magnification of the mirror or lens is

the ratio of the image height to the object height

Denoted by M

54

Images

Images are always located by extending

diverging rays back to a point at which they

intersect

Images are located either at a point from

which the rays of light actually diverge or at a

point from which they appear to diverge

55

Types of Images

A real image is formed when light rays pass

through and diverge from the image point

Real images can be displayed on screens

A virtual image is formed when light rays do

not pass through the image point but only

appear to diverge from that point

Virtual images cannot be displayed on screens

56

Images Formed by Flat Mirrors

Simplest possible mirror

Light rays leave the source and are reflected from the mirror

Point I is called the image of the object at point O

The image is virtual

57

Images Formed by Flat

Mirrors, 2

A flat mirror always produces a virtual image

Geometry can be used to determine the

properties of the image

There are an infinite number of choices of

direction in which light rays could leave each

point on the object

Two rays are needed to determine where an

image is formed

58


Mirrors, 3

One ray starts at point P, travels to Q and reflects back on itself

Another ray follows the path PR and reflects according to the law of reflection

The triangles PQR and P’QR are congruent

59

Active Figure 36.2

Use the active

figure to move

the object

Observe the

effect on the

image

PLAY

ACTIVE FIGURE

60




Mirrors, 4

To observe the image, the observer would

trace back the two reflected rays to P’

Point P’ is the point where the rays appear to

have originated

The image formed by an object placed in

front of a flat mirror is as far behind the mirror

as the object is in front of the mirror

|p| = |q|

61

Lateral Magnification

Lateral magnification, M, is defined as

This is the general magnification for any type of

mirror

It is also valid for images formed by lenses

Magnification does not always mean bigger, the

size can either increase or decrease

M can be less than or greater than 1

Image height

Object height

'hM

h

62

Lateral Magnification of a Flat

Mirror

The lateral magnification of a flat mirror is +1

This means that h’ = h for all images

The positive sign indicates the object is

upright

Same orientation as the object

63

Reversals in a Flat Mirror

A flat mirror produces

an image that has an

apparent left-right

reversal

For example, if you raise

your right hand the

image you see raises its

left hand

64

Reversals, cont.

The reversal is not actually a left-right

reversal

The reversal is actually a front-back reversal

It is caused by the light rays going forward toward

the mirror and then reflecting back from it

65

Properties of the Image Formed by

a Flat Mirror – Summary

The image is as far behind the mirror as the object is

in front

|p| = |q|

The image is unmagnified

The image height is the same as the object height

h’ = h and M = +1


The image is upright

It has the same orientation as the object

There is a front-back reversal in the image

66

Spherical Mirrors

A spherical mirror has the shape of a section of a

sphere

The mirror focuses incoming parallel rays to a point

A concave spherical mirror has the silvered surface

of the mirror on the inner, or concave, side of the

curve

A convex spherical mirror has the silvered surface of

the mirror on the outer, or convex, side of the curve

67

Concave Mirror, Notation

The mirror has a radius of curvature of R

Its center of curvature is the point C

Point V is the center of the spherical segment

A line drawn from C to V is called the principal axis of the mirror

68

Paraxial Rays

We use only rays that diverge from the object

and make a small angle with the principal

axis

Such rays are called paraxial rays

All paraxial rays reflect through the image

point

69

Spherical Aberration

Rays that are far from the principal axis converge to other points on the principal axis

This produces a blurred image

The effect is called spherical aberration

70

Image Formed by a Concave

Mirror

Geometry can be used to determine the magnification of the image

h’ is negative when the

image is inverted with respect to the object

'h qM

h p

71

Image Formed by a Concave

Mirror

Geometry also shows the relationship between the image and object distances

This is called the mirror equation

If p is much greater than R, then the image point is half-way between the center of curvature and the center point of the mirror p → ∞ , then 1/p 0 and q R/2

1 1 2

p q R

72

Focal Length

When the object is very far away, then p → ∞ and the incoming rays are essentially parallel

In this special case, the image point is called the focal point

The distance from the mirror to the focal point is called the focal length

The focal length is ½ the radius of curvature

73

Focal Point, cont.

The colored beams are traveling parallel to the principal axis

The mirror reflects all three beams to the focal point

The focal point is where all the beams intersect It is the white point

74

Focal Point and Focal Length,

cont.

