26.4 Polarization and the Reflection and Refraction of Light

1

21tannn

B−=θ

Brewster’s law

1

When the incident angle is equal to

(b) The reflected ray is completely polarized in the horizontal plane.

(a) The reflected ray is perpendicular to the refracted ray

For air to glass: °== − 560.15.1tan 1

Bθ

If the incident ray is linearly polarized in the incident plane 100% transmission

Dispersion The index of refraction of visible light changes slightly with wavelength. In glass n is an decreasing function of λ (and an increasing function of f )

1n

2n

26.5 The Dispersion of Light: Prisms and Rainbows

2

The reflection is NOT total internal reflection

In seeing a rainbow, we are looking at sunlight that has gone through a refraction-reflection-refraction interaction in the water droplets suspended in air

Common Misconception

3

30.0 cm

30.0 cm

50.0 cm

x

θ1

θ2

n1=1.50

n2=1.00

A

B

C

Fermat’s Principle A ray originates from point A (0,0) inside a piece of glass, and crosses the glass-air boundary at point B (x, 30.0 cm), and is detected by a sensor at point C (50.0 cm, 60.0 cm).

Glass has refractive index n1=1.50

Find x, where the ray crosses the boundary An interesting solution: Fermat’s Principle: The ray from A to C takes the path of least time

2222 )30()50(BC ,)30(AB +−=+= xx

cm/ns 0.3000.1

0.30 ,cm/ns 0.2050.1

0.30 cm/ns, 30.02

21

1 =======ncv

ncvc

30)30()50(

20)30(BCAB

2222

21

+−+

+=+=

xxvv

t

Minimum at x=17.0 cm

493.0)30()17(

17sin 221 =

+=θ

740.0)30()33(

33sin 222 =

+=θ

!!! 500.1493.0740.0

sinsin

2

1

1

2

nn

===θθ

26.6 Lenses

Refractive Lenses are typically formed by grinding and polishing the surface of glass or special plastics to spherical and/or planar shapes. They come in two general types: With a converging lens,

paraxial rays that are parallel to the principal axis converge to the focal point.

4

With a diverging lens, paraxial rays that are parallel to the principal axis appear to originate from the (virtual) focal point.

−

26.6 Lenses

5

For lenses, the object and the viewer are on opposite sides (for mirrors they were on the same side) :

For ray tracing we commonly use three rays.

[1] from tip of object, parallel to the principle axis on the object sides, through the lens and then through the real focal point on the viewer side

[2] from tip of object, through the real focal point on the object side, then through the lens and emerge parallel to the principle axis on the viewer side

[3] from tip of object straight through the middle of the lens

4

Converging lenses are analogous to concave mirrors.

26.7 The Formation of Images by Lenses Image Formation by a Converging Lens

6

Case 1: (Example: ordinary camera taking a picture of a distant object)

Object is placed beyond 2F (object side): 2F for a converging lens is analogous to C Shrunken, inverted REAL image located between F and 2F on observer side.

fd 20 > fdf i 2<<

Real images are always inverted and are always on the viewer side of the lens

fdd io

111=+

o

i

o

i

dd

hhm −==and

The same equations that applies to mirrors also apply to lenses

fdo 2>fdo 2

11<→

oi dfd111

−=ffdi 2

111−>→

fdi 211

>→ fdi 2 <→

0>od 01>→

od oi dfd111

−=fdi

11<→ fdi >→ fdf i 2 <<→

26.7 The Formation of Images by Lenses

7

Case 2: (Example: projection of a downward nearby object onto a upward REAL image on screen)

fdf 20 << fdi 2>

enlarged, inverted REAL image beyond 2F on observer side. Object placed between F (object side) and 2F (analogous to C for mirrors)

fdo 2<fdo 2

11>→

oi dfd111

−=ffdi 2

111−<→

fdi 211

<→ fdi 2 >→

enlarged, upright VIRTUAL image on object side

Object placed inside F (object side)

fdo <<0fdo

11>→ 0111

<−=→oi dfd

0 <→ id

fd <0

0<id Case 3: (Example: magnifying glass)

1>−

=−=oo

i

dff

ddm

26.8 The Thin-Lens Equation and the Magnification Equation

Example: The Real Image Formed by a Camera Lens A 1.70-m tall person is standing 2.50 m in front of a camera. The

camera uses a converging lens whose focal length is 0.0500 m.

