1

The human eye operates very much like a modern electronic camera

Iris: controls the amount of light energy entering the lens

Lens: focus light onto retina (adjustable)…refraction also provided by cornea + A.H.

Retina: Layer of electronic (ok…neural) pixel elements

Monocular Vision The lens focuses some of the rays (“emitted” in all directions) from points on the pencil (the object) on to individual points (the image) on the retina

The electrical impulses are carried by the optic nerve into the brain for processing shapes and colors

26.10 The Human Eye

2

Muscles in the eye changes the shape (focal length) of the lens in response to near and far objects depth perception with just one eye

This is a skill learned by a baby in the first few days after birth. It is difficult to demonstrate—it is so automatic (1) Cover one eye. Stare at this screen with other eye. Then move a finger into field of view. (2) Cover one eye. Look down at one finger. Raise your head until this screen comes into field of view

Binocular Vision (a) Eyeballs rotate to center the object in each eye (conscious but fairly automatic response by the brain) more depth perception (1) Put one finger from each hand in front of you—one at twice the distance of the other. (2) Alternately focus on one finger—the other will be seen in “double”

(b) The slightly different images seen in the two eyes are interpreted by the brain to given even more depth perception – 3D glasses!

relaxed lens

tensed lens

26.10 The Human Eye

Far Point of nearsighted eye

Relaxed Eye Lens Distant Object

Image formed in front of retina

FD (non-standard notation)

26.10 The Human Eye Nearsightedness (myopia)

3

Ideally, the lens of the eye should be able to adjust to objects at any distance.

But the nearsighted eye has a lens-retina combination that cannot relax itself enough to focus objects out to infinity. A distant object focus to a real image in front of (but missing) the retina.

Corrective Lens The patient is prescribed a diverging lens to compensate for the over-convergence Far Point of

nearsighted eye

Distant Object

Image formed on the retina

Diverging Lens

Usually there is a maximum object distance, called the far point, to which the eye can focus

Prescription We want to put the virtual image made by the diverging lens of a distant object (i.e. do = ∞) at the far point: DF.

∞=od

Distant Object

Far Point of nearsighted eye

Virtual Image formed by diverging lens

LD

Remember that the corrective lens is worn at a small distance DL in front of the eye (DL=0 for a contact lens)

( )LFi DDd −−=io ddf

111+=

)(11

LF DD −−+

∞= )( LF DDf −−=→

26.10 The Human Eye

Example : Eyeglasses for the Nearsighted Person

A nearsighted person has a far point located only 521 cm from the eye. Assuming that eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lens of the glasses so the person can see distant objects.

With this prescription, objects at finite, but far distances are mapped into virtual images located between the corrective lens (at distance DL from the eye) and the far point (at distance DF from the eye)

4

26.10 The Human Eye

Example : Eyeglasses for the Nearsighted Person

A nearsighted person has a far point located only 521 cm from the eye. Assuming that eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lens of the glasses so the person can see distant objects.

LFio DDddf −−

∞=+=

11111cm 519 −=→ f

With this prescription, objects at finite, but far distances are mapped into virtual images located between the corrective lens (at distance DL from the eye) and the far point (at distance DF from the eye)

cm 2cm 5211−

−=

LF DD −−

∞=

11

Optometrists who prescribe correctional lenses and the opticians who make the lenses do not specify the focal length. Instead they use the concept of refractive power.

THE REFRACTIVE POWER OF A LENS – THE DIOPTER

( )meters in 1diopters) in :( Power Refractive

fRP =

dpt 1930 m 1930

cm 5191-

..RP

f

−=−=→

−=

RP is not a standard notation, and diopter is not an SI unit. 5

Near Point of nearsighted eye

Converging Lens

Sharp image on retina

Close-by object

26.10 The Human Eye

Tensed Eye Lens Near Point of nearsighted eye

Sharp image formed behind

the retina

Close-by object

ND (non-standard notation)

Farsightedness (hyperopia)

Ideally, the lens of the eye should be able to adjust to objects at any distance.