The focal point is dependent solely on the

curvature of the mirror, not on the location of

the object

It also does not depend on the material from

which the mirror is made

ƒ = R / 2

The mirror equation can be expressed as

1 1 1

ƒp q

75

Focal Length Shown by

Parallel Rays

76

77

Example 1: Assume that a certain spherical mirror has a

focal length of +10 cm. Locate and describe the image for

object

distances of

(A) 25 cm, (B) 10 cm, and (C) 5 cm.

Solution:

A) Because the focal length is positive f=+10cm, we

expect the image to be real.

1 1 1

1 1 116.7

25 10

p q f

q cmcm q cm

The magnification of the image is given by

M is less than unity tells us that

the image is smaller than the object, and

the negative sign for M tells us that the image is inverted

16.70.668

25

q cmM

p cm

78

B) When the object distance is 10 cm, the object is

located at the focal point

1 1 1

1 1 1

10 10

p q f

qcm q cm

so that the image is formed at an infinite distance from the mirror;

that is, the rays travel parallel to one another after reflection. This

is the situation in a flashlight, where the bulb filament is placed at

the focal point of a reflector, producing a parallel beam of light.

79

C) When the object is at p=5 cm, it lies halfway between the focal

point and the mirror surface

1 1 1

1 1 110

5 10

p q f

q cmcm q cm

The image is virtual because it is located behind the mirror.

The magnification of the image is

10( ) 2

5

q cmM

p cm

The image is twice as large as the object, and

the positive sign for M indicates that the image is upright

Convex Mirrors

A convex mirror is sometimes called a

diverging mirror

The light reflects from the outer, convex side

The rays from any point on the object diverge

after reflection as though they were coming

from some point behind the mirror

The image is virtual because the reflected

rays only appear to originate at the image

point 80

Image Formed by a Convex

Mirror

In general, the image formed by a convex mirror is

upright, virtual, and smaller than the object

81

Sign Conventions

These sign conventions

apply to both concave

and convex mirrors

The equations used for

the concave mirror also

apply to the convex

mirror

82

Sign Conventions, Summary

Table

83

Ray Diagrams

A ray diagram can be used to determine the

position and size of an image

They are graphical constructions which

reveal the nature of the image

They can also be used to check the

parameters calculated from the mirror and

magnification equations

84

Drawing a Ray Diagram

To draw a ray diagram, you need to know:

The position of the object

The locations of the focal point and the center of curvature

Three rays are drawn

They all start from the same position on the object

The intersection of any two of the rays at a point locates the image

The third ray serves as a check of the construction

85

The Rays in a Ray Diagram –

Concave Mirrors

Ray 1 is drawn from the top of the object

parallel to the principal axis and is reflected

through the focal point, F


through the focal point and is reflected

parallel to the principal axis

Ray 3 is drawn through the center of

curvature, C, and is reflected back on itself

86

Notes About the Rays

The rays actually go in all directions from the

object

The three rays were chosen for their ease of

construction

The image point obtained by the ray diagram

must agree with the value of q calculated

from the mirror equation

87

Ray Diagram for a Concave

Mirror, p > R

The center of curvature is between the object and the concave mirror surface

The image is real

The image is inverted

The image is smaller than the object (reduced)

88

Ray Diagram for a Concave

Mirror, p < f

The object is between the mirror surface and the focal point



The image is larger than the object (enlarged)

