Image formationLenses and Mirrors

Object:pImage: qFocal length : fMagnification :MReal : light at imageVirtual : no light

1 1 2 2

1 2

sin sin

1 1 1( 1)

n n

nf R R

Mirrors Lenses

/

1 1 1

image

object

n c v

p q f

h qM

h p

1 2

2

Rf

+p

+q

+R

+f

Ray1- || to axis, then thru f

Ray2-thru f then ||

Ray3-thru R and back

(Ray-4- to center of mirror then reflect at same angle)

Ray1- || to axis, then thru f on back

Ray2-thru f on front then ||

Ray3-thru center of lens

1 2 1 2

1

2

1n n n n

p q R f

n qM

n p

Refraction at curved surface

+p

+q

+R

+f Convex-f Concave

Simple optical componentsImagine a spherical mirror

A horizontal ray gets reflected according to…

But if they area close to the axis they almost meet

The rays don’t really meet

Horizontal rays close to the axis focus

So, for• Optical components that Symmetric about an axis • With rays close to the axis• And not very thick when transparent

Horizontal rays are focused to a point

Axis of symmetry orprincipal axis

f

Focal pointRays form a point close to the axis are also focused

Image

A pinhole camera

oatmeal boxsmall hole

A goldfish in a bowl

Mirrors

p q

object

image

Flat Mirrors Light rays appear to diverge from pointbehind mirror

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

Law of Reflection: p =|q|independent of viewing angle

Image is Virtual; behind mirror, can’t project on screen, light rays don’t actually pass through image point

h h’

Principle axes

p

q

q

Magnification: M = h’/h = 1Image is Upright

p qp q

• In the overhead view of Figure 26.4, the image of the stone seen by observer 1 is at C. At which of the five points A, B, C, D, or E does observer 2 see the image?

’

x

x’

p

q

p-R

R-q

R

'

1sin sin x x

p R R

2

' 'sin sin '

x x

R q R

1 1

2 2

sin

sin

R q q

p R p

1 2p u q

u

1 1 2 1

p q R f

Just geometry…

Reflection by spherical mirrors

R

pq

Look at the mirror formula:

1 1 2 1

p q R f

q=R/2 when p is very large

All rays converge at R/2, the focal point

For concave mirror the image is in front of the mirror

The ray is reflected straight back because it move in the radial direction and so it is normal to the mirror

The image is inverted

h

h’

'h qM

h p

q

p

For convex mirror the image appears to be behind the mirror, it is virtual

The ray is reflected straight back because it move in the radial direction and so it is normal to the mirror

The image is not inverted

h

h’

'h qM

h p

qp

The mirror formula still holds, but with q<0 and R<0

The image is virtual The mirror is convex

• You wish to reflect sunlight from a mirror onto some paper under a pile of wood to start a fire. Which would be the best choice for the type of mirror? (a) flat (b) concave (c) convex

• Ray tracing for lenses (and mirrors)– ray parallel to the axis refracts (reflects) through

the focal point– ray through the focal point emerges parallel to

the axis– ray through the center of the lens (mirror)

• does not bend (lenses)• goes out at an equal angle (mirror)

Consequences of Snell’s law and the laws of reflection PLUS geometry

f’

f

p

q

h

-h’

26.2

• In a church choir loft, two parallel walls are 5.3 m apart. The singers stand against the north wall. The organist faces the south wall sitting 0.8 m away from it. To enable her to see the choir, a flat mirror 0.6 m wide is mounted on the south wall, straight in front of her. What is the width of the north wall she can see.

Exercise 26.2Sounds like ray tracingWill need the formula for reflected rays form mirrors

0.8m

organist

5.3m

choir0.6mL m 575.4m 6.02

m 9875.1 8.0

3.0

3.5

lL

ll

l

Exercise 26.11Sounds like an application of the formula

1 1 1

'

p q f

h pM

h q

/ 2 20 cm qpf

f Rp f

30 ( 20) 12) p 30 cm q 12 cm .4

30 20 30a M

60 ( 20) 15) p 60 cm q 15 cm 0.25

60 20 60b M

) 0 upright images in both casesc M

A spherical convex mirror has radius of curvature with magnitude 40 cm.Find q and M when p=30 cm and p=60 cm

http://www.phy.ntnu.edu.tw/java/Lens/lens_e.htmlcorrected apr 25

+ side - side

So, to make an image just do some ray-tracing

A ray trough a focal point is refracted to a horizontal ray

’

f’

f

A horizontal ray is refracted towards a focal point

p

q

h

-h’' ' '

' '

' ' '

'

' '

'

h f h f

p h h qθ

p f f q

p f f

f q f

Let f f

1 1 1

'

p q f

h qM

h p

The optical component may not be back-front symmetric so in general f≠f’

f=f’

Specific cases:

1 1 1

p q f

Spherical concave mirror of radius R

Spherical convex mirror of radius R

Spherical boundary of radius R between a medium of refraction index n1 (left) and refraction index n2 (right)

2

Rf

2

Rf

2112 /1'

1/ nn

Rf

nn

Rf

Thin lens of curvature radii Rleft, Rright and of material with refraction index n

left right

1 1 1( 1)n

f R R

1 2 2 1+ =n n n n

p q R

An object is:•REAL if light diverges from it•VIRTUAL if light converges toward it