(a) Find the image distance and determine whether the image is real or virtual.

(b) Find the magnification and height of the image on the film.

1m 6.19m 50.2

1m 0500.0

1111 −=−=−=oi dfd

(a)

m 0510.0=id real image

(b) 0204.0m 50.2m 0510.0

−=−=−=o

i

ddm

( )( ) m 0347.0m 70.10204.0 −=−== oi mhh 8

fdd io

111=+

o

i

o

i

dd

hhm −==and

26.6 Lenses

9

Diverging Lenses are analogous to convex mirrors

Ray-tracing for diverging lenses

[1] from tip of object, parallel to the principle axis on the object sides, through the lens and bending outward as if it came from the virtual focal point on the object side

[2] from tip of object, toward the real focal point on the viewer side, then through the lens and emerge parallel to the principle axis on the viewer side

[3] from tip of object straight through the middle of the lens

4

26.7 The Formation of Images by Lenses

Image Formation by a Diverging Lens

10

00 >d

0<id

0<f

Virtual images are always upright and are always on the object side of the lens

A diverging lens always forms a shrunken, upright, VIRTUAL, image.

A diverging lens has a VIRTUAL focus on the object side of the lens

0 ,0 <> fd 011111<−−=−=→

ooi dfdfd0 <→ id

oo

i

dff

ddm

−=−=

odff−−

−=

odff+

= 1 <→ m

26.8 The Thin-Lens Equation and the Magnification Equation

Textbook Summary of Sign Conventions for Lenses

lens. converging afor is +f

lens. diverging afor is −f

lens. theofleft the toisobject theif is +od

lens. theofright the toisobject theif is −od

image). (real lens theofright the toformed imagean for is +id

image). (virtual lens theofleft the toformed imagean for is −id

image.upright an for is +m

image. invertedan for is −m

11

12

image andobject focus, REALafor are and , +io ddf

images and objects upwardfor are and +io hh

Prof. Jui’s Conventions for Thin Lenses

image andobject focus, VIRTUALafor are and , −io ddf

images and objects downwardfor are and −io hh

images realfor is −=o

i

hhm

images lfor virtua is +=o

i

hhm

26.8 The Thin-Lens Equation and the Magnification Equation

Real objects are on the object side Real image are on the observer side Virtual objects are on the observer side Virtual image are on the object side Converging lenses have REAL foci on both sides of the lens. Diverging lenses have VIRTUAL foci on both sides of the lens.

26.9 Lenses in Combination

The image produced by one lens can serves as the object for the next lens.

13

The pair of lenses shown here is functionally identical to a compound microscope

Example: The objective and eyepiece of the compound microscope in the above figure are both converging lenses and have focal lengths of fo = 15.0 mm and fe = 25.5 mm. A distance of 61.0 mm separates the lenses. The microscope is being used to examine an object placed do1 = 24.1 mm in front of the objective. Find the final image distance.

First look at the action of the objective lens: both do1, di1 are relative to this lens, and expressed in mm

1-

11

mm 02517.01.24

10.15

11

o

11=−=−=

oi dfd mm 7.39 mm 02517.0

11-1 ==id

14

mm 7.391 =id

mm 21.3mm 7.39mm 0.61

mm 0.61 12

=−=

−= io dd

mm 5.25e =f

1-

22

mm 007733.03.21

15.25

11

e

11−=−=−=

oi dfdmm 129

)mm 007733.0(1

1-2 −=−

=id

15

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

16

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)

17

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.

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. 18

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 −−

−= 19

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=ND20