89

The Rays in a Ray Diagram –

Convex Mirrors


parallel to the principal axis and is reflected

away from the focal point, F


toward the focal point and is reflected parallel

to the principal axis

Ray 3 is drawn through the center of

curvature, C, on the back side of the mirror

and is reflected back on itself 90

Ray Diagram for a Convex

Mirror

The object is in front of a convex mirror



The image is smaller than the object (reduced) 91

Active Figure 36.13

Use the active figure to Move the object

Change the focal length

Observe the effect on the images

PLAY

ACTIVE FIGURE

92


Notes on Images

With a concave mirror, the image may be either real

or virtual

When the object is outside the focal point, the image is real

When the object is at the focal point, the image is infinitely

far away

When the object is between the mirror and the focal point,

the image is virtual

With a convex mirror, the image is always virtual

and upright

As the object distance decreases, the virtual image

increases in size

93

The Image from a Convex Mirror

Example 2: An anti-shoplifting mirror, as shown in Figure, shows an

image of a woman who is located 3 m from the mirror. The focal length

of the mirror is -0.25 m. Find

(A) the position of her image and

(B) the magnification of the image

Solution: (A) We should expect to find

an upright, reduced, virtual image

94

1 1 1

1 1 10.23

3 0.25

p q f

q mm q m

(B) The magnification of the image is

0.23( ) 0.077

3

q mM

p m

The image is much smaller than the woman, and it is upright because M is positive

Images Formed by Refraction

Consider two transparent media having indices of refraction n1 and n2

The boundary between the two media is a spherical surface of radius R

Rays originate from the object at point O in the medium with n = n1

95

Images Formed by Refraction, 2

We will consider the paraxial rays leaving O

All such rays are refracted at the spherical

surface and focus at the image point, I

The relationship between object and image

distances can be given by

1 2 2 1n n n n

p q R

96

Images Formed by Refraction, 3

The side of the surface in which the light rays

originate is defined as the front side

The other side is called the back side

Real images are formed by refraction in the

back of the surface

Because of this, the sign conventions for q and R

for refracting surfaces are opposite those for

reflecting surfaces

97

Sign Conventions for

Refracting Surfaces

98

Flat Refracting Surfaces

If a refracting surface is flat, then R is infinite

Then q = -(n2 / n1)p The image formed by a flat

refracting surface is on the same side of the surface as the object

A virtual image is formed

99

Active Figure 36.18


to move the object

Observe the effect

on the location of

the image

PLAY

ACTIVE FIGURE

100


101

Example 3: A set of coins is embedded in a spherical plastic

paperweight having a radius of 3 cm. The index of refraction of

the plastic is n1=1.5. One coin is located 2 cm from the edge of

the sphere (Figure). Find the position of the image of the coin.

Solution Because n1 > n 2, where n2=1 is the index of refraction for

air, the rays originating from the coin are refracted away from the

normal at the surface and diverge outward. Hence, the image is

formed inside the paperweight and is virtual. Note that: (R=-3cm)

1 2 2 1

1.5 1 1 1.51.7

2 3

n n n n

p q R

q cmq

The negative sign for q indicates that the image is in

front of the surface the image must be virtual. The

coin appears to be closer to the paperweight surface

than it actually is.

102

Example 4: A small fish is swimming at a depth d below the

surface of a pond (Figure). What is the apparent depth of the

fish, as viewed from directly overhead?

Solution Because the refracting surface is flat, R is infinite.

Hence, we can use the Equation to determine the location

of the image with p =d

2

1

10.752

1.33

nq p d d

n

Because q is negative, the image is

virtual, as indicated by the dashed lines in

Figure.

The apparent depth is approximately

three-fourths the actual depth.

