Geometrical Optics

• The objects of our daily life (>mm) are much larger than the wavelengths of light (500 nm) with which we observe them

• The spatial resolution of our eye, in resolving images or boundaries between light and shadow is also much coarser than the wavelengths of visible light.

• This separation of scale leads to a number of simplifications, described by the title Geometrical Optics.

• This is not true of sound waves in our daily lives.– Speed of sound = 340 m/s– Concert A, f = 440 Hz, = v/f = (340 m/s) /(440 /s) = 0.77 m– We can hear around corners we cannot see around!

Geometrical Optics

In describing the propagation of light as a wave we need to understand:

wavefronts: a surface passing through points of a wave that have the same phase.

rays: a ray describes the direction of wave propagation. A ray is a vector perpendicular to the wavefront.

Light Rays

• The propagation of the wavefronts can be described by light rays.

• In free space, the light rays travel in straight lines, perpendicular to the wavefronts.

Shadows

• In geometrical optics, apertures cast sharp shadows

Light rays from point source screen

Image formation with a Pinhole camera

Image is fuzzy if pinhole is larger

Image is sharper if pinhole is smaller

Image is dimmer is pinhole is smaller

o

i

o

i

d

d

h

hm

heightobject

height image

Light source

Dark room = Camera obscura

do = distance from object to pinhole

di = distance from pinhole to image

ho = height of object

hi = height of image

Object

Image

Fuzziness of image from size of pinhole

• A larger pinhole brings in more light, but produces a fuzzy image.

Light source

Dark room = Camera obscura

Object

Image

Fuzzy Image,from finite aperture

Reflection and RefractionWhen a light ray travels from one medium to

another, part of the incident light is reflected and part of the light is transmitted at the boundary between the two media.

The transmitted part is said to be refracted in the second medium.

incident ray reflected ray

refracted ray

Types of Reflection

If the surface off which the light is reflected is smooth, then the light undergoes specular reflection (parallel rays will all be reflected in the same directions).

If, on the other hand, the surface is rough, then the light will undergo diffuse reflection (parallel rays will be reflected in a variety of directions)

The Law of Reflection

For specular reflection the incident angle i equals the reflected angle r

i r

The angles are measured relative to the normal, shown here as a dotted line.

Forming Images with a Plane MirrorA mirror is an object that reflects light. A plane mirror

is simply a flat mirror. Consider an object placed at point P in front of a plane

mirror. An image will be formed at point P´ behind the mirror.

do diFor a plane mirror:

do = -di and ho = hi

hohi

do = distance from object to mirror

di = distance from image to mirror

ho = height of object

hi = height of image

Image is behind mirror: di < 0

Images

An image is formed at the point where the rays of light leaving a single point on an object either actually intersect or where they appear to originate from.

If the light rays actually do intersect, then the image is a real image. If the light only appears to be coming from a point, but is not physically there, then the image is a virtual image.

We define the magnification, m, of an image to be:

o

i

o

i

d

d

h

hm

heightobject

height image

If m is negative, the image is inverted (upside down).

Plane MirrorsA plane mirror image has the following properties:• The image distance equals the object distance ( in magnitude )

• The image is unmagnified

• The image is virtual: negative image distance di < 0

• m>0, The image is not inverted do di

o

i

o

i

d

d

h

hm

heightobject

height image

To save expenses, you would like to buy the shortest mirror that will allow you to see your entire body. Should the mirror be

(a) half your height

(b) two-thirds your height, or

(c) equal to your height?

To save expenses, you would like to buy the shortest mirror that will allow you to see your entire body. Should the mirror be

(a) half your height

(b) two-thirds your height, or

(c) equal to your height?

To save expenses, you would like to buy the shortest mirror that will allow you to see your entire body. Should the mirror be half your height two-thirds your height, or equal to your height?

Does the answer depend on how far away from the mirror you stand?

a) Yes b) No

To save expenses, you would like to buy the shortest mirror that will allow you to see your entire body. How tall should the mirror be?

Does the answer depend upon the distance from your eyes to the top of your head?

a) Yes b) No

To save expenses, you would like to buy the shortest mirror that will allow you to see your entire body. How tall a mirror do you need?

Does the answer depend upon how high you hang the mirror?

a) Yes b) No

Spherical Mirrors

A spherical mirror is a mirror whose surface shape is spherical with radius of curvature R. There are two types of spherical mirrors: concave and convex.

We will always orient the mirrors so that the reflecting surface is on the left. The object will be on the left.

concave

convex

Focal PointWhen parallel rays (e.g. rays from a

distance source) are incident upon a spherical mirror, the reflected rays intersect at the focal point F, a distance R/2 from the mirror. (Locally, the mirror is a flat surface, perpendicular to the radius drawn from C, at an angle from the axis of symmetry of the mirror).