But the Farsighted eye has a lens-retina combination that cannot tense itself enough to focus objects close by. A close-by object focus to a sharp, real image behind (but missing) the retina. Usually there is a minimum object distance, called the near point, to which the eye can focus

Corrective Lens The patient is prescribed a converging lens to compensate for the under-convergence

Prescription Put the virtual image made by the converging lens of the nearest object you want to see (typically at DMIN = 25 cm) to the near point: DN.

Converging Lens

Near Point of nearsighted eye

Close-by object

Virtual Image formed by converging lens

LD

io ddfRP 111

+==)(

1)(

1

LNLMIN DDDD −−

−=

6

Example of corrective lens for farsightedness: this is a pathology everyone gets as they get older – starting at ~40 yrs of age (nearsightedness improves somewhat in combination with this)

Your professor wears reading glasses with refractive power of RP = 1.75 dpt = 1.75 m-1. Where is his near point (inside of which he cannot see). Assume the glasses to correct for objects as near as 25 cm, and that the glasses are worn 2 cm from the eyes.

7

io ddfRP 111

+==)(

1)(

1

LNLMIN DDDD −−

−=

Example of corrective lens for farsightedness: this is a pathology everyone gets as they get older – starting at ~40 yrs of age (nearsightedness improves somewhat in combination with this)

Your professor wears reading glasses with refractive power of RP = 1.75 dpt = 1.75 m-1. Where is his near point (inside of which he cannot see). Assume the glasses to correct for objects as near as 25 cm, and that the glasses are worn 2 cm from the eyes.

m) 02.0(1

m) 02.0m 25.0(1m 75.1 1

−−

−=−

ND

11 m 60.2m 75.1m 23.0

1m) 02.0(

1 −− =−=−ND

m 38.0m 60.2

1m 02.0 1 ==− −ND

m 40.0=ND8

26.11 Angular Magnification and the Magnifying Glass The eye is basically like a camera The size of the image on the retina determines how large an object appears to be.

( )o

o

dh

≈= sizeAngular radiansin θ

9

26.11 Angular Magnification and the Magnifying Glass

Example: A Penny and the Moon

Compare the angular size of a penny held at arms length with that of the moon.

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26.11 Angular Magnification and the Magnifying Glass

Example: A Penny and the Moon

Compare the angular size of a penny held at arms length with that of the moon.

rad 027.0cm 71cm 9.1 ==≈

o

o

dhθPenny

Moon rad 0090.0m 103.9m 105.3 8

6

=××

=≈o

o

dhθ

11

26.11 Angular Magnification and the Magnifying Glass

Angular magnification

θθ ′

=M

Ndf

Mi

−≈

11

Angular magnification of a magnifying glass

12

26.12 The Compound Microscope

To increase the angular magnification beyond that possible with a magnifying glass, an additional converging lens can be included to “premagnify” the object.

Angular magnification of a compound microscope

( )eo

e

ffNfLM −

−≈

13

Chapter 27

Interference and the Wave Nature of Light

14

27.1 The Principle of Linear Superposition When two or more light waves pass through a given point, their electric (and magnetic) fields combine (interfere) according to the principle of superposition.

The waves emitted by the sources start out in phase and arrive at point P in phase, leading to constructive interference.

,3,2,1,0 12

==−

mmλ

The waves emitted by the sources start out in phase and arrive at point P in opposite phase, leading to destructive interference.

( )

,3,2,1,0 2

112

=

+=−

mm λ

Special Case 1 Special Case 2

If constructive or destructive interference is to continue ocurring at a point, the sources of the waves must be coherent sources.

Two sources are coherent if the waves they emit maintain a constant phase relation. 15

27.2 Young’s Double Slit Experiment Two independent, coherent light sources are virtually impossible to construct .

Light waves from these slits interfere both constructively and destructively on the projection screen.