Lenses

Lenses are commonly used to form images

by refraction

Lenses are used in optical instruments

Cameras

Telescopes

Microscopes

103

Images from Lenses

Light passing through a lens experiences

refraction at two surfaces

The image formed by one refracting surface

serves as the object for the second surface

104

Locating the Image Formed by

a Lens

The lens has an index of

refraction n and two spherical

surfaces with radii of R1 and

R2

R1 is the radius of curvature

of the lens surface that the

light of the object reaches

first

R2 is the radius of curvature

of the other surface

The object is placed at point

O at a distance of p1 in front

of the first surface

105


a Lens, Image From Surface 1

There is an image formed by surface 1

Since the lens is surrounded by the air, n1 = 1

and

If the image due to surface 1 is virtual, q1 is

negative, and it is positive if the image is real

1 2 2 1

1 1 1

1 1n n n n n n

p q R p q R

106


a Lens, Image From Surface 2

For surface 2, n1 = n and n2 = 1

The light rays approaching surface 2 are in the

lens and are refracted into air

Use p2 for the object distance for surface 2

and q2 for the image distance

1 2 2 1

2 2 2

1 1n n n n n n

p q R p q R

107

Image Formed by a Thick Lens

If a virtual image is formed from surface 1,

then p2 = -q1 + t

q1 is negative

t is the thickness of the lens

If a real image is formed from surface 1, then

p2 = -q1 + t

q1 is positive

Then

1 2 1 2

1 1 1 11n

p q R R

108

Image Formed by a Thin Lens

A thin lens is one whose thickness is small

compared to the radii of curvature

For a thin lens, the thickness, t, of the lens

can be neglected

In this case, p2 = -q1 for either type of image

Then the subscripts on p1 and q2 can be

omitted

109

Lens Makers’ Equation

The focal length of a thin lens is the image

distance that corresponds to an infinite object

distance

This is the same as for a mirror

The lens makers’ equation is

1 2

1 1 1 1 1( 1)

ƒn

p q R R

110

Thin Lens Equation

The relationship among the focal length, the

object distance and the image distance is the

same as for a mirror

1 1 1

ƒp q

111

Notes on Focal Length and

Focal Point of a Thin Lens

Because light can travel in either direction

through a lens, each lens has two focal points

One focal point is for light passing in one direction

through the lens and one is for light traveling in

the opposite direction

However, there is only one focal length

Each focal point is located the same distance

from the lens

112

Focal Length of a Converging

Lens

The parallel rays pass through the lens and

converge at the focal point

The parallel rays can come from the left or right of

the lens

113

Focal Length of a Diverging

Lens

The parallel rays diverge after passing through the

diverging lens

The focal point is the point where the rays appear to

have originated

114

Determining Signs for Thin

Lenses

The front side of the

thin lens is the side of

the incident light

The light is refracted

into the back side of the

lens

This is also valid for a

refracting surface

115

Sign Conventions for Thin

Lenses

116

Magnification of Images

Through a Thin Lens

The lateral magnification of the image is

When M is positive, the image is upright and

on the same side of the lens as the object

When M is negative, the image is inverted

and on the side of the lens opposite the

object

'h qM

h p

117

Thin Lens Shapes

These are examples of

converging lenses

They have positive

focal lengths

They are thickest in the

middle

118

More Thin Lens Shapes

These are examples of

diverging lenses

They have negative

focal lengths

They are thickest at the

edges

119

Ray Diagrams for Thin Lenses

– Converging

Ray diagrams are convenient for locating the images formed by thin lenses or systems of lenses

For a converging lens, the following three rays are drawn:

Ray 1 is drawn parallel to the principal axis and then passes through the focal point on the back side of the lens

Ray 2 is drawn through the center of the lens and continues in a straight line

Ray 3 is drawn through the focal point on the front of the lens (or as if coming from the focal point if p < ƒ) and emerges from the lens parallel to the principal axis

120

Ray Diagram for Converging

Lens, p > f

The image is real

The image is inverted

The image is on the back side of the lens

121

Ray Diagram for Converging

Lens, p < f



The image is larger than the object

The image is on the front side of the lens 122

Ray Diagrams for Thin Lenses

– Diverging

For a diverging lens, the following three rays are drawn:

Ray 1 is drawn parallel to the principal axis and emerges

directed away from the focal point on the front side of the lens

Ray 2 is drawn through the center of the lens and continues in

a straight line

Ray 3 is drawn in the direction toward the focal point on the

back side of the lens and emerges from the lens parallel to the

principal axis

123

Ray Diagram for Diverging

Lens



The image is smaller

The image is on the front side of the lens

124

Active Figure 36.26

Use the active

figure to

Move the object

Change the

focal length of

the lens

Observe the

effect on the

image

PLAY

ACTIVE FIGURE

125


Image Summary

For a converging lens, when the object distance is

greater than the focal length, (p > ƒ)

The image is real and inverted

For a converging lens, when the object is between

the focal point and the lens, (p < ƒ)

The image is virtual and upright

For a diverging lens, the image is always virtual and

upright

This is regardless of where the object is placed

126

127

Example 5: A converging lens of focal length 10 cm forms images of

objects placed at

(A) 30 cm, (B) 10 cm, and (C) 5 cm from the lens.

In each case, construct a ray diagram, find the image distance and

describe the image.

Solution: (A) we construct a ray diagram as shown in Figure. The

diagram shows that we should expect a real, inverted, smaller image

to be formed on the back side of the lens.