For a concave mirror, the focal point is in front of the mirror (real).

For a convex mirror, the focal point is behind the mirror (virtual).

The incident rays diverge from the convex mirror, but they trace back to a virtual focal point F.

Focal LengthThe focal length f is the distance

from the surface of the mirror to the focal point.

CF = FA = radius = FM

A

M

The focal length FM is half the radius of curvature of a spherical mirror.

Sign Convention: the focal length is negative if the focal point is behind the mirror.

For a concave mirror, f = ½R

For a convex mirror, f = ½R (R is always positive)

Ray TracingIt is sufficient to use two

of four principal rays to determine where an image will be located.

The parallel ray (P ray) reflects through the focal point. The focal ray (F ray) reflects parallel to the axis, and the center-of-curvature ray (C ray) reflects back along its incoming path. The Mid ray (M ray) reflects with equal angles at the axis of symmetry of the mirror.

M ray

Ray Tracing – Examples

concave convex

Virtual imageReal image

Put film here for

Sharp image.

The Mirror EquationThe ray tracing technique shows

qualitatively where the image will be located. The distance from the mirror to the image, di, can be found from the mirror equation:

fdd io

111

do = distance from object to mirror

di = distance from image to mirror

f = focal length

m = magnification

Sign Conventions:

do is positive if the object is in front of the mirror (real object)

do is negative if the object is in back of the mirror (virtual object)

di is positive if the image is in front of the mirror (real image)

di is negative if the image is behind the mirror (virtual image)

f is positive for concave mirrors

f is negative for convex mirrors

m is positive for upright images

m is negative for inverted imageso

id

dm

Example 1An object is placed 30 cm in front of a concave mirror of radius 10

cm. Where is the image located? Is it real or virtual? Is it upright or inverted? What is the magnification of the image?

cmd

cmcmcmcmd

cmcmdfd

cmd

fdd

cmRf

i

i

i

o

i

6

6

1

30

5

30

1

30

61

30

1

5

1111

30

111

52/

0

0

di>0 Real Image

m = di / do = 1/5

Example 2An object is placed 3 cm in front of a concave mirror of radius 20

cm. Where is the image located? Is it real or virtual? Is it upright or inverted? What is the magnification of the image?

43.1/

29.4

30

7

30

10

30

31

3

1

10

1111

3

111

102/

0

0

oi

i

i

i

o

i

ddm

cmd

cmcmcmd

cmcmdfd

cmd

fdd

cmRf

Virtual image, di <0

Magnified, |m| > 1,

not inverted. m > 0

Example 3An object is placed 5 cm in front of a convex mirror of focal length

10 cm. Where is the image located? Is it real or virtual? Is it upright or inverted? What is the magnification of the image?

66.0/

33.3

10

3

10

2

10

11

5

1

10

1111

5

111

102/

0

0

oi

i

i

i

o

i

ddm

cmd

cmcmcmd

cmcmdfd

cmd

fdd

cmRf

Virtual image, di <0

De-Magnified, |m| < 1,

not inverted. m > 0

Example 4A concave mirror

produces a virtual image that is three times as tall as the object. (a) If the object is 22 cm in front of the mirror, what is the image distance? (b) What is the focal length of this mirror?

cmf

cmcmcmcmcmf

cmdd

df

ddf

f

cmd

h

h

d

d

h

h

oi

i

io

o

o

i

i

o

o

i

11

22

2

22

3

22

1

33.7

1

22

11

33.73/

negative bemust negative, is Since

111

0

22

3

3

The Refraction of LightThe speed of light is different in different materials. We

define the index of refraction, n, of a material to be the ratio of the speed of light in vacuum to the speed of light in the material:

n = c/vWhen light travels from one medium to another its

velocity and wavelength change, but its frequency remains constant.

If a dielectric medium has dielectric constant , then • v = 1/()1/2 =• n = 1/

Refraction of waves, Marching band illustration

Snell’s LawIn general, when light enters a new material its direction will change.

The angle of refraction is related to the angle of incidence 1 by Snell’s Law:

where v is the velocity of light in the medium.

Snell’s Law can also be written as

n1sin1 = n2sin2

constantv

sin sin

2

2

1

1

v

Air

GlassThe angles 1 and 2 are measured relative to the line normal to the surface between the two materials.