In Young’s experiment (Young also participated in the deciphering of the Rosetta Stone), two slits illuminated by a single source acts as two separate but coherent sources of light.

The waves coming from the slits interfere constructively or destructively, depending on the difference in distances between the slits and the screen. 16

For best results, light of a single color (e.g. produced by a laser) is used.

27.2 Young’s Double Slit Experiment

λθ md ==∆ sin

17

In the typical set up, the screen is very far away compared to the separation of the slits. We treat the rays from the slits as if they were parallel

Bright fringes

Dark fringes λθ )(sin 21+==∆ md

θsind=∆

,3,2,1,0=m

Bright fringes from 2 slits

Dark fringes from 2 slits d

m λθ =sin ( )d

m λθ 21sin +=

m

http://www.youtube.com/watch?v=5PmnaPvAvQY

Bright fringe

Bright fringe

Bright fringe

Dark fringe

Dark fringe

27.2 Young’s Double Slit Experiment

Example: Young’s Double-Slit Experiment

Red light (664 nm) is used in Young’s experiment with slits separated by d=0.000120 m. The screen is located a distance 2.75 m from the slits.

Find the distance on the screen between the central bright fringe and the third-order bright fringe.

18

951.0m101.20m106643sin sin 4

911 =

××

=

= −

−−−

dm λθ

( ) ( ) m 0456.0 951.0tanm 75.2tan === θLy

Note: Double slit interference also separates out the colors from a white or multi-colored source.

The angle of the bright fringes is different for each wavelength the bright fringes for different colors form a rainbow on screen.

27.3 Thin Film Interference Because of reflection and refraction, (at least) two light waves enter the eye when light shines on a thin film of gasoline floating on a thick layer of water.

film

vacuumfilm n

λλ =

19

Because of the extra distance traveled by ray [2] relative to ray [1], there can be interference between the two waves.

For the comparison of the extra path to wavelength, we need to use the wavelength in the medium

When light travels through a material with a smaller refractive index towards a material with a larger refractive index, (partial) reflection at the boundary occurs along with a phase change that is equivalent to one-half of a wavelength in the film.

When light travels from a larger towards a smaller refractive index, there is no phase change upon reflection.

27.3 Thin Film Interference

Example Color of a Thin Film of Gasoline A thin film of gasoline floats on a puddle of water. Sunlight falls perpendicularly on the film and reflects into your eyes. The film has a yellow hue because destructive interference eliminates the color of blue (469 nm) from the reflected light. The refractive indices of the blue light in gasoline and water are 1.40 and 1.33. Determine the minimum non-zero thickness of the film.

20

... , ,2

film25

film23

film21

film21

λλλλ

=

−=∆ t

...3 ,2 ,2 filmfilmfilm λλλ=→ t

nm 168 40.1nm 469

21

2film =⋅==

λt

A

B Ray [1]: reflection at point A from lower to higher index: phase shift equivalent to ½ λfilm film2

11 λ=

Ray [2]: reflection at point B from higher to lower no phase shift ; but roundtrip through film:

t22 =

film21

12 2 λ−=−=∆ t

Effective path difference between Ray [1] and Ray [2] (including phase shift for reflection)

Destructive interference between Ray [1] and Ray [2]: difference of half-integer wavelengths

Minimum non-zero thickness t :

27.3 Thin Film Interference

The wedge of air formed between two glass plates causes an interference pattern of alternating dark and bright fringes.

21

y

λ212 −=∆ y

Dark Fringes λ)( 21+=∆ m

Bright Fringes ...3,2,1,0 ==∆ mmλ

Thin-film band-pass optical filter

As long as the film has the highest refractive index of the three regions, then neither of the two reflections incur a phase shift.

t2=∆→

Constructive interference between transmitted ray [1] and ray [2] occurs for the minimum thickness of

1

2

film21 λ=t