1 1 1

1 1 115

30 10

p q f

q cmq


150.5

30

qM

p

The positive sign for the image distance tells

us that the image is indeed real and on the

back side of the lens.

Thus, the image is reduced in height by one half, and the negative sign for M tells us

that the image is inverted

128

(B) the object is placed at the focal point, the image is formed at infinity.

This is readily verified by substituting p=10 cm into the thin lens equation.

(C) The ray diagram in Figure shows that in this case the lens acts as a magnifying

glass; that is, the image is magnified, upright, on the same side of the lens as the

object, and virtual. Because the object distance is 5 cm.

1 1 1

1 1 110

5 10

p q f

q cmq

and the magnification of the image is

10( ) 2

5

qM

p

The negative image distance tells us that the image

is virtual and formed on the side of the lens from

which the light is incident, the front side. The image

is enlarged, and the positive sign for M tells us that

the image is upright

129

Example 6: A diverging lens of focal length 10 cm forms images of objects

placed at

(A) 30 cm, (B) 10 cm, and (C) 5 cm from the lens.

In each case, construct a ray diagram, find the image distance and

describe the image.

Solution: (A) constructing a ray diagram as in Figure taking the object distance to

be 30 cm. The diagram shows that we should expect an image that is virtual, smaller

than the object, and upright


7.5( ) 0.25

30

qM

p

This result confirms that the image is virtual, smaller than the object, and upright

ƒ

.

1 1 1

1 1 175

30 10

p q

q cmq

130

(B) When the object is at the focal point, the ray diagram appears as in

Figure

1 1 1

1 1 15

10 10

p q f

q cmq


5( ) 0.510

qM

p

Notice the difference between this situation and that for

a converging lens. For a diverging lens, an object at the

focal point does not produce an image infinitely far

away.

131

(C) When the object is inside the focal point, at p =5 cm, the ray

diagram in Figure, shows that we expect a virtual image that is smaller

than the object and upright

1 1 1

1 1 13.33

5 10

p q f

q cmq

3.33( ) 0.667

5

qM

p

and the magnification of the image is

This confirms that the image is virtual, smaller than the object, and upright

132

Example 7: A converging glass lens (n =1.52) has a focal length of 40 cm

in air. Find its focal length when it is immersed in water, which has an index

of refraction of 1.33.

Solution We can use the lens makers’ equation in both cases, noting that

R1 and R2 remain the same in air and water:

1 2

1 2

1 1 1( 1)( )

1 1 1( 1)( )

air

water

nf R R

nf R R

where is the ratio of the index of refraction of glass to that of water:

=1.52/1.33=1.14. Dividing the first equation by the second gives

1 1.52 13.71

1 1.14 1

3.71 3.71 (40) 148

water

air

water air

f n

f n

f f cm

1

1

n

n

The focal length of any lens is increased by a factor

when the lens is immersed in a fluid, where is the ratio of the index of

refraction n of the lens material to that of the fluid.

Fresnal Lens

Refraction occurs only at the surfaces of the lens

A Fresnal lens is designed to take advantage of this fact

It produces a powerful lens without great thickness

133

Fresnal Lens, cont.

Only the surface curvature is important in the

refracting qualities of the lens

The material in the middle of the Fresnal lens

is removed

Because the edges of the curved segments

cause some distortion, Fresnal lenses are

usually used only in situations where image

quality is less important than reduction of

weight 134

Combinations of Thin Lenses

The image formed by the first lens is located

as though the second lens were not present

Then a ray diagram is drawn for the second

lens

The image of the first lens is treated as the

object of the second lens

The image formed by the second lens is the

final image of the system

135

Combination of Thin Lenses, 2

If the image formed by the first lens lies on

the back side of the second lens, then the

image is treated as a virtual object for the

second lens

p will be negative

The same procedure can be extended to a

system of three or more lenses

The overall magnification is the product of the

magnification of the separate lenses 136

Two Lenses in Contact

Consider a case of two lenses in contact with

each other

The lenses have focal lengths of ƒ1 and ƒ2

For the first lens,

Since the lenses are in contact, p2 = -q1

1 1

1 1 1

ƒp q

137

Two Lenses in Contact, cont.