Normal line

Principle of least time

Which of these rays can be the refracted ray?

n = 1.3 n = 2a)

b)

c)

d)

Heavy glasswatern1sin1 = n2sin2

[c) Is straight ahead]

Which way will the rays bend? Which of these rays can be the refracted ray?

n = 1.2

n = 1.6

n1sin1 = n2sin2

a)b)

c)

d)

[c) Is straight ahead]

ExampleYou have a semicircular disk of glass with an index of refraction of n = 1.52. Find

the incident angle for which the beam of light in the figure will hit the indicated point on the screen.

1 sin() = n sin(1) = n (5cm)/[(5cm)2+(20cm)2]1/2 = (1.52)(5)/20.62=0.369

= 21.6 deg

Total Internal ReflectionWhen light travels from a medium with n1 >

n2, there is an angle, called the critical angle c, at which all the light is reflected and none is transmitted. This process is known as total internal reflection. The critical angle occurs when 2= 90 degrees:

The incident ray is both reflected and refracted.

Total Internal Reflection

1

2sinn

nc Total internal reflection if sin2>1

Example

A ray of light enters the long side of a 45°-90°-45° prism and undergoes two total internal reflections, as indicated in the figure. The result is a reversal in the ray’s direction of propagation. Find the minimum value of the prism’s index of refraction, n, for these internal reflections to be total.

Lenses

Light is reflected from a mirror. Light is refracted through a lens.

Focal PointThe focal point of a lens is the place where

parallel rays incident upon the lens converge.

converging lens diverging lens

Ray Tracing for Lenses

The P ray propagates parallel to the principal axis until it encounters the lens, where it is refracted to pass through the focal point on the far side of the lens. The F ray passes through the focal point on the near side of the lens, then leaves the lens parallel to the principal axis. The M ray passes through the middle of the lens with no deflection (in thin lens limit).

Just as for mirrors we use three “easy” rays to find the image from a lens. The lens is assumed to be thin.

Ray Tracing Examples

M ray

As the thickness of glass decreases to zero, so does the sideways displacement

The Thin Lens EquationThe ray tracing technique shows qualitatively

where the image from a lens will be located. The distance from the lens to the image, di, can be found from the thin-lens equation: fdd io

111

Sign Conventions: do is positive for real objects (from which light diverges)

do is negative for virtual objects (toward which light converges)

di is positive for real images (on the opposite side of the lens from the object)

di is negative for virtual images (same side as object)

f is positive for converging (convex) lenses

f is negative for diverging (concave) lenses

m is positive for upright images

m is negative for inverted images

Example 4

An object is placed 20 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image? fdd io

111

cmd

cmcmcmd

cmcmdfd

cmf

cmd

i

i

oi

o

20

20

1

20

1

20

21

20

1

10

1111

10

20

Real image,

magnification =

Example 5An object is placed 5 cm in front of a converging lens of focal

length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?

cmd

cmcmcmd

cmcmdfd

cmf

cmd

i

i

oi

o

10

10

1

10

2

10

11

5

1

10

1111

10

5

fdd io

111

Virtual image, as viewed from the right, the light appears to be coming from the (virtual) image, and not the object.

Magnification = +2

Example 6An object is placed 8 cm in front of a diverging lens of focal length

4 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?

05.0/

02

4

1

4

1111

4

111

(concave) 4

0

0

oi

i

i

o

i

ddm

cmd

cmcmdfd

cmd

fdd

cmf

Walker Problem 69, pg. 885

(a) Determine the distance from lens 1 to the final image for the system shown in the figure. (b) What is the magnification of this image?

When you have two lenses, the image of the first lens is the object for the second lens.

Two Lenses

226.0/

42.5

111

24

111

(concave) 7

0

0

oi

i

i

o

i

ddm

cmd

dfd

cmd

fdd

cmf Image

First lens

2nd Lens53.0/

42.21 111

42.4042.50.35

(concave) 14

0

oi

ii

o

ddm

cmddfd

cmcmd

cmf

Illusions, Mirages

DispersionIn a material, the velocity of light (and therefore the index of

refraction) can depend on the wavelength. This is known as dispersion. Blue light travels slower in glass and water than does red light.

As a result of dispersion, different colors entering a material will be refracted into different angles.

Dispersive materials can be used to separate a light beam into its spectrum (the colors that make up the light beam). Example: prism

The Rainbow

No two people ever see the same rainbow

The index of refraction for red light in a certain liquid is 1.320; the index of refraction for violet light in the same liquid is 1.332. Find the dispersion (v – r) for red and violet light when both are incident on the flat surface of the liquid at an angle of 45.00° to the normal.

45

1.332 sinv = 1*sin(45°) = 0.7071

sinv = 0.7071/1.332 =0.5309

v = 32.1°

1.320 sinr = sin(45°)

sinr = 0.7071 / 1.320

r = 32.4 °