Definitions and sign conventions

An image is:•VIRTUAL if light diverges from it•REAL if light converges toward it

f

f

p

qh h’

p (object) q (image) (object) ' (image) (image)

convex real real upright upright upright

concave virtual virtual inverted inverted inverted

f h h M

+p

+q

+R

+f Convex-f Concave

+p

+q

+R

+f

mirror

lens

For example:

p (object) q (image) (object) ' (image) (image)

convex real real upright upright upright

concave virtual virtual inverted inverted inverted

f h h M

An image image is:VIRTUAL if light diverges from itREAL if light converges toward it

An objectobject is:REAL if light diverges from itVIRTUAL if light converges toward it

+p

+q

+R

+f Convex-f Concave

1 1 1

'

p q f

h pM

h q

positive real image concave concave upright image

negative virtual image convex convex inverted image

q R f M

Sign conventions for mirrors:

+p

+q

+R

+f

Exercise Sounds like another application of the formulaThe tricky part are the signs:•M=+2 since the image is not inverted•The object and the image are on same sides of the lens so p and q have the different signs

1 1 1

qM

p q f p

2 q 2 84 cm

1 1 1 1 1

2.84 cm1

qM .

p

M M

f p q q q q

qf

M

Magnifying looks at a map. The image is virtual, of twice the size and |q|=2.84 cm. Find f

+p

+q

+R

+f Convex-f Concave

A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the transparent cornea. Assume that this outer surface has a 6mm surface of curvature and assume that the eyeball contains just one fluid with index of refraction 1.4. Where is the image? Does it fall on the retina? Describe it.

1 2 2 1n n n n

p q R

1.4(6 )21

1 .4

nR mmp so q mm

n

real, inverted

two lens system

Fig 36-32a, p.1151

first lens

1 1 1

10 20q

1 1 1

15 10q

q=30

1 1 1 qM

p q f p

second lens

p=40-30=10cm

q=-20 cm

M1=-30/15=-2

15 cm 40

http://www.hazelwood.k12.mo.us/~grichert/optics/intro.html

M=M1M2=-8

M2=-(-20)/(10)=4

real inverted

virtual, upright

Total

inverted

+p

+q

+R

Microscope

Fig 36-44a, p.1161

f=2cmf=20 cm

f=2cmf=20 cm

virtual object?

first lens

1 1 1

15 2q

1 1 1

20q

q=20 cm

second lens

p=5-20=-15cmVIRTUAL OBJECT!

q=1.8 cm

double lens q=20 cm

1 1 1

p q f +p

+q

+R

Fig 36-40, p.1157

Fig 36-41, p.1158

http://webphysics.davidson.edu/physlet_resources/dav_optics/Examples/eye_demo.html

• 6. If you cover the top half of the lens in Active Figure 26.24a with a piece of paper, which of the following happens to the appearance of the image of the object? (a) The bottom half disappears. (b) The top half disappears. (c) The entire image is visible but dimmer. (d) There is no change. (e) The entire image disappears.

http://www.phy.ntnu.edu.tw/java/Lens/lens_e.html

http://www.hazelwood.k12.mo.us/~grichert/optics/intro.html

http://webphysics.davidson.edu/physlet_resources/dav_optics/Examples/eye_demo.html

Rope with total mass m = 2 kg, L = 80 mand mass M = 20 kg at end

L = 80 m

M = 20 kg

Tension in rope F and wave speed v

2 kg

80 m0.025 kg /m Tension F Mg (20)(9.8) 196 N

v F

1960.025

88.5 m / s

If end of rope driven in SHM with f=0.056 Hz then wavelength of travelingwave up rope is

v

f

88.5

0.0561590 m L 80 m

If end of rope driven in SHM with f=0.056 Hz then wavelength of travelingwave up rope is ?

2

12

F=-kx x=Acos( t+ )

1

2spring

f fT

kSHO

m

gPendulum

L

PE kx Energy PE KE

total 0

0

2 2 2 20

2 20

sin( )

/

( / ) ( )

/tan

ma F kx bv F t

F mA

b m

b m

2

)2/(

2

)cos(

m

b

m

k

tAex tmb

/0

t mE E e where

b

( , ) sin( )

2

D x t A kx t

k vk

damped SHO

driven and dampled SHO

traveling waves

2 2

2 2 2

1d d

dt c dx

E E

8

0 0

13 10 m/sc

0

0

0 0

cos( )

cos( )

E kx t

B kx t

E cB

E x

B z

^

^

2av 0 0

1

2u E

0

1S E B

2

0 0

~ /EB E

Power areac

S

20

02avg

EPowerI S

Area c p=E/c

1 1 / 1 PowerP

F p E c I

A A t A t A c c

I=I0 cos2

Tv

LL S

S

v vf f

v v

positive direction is when one is moving toward the other

Object:pImage: qFocal length : fMagnification :MReal : light at imageVirtual : no light

1 1 2 2

1 2

sin sin

1 1 1( 1)

n n

nf R R

Mirrors Lenses

/

1 1 1

image

object

n c v

p q f

h qM

h p

1 2

2

Rf

+p

+q

+R

+f

Ray1- || to axis, then thru f

Ray2-thru f then ||

Ray3-thru R and back

(Ray-4- to center of mirror then reflect at same angle)

Ray1- || to axis, then thru f on back

Ray2-thru f on front then ||

Ray3-thru center of lens

1 2 1 2

1

2

1n n n n

p q R f

n qM

n p

Refraction at curved surface

+p

+q

+R

+f Convex-f Concave

virtual object?

Fig 36-32a, p.1151

first lens

1 1 1

p q f

1 1 1

10 20q

1 1 1

15 10q

q=30 cm

1 1 1

p q f

second lens

q=20-30=-10cmVIRTUAL OBJECT!

q=6.7 cm

15 cm distance

20.0 cm