For the second lens,

For the combination of the two lenses

Two thin lenses in contact with each other are equivalent to a single thin lens having a focal length given by the above equation

2 2 2 1

1 1 1 1 1

ƒp q q q

21ƒ

1

ƒ

1

ƒ

1

138

Combination of Thin Lenses,

example

139

Combination of Thin Lenses,

example

Find the location of the image formed by lens 1

Find the magnification of the image due to lens

1

Find the object distance for the second lens

Find the location of the image formed by lens 2

Find the magnification of the image due to lens

2

Find the overall magnification of the system

140

141

Example 8: Two thin converging lenses of focal lengths f1=10 cm and

f2=20 cm are separated by 20 cm, as illustrated in Figure. An object is

placed 30 cm to the left of lens 1. Find the position and the magnification

of the final image.

Solution: To analyze the problem, we first draw a ray

diagram showing where the image from the first lens

falls and how it acts as the object for the second lens.

The location of the image formed by lens 1 is found

from the thin lens equation:

1 1 1

1

1

1 1 1

1 1 115

30 10

p q f

q cmq

The magnification of this image is 11

1

150.5

30

qM

p

The image formed by this lens acts as the object for the second lens. Thus,

the object distance for the second lens is 20 cm - 15 cm =5 cm. We again

apply the thin lens equation to find the location of the final image:

142

2 2 2

2

2

1 1 1

1 1 16.67

5 20

p q f

q cmq

The magnification of the second image is 2

2

2

6.67( ) 1.33

5

qM

p

Thus, the overall magnification of the system is

1 2 ( 0.5) (1.33) 0.667M M M Note that the negative sign on the overall magnification indicates that the final image is

inverted with respect to the initial object. The fact that the absolute value of the magnification

is less than one tells us that the final image is smaller than the object. The fact that q2 is

negative tells us that the final image is on the front, or left, side of lens 2.

Lens Aberrations

Assumptions have been:

Rays make small angles with the principal axis

The lenses are thin

The rays from a point object do not focus at a single

point

The result is a blurred image

This is a situation where the approximations used in the

analysis do not hold

The departures of actual images from the ideal

predicted by our model are called aberrations

143

Spherical Aberration

This results from the focal

points of light rays far from

the principal axis being

different from the focal points

of rays passing near the axis

For a camera, a small

aperture allows a greater

percentage of the rays to be

paraxial

For a mirror, parabolic

shapes can be used to

correct for spherical

aberration

144

Chromatic Aberration

Different wavelengths of light

refracted by a lens focus at

different points

Violet rays are refracted

more than red rays

The focal length for red light

is greater than the focal

length for violet light

Chromatic aberration can be

minimized by the use of a

combination of converging

and diverging lenses made of

different materials

145

The Camera

The photographic camera is a simple optical instrument

Components Light-tight chamber

Converging lens

Produces a real image

Film behind the lens

Receives the image

146

Camera Operation

Proper focusing will result in sharp images

The camera is focused by varying the

distance between the lens and the film

The lens-to-film distance will depend on the object

distance and on the focal length of the lens

The shutter is a mechanical device that is

opened for selected time intervals

The time interval that the shutter is opened is

called the exposure time

147

Camera Operation, Intensity

Light intensity is a measure of the rate at

which energy is received by the film per unit

area of the image

The intensity of the light reaching the film is

proportional to the area of the lens

The brightness of the image formed on the

film depends on the light intensity

Depends on both the focal length and the

diameter of the lens

148

Digital Camera

Digital cameras are similar in operation

The image does not form on photographic

film

The image does form on a charge-coupled

device (CCD)

This digitizes the image and turns it into a binary

code

The digital information can then be stored on a

memory chip for later retrieval

149

The Eye

The normal eye focuses

light and produces a sharp

image

Essential parts of the eye:

Cornea – light passes

through this transparent

structure

Aqueous Humor – clear

liquid behind the cornea

150

The Eye – Parts, cont.

The pupil

A variable aperture

An opening in the iris

The crystalline lens

Most of the refraction takes place at the outer

surface of the eye

Where the cornea is covered with a film of tears

151

The Eye – Close-up of the

Cornea

152

The Eye – Parts, final

The iris is the colored portion of the eye

It is a muscular diaphragm that controls pupil size

The iris regulates the amount of light entering the

eye

It dilates the pupil in low light conditions

It contracts the pupil in high-light conditions

The f-number of the eye is from about 2.8 to 16

153

The Eye – Operation

The cornea-lens system focuses light onto

the back surface of the eye

This back surface is called the retina

The retina contains sensitive receptors called rods

and cones

These structures send impulses via the optic

nerve to the brain

This is where the image is perceived

154

The Eye – Operation, cont.

Accommodation

The eye focuses on an object by varying the

shape of the pliable crystalline lens through this

process

Takes place very quickly

Limited in that objects very close to the eye

produce blurred images

155

The Eye – Near and Far Points

The near point is the closest distance for which the

lens can accommodate to focus light on the retina

Typically at age 10, this is about 18 cm

The average value is about 25 cm

It increases with age

Up to 500 cm or greater at age 60

The far point of the eye represents the largest

distance for which the lens of the relaxed eye can

focus light on the retina

Normal vision has a far point of infinity

156

The Eye – Seeing Colors

Only three types of color-sensitive cells are present in the retina They are called red,

green and blue cones

What color is seen depends on which cones are stimulated

157

Conditions of the Eye

Eyes may suffer a mismatch between the focusing power of the lens-cornea system and the length of the eye

Eyes may be:

Farsighted

Light rays reach the retina before they converge to form an image

Nearsighted

Person can focus on nearby objects but not those far away

158

Farsightedness

Also called hyperopia

The near point of the farsighted person is much farther away than that of the normal eye

The image focuses behind the retina

Can usually see far away objects clearly, but not nearby objects

159

Correcting Farsightedness

A converging lens placed in front of the eye can correct the condition

The lens refracts the incoming rays more toward the principal axis before entering the eye This allows the rays to converge and focus on the retina

160

Nearsightedness

Also called myopia

The far point of the nearsighted person is not infinity and may be less than one meter

The nearsighted person can focus on nearby objects but not those far away

161

Correcting Nearsightedness

A diverging lens can be used to correct the condition

The lens refracts the rays away from the principal axis

before they enter the eye

This allows the rays to focus on the retina

162

Presbyopia and Astigmatism

Presbyopia (literally, “old-age vision”) is due to a

reduction in accommodation ability

The cornea and lens do not have sufficient focusing power

to bring nearby objects into focus on the retina

Condition can be corrected with converging lenses

In astigmatism, light from a point source produces

a line image on the retina

Produced when either the cornea or the lens or both are

not perfectly symmetric

Can be corrected with lenses with different curvatures in

two mutually perpendicular directions

163

Diopters

Optometrists and ophthalmologists usually

prescribe lenses measured in diopters

The power P of a lens in diopters equals the

inverse of the focal length in meters

P = 1/ƒ

164

165

Example 9: A particular nearsighted person is unable to

see objects clearly when they are beyond 2.5 m away

(the far point of this particular eye). What should the

focal length be in a lens prescribed to correct this

problem?

Solution The purpose of the lens in this instance is to “move” an object from

infinity to a distance where it can be seen clearly. This is accomplished by

having the lens produce an image at the far point. From the thin lens

equation, we have

1 1 1

1 1 1

2.5

2.5

p q f

m f

f m

We use a negative sign for the image distance

because the image is virtual and in front of the

eye. As you should have suspected, the lens

must be a diverging lens (one with a negative

focal length) to correct nearsightedness.

Simple Magnifier

A simple magnifier consists of a single

converging lens

This device is used to increase the apparent

size of an object

The size of an image formed on the retina

depends on the angle subtended by the eye

166

The Size of a Magnified Image

When an object is placed at the near point, the angle subtended is a maximum

The near point is about 25 cm

When the object is placed near the focal point of a converging lens, the lens forms a virtual, upright, and enlarged image

167

Angular Magnification

Angular magnification is defined as

The angular magnification is at a maximum

when the image formed by the lens is at the

near point of the eye

q = - 25 cm

Calculated by

angle with lens

angle without lenso

θm

θ

max

25 cm1 m

f

168

Angular Magnification, cont.

The eye is most relaxed when the image is at

infinity

Although the eye can focus on an object

anywhere between the near point and infinity

For the image formed by a magnifying glass

to appear at infinity, the object has to be at

the focal point of the lens

The angular magnification is min

25 cm

ƒo

θm

θ

169

170

Example 10: What is the maximum magnification that

is possible with a lens having a focal length of 10 cm,

and what is the magnification of this lens when the eye

is relaxed?

Solution The maximum magnification occurs when

the image is located at the near point of the eye

max

25 251 1 3.5

10

cm cmm

f cm

When the eye is relaxed, the image is at infinity

min

25 252.5

10

cm cmm

f cm

Magnification by a Lens

With a single lens, it is possible to achieve

angular magnification up to about 4 without

serious aberrations

With multiple lenses, magnifications of up to

about 20 can be achieved

The multiple lenses can correct for aberrations

171

Compound Microscope

A compound microscope consists of two lenses Gives greater

magnification than a single lens

The objective lens has a short focal length,

ƒo< 1 cm

The eyepiece has a focal length, ƒe of a few cm

172

Compound Microscope, cont.

The lenses are separated by a distance L

L is much greater than either focal length

The object is placed just outside the focal point of

the objective

This forms a real, inverted image

This image is located at or close to the focal point of the

eyepiece

This image acts as the object for the eyepiece

The image seen by the eye, I2, is virtual, inverted and very

much enlarged

173

Active Figure 36.41


to adjust the focal

lengths of the

objective and

eyepiece lenses

Observe the effect on

the final image

PLAY

ACTIVE FIGURE

174


Magnifications of the

Compound Microscope

The lateral magnification by the objective is

Mo = - L / ƒo

The angular magnification by the eyepiece of

the microscope is

me = 25 cm / ƒe

The overall magnification of the microscope

is the product of the individual magnifications

25 cm

ƒ ƒo e

o e

LM M m

175

Other Considerations with a

Microscope

The ability of an optical microscope to view

an object depends on the size of the object

relative to the wavelength of the light used to

observe it

For example, you could not observe an atom (d

0.1 nm) with visible light (λ 500 nm)

176

Telescopes

Telescopes are designed to aid in viewing distant objects

Two fundamental types of telescopes Refracting telescopes use a combination of lenses to

form an image

Reflecting telescopes use a curved mirror and a lens to form an image

Telescopes can be analyzed by considering them to be two optical elements in a row The image of the first element becomes the object of the

second element

177

Refracting Telescope

The two lenses are arranged so that the objective forms a real, inverted image of a distant object

The image is formed at the focal point of the eyepiece p is essentially infinity

The two lenses are separated by the distance ƒo + ƒe which corresponds to the length of the tube

The eyepiece forms an enlarged, inverted image of the first image

178

Active Figure 36.42


to adjust the focal

lengths of the

objective and

eyepiece lenses

Observe the effects

on the image

PLAY

ACTIVE FIGURE

179


Angular Magnification of a

Telescope

The angular magnification depends on the focal

lengths of the objective and eyepiece

The negative sign indicates the image is inverted

Angular magnification is particularly important for

observing nearby objects

Nearby objects would include the sun or the moon

Very distant objects still appear as a small point of light

ƒ

ƒ

o

o e

θm

θ

180

Disadvantages of Refracting

Telescopes

Large diameters are needed to study distant

objects

Large lenses are difficult and expensive to

manufacture

The weight of large lenses leads to sagging

which produces aberrations

181

Reflecting Telescope

Helps overcome some of the disadvantages of refracting telescopes

Replaces the objective lens with a mirror

The mirror is often parabolic to overcome spherical aberrations

In addition, the light never passes through glass

Except the eyepiece

Reduced chromatic aberrations

Allows for support and eliminates sagging

182

Reflecting Telescope,

Newtonian Focus The incoming rays are reflected

from the mirror and converge toward point A

At A, an image would be formed

A small flat mirror, M, reflects the light toward an opening in the side and it passes into an eyepiece

This occurs before the image is formed at A

183

Examples of Telescopes

Reflecting Telescopes

Largest in the world are the 10-m diameter Keck

telescopes on Mauna Kea in Hawaii

Each contains 36 hexagonally shaped, computer-

controlled mirrors that work together to form a large

reflecting surface

Refracting Telescopes

Largest in the world is Yerkes Observatory in

Williams Bay, Wisconsin

Has a diameter of 1 m

184

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