Multiple Lenses • We determine the effect of a system of lenses by considering

the image of one lens to be the object for the next lens.

For the first lens: s1 = +1.5m, f1 = +1m

For the second lens: s2 = +1m, f2 = -4m

∴

∴

f = +1m f = -4m

-1 +3+10 +2 +6+5+4

Draw Rays !

m3'1 =s 2

1

'1

1 −=−=ssm

m31

m5.11

m11111

11'1

=−=−=sfs

m8.0'2 −=s 5

4

2

'2

2 +=−=ssm

m45

m11

m41111

22'2

−=−−

=−=sfs

58

21 −== mmm

Multiple Lenses • Objects of the second lens can be virtual. Let’s move the

second lens closer to the first lens (in fact, to its focus):

∴

∴

Note the negative object distance for the 2nd lens.

f = +1m f = -4m

-1 +3+10 +2 +6+5+4

421 −== mmm2

2

'2

2 +=−=ssmm4'

2 +=s

m41

m21

m41111

22'2

=−

−−

=−=sfs

For the first lens: s1 = +1.5m, f1 = +1mm31

m5.11

m11111

11'1

=−=−=sfs

m3'1 =s 2

1

'1

1 −=−=ssm

For the second lens: s2 = -2m, f2 = -4m

The eye

The Eye• The “Normal Eye”

– Far Point ≡ distance that relaxed eye can focus onto retina = ∞– Near Point ≡ closest distance that can be focused on to the retina

~ 25 cm

2.5cm

25cm

This is called “accommodation”Diopter: 1/f Eye = 40 diopters, accommodates by about 10%, or 4 diopters

cm 5.2=f

cm 3.2=f

cm 5.210111

' +=+=ssf

5.21

251111

' +=+=ssf

Therefore the normal eye acts as a lens with a focal lengthwhich can vary from 2.5 cm (the eye diameter) to 2.3 cm which

allows objects from 25 cm → ∞ to be focused on the retina!

Eye corrections (glasses, contacts)Near-sighted eye is elongated, image of distant object forms in front of retina

Add diverging lens, image forms on retina

Far-sighted eye is short, image of close object forms behind retina

Add converging lens, image forms on retinapower = 1/f; f in meters

Near vision Near point changes with age:

7 cm → 200 cm7 years 60 years

Most distinct vision is at near point. Image is largest.

Example: Reading glasses. The near point of a person’seye is 75 cm. What power reading glasses should be usedto bring the near point to 25 cm?

dioptersmf

cmcmfss

cmscms

67.2375.0

11751

251111

7525

==

−+==

′+

−=′=

ss’

oi F F

Assumes eye very close. Results are slightly different when distancebetween them is taken ino consideration.

virtual

A person with normal vision (near point 28cm) is standing in front of a plane mirror. What is the closest distance to the mirror the person can stand and still see himself in focus?

a) 14 cmb) 28 cmc) 56 cm

Object for eye = Image of self• Distance from eye to object = s+s’• Set s+s’ = near point

s s’

Old Preflight

Two people who wear glasses are camping. One of them is nearsighted and the other is farsighted. Which person's glasses will be useful in starting a fire with the sun's rays?

Old Preflight

a) the farsighted person's glasses

b) the nearsighted person's glasses

What do you need to start a fire?• REAL IMAGE ! (light is focused to a point)

• Converging lens gives REAL IMAGE• Far-sighted people need converging lenses!

Special Lens CombinationsIf two thin lenses are close together, they act effectively as a single lens. The focal length of the “doublet” is given by

1 2

1 1 1doubletf f f

= +f1 f2

fdoublet

Note the power (=1/f) of the combination is just Pdoublet = P1 + P2

Lecture 27, ACT 2• Hildegard’s retina is 2.5 cm behind

the lens, which has a minimum focal length of 2.6 cm.

1. What does the focal length fcl of her contact lens need to be?

(a) 65 cm (c) -0.1 cm(b) -65 cm

2. What is the power Pcl of the contact lens?

(a) 1.5 D (c) 1000 D(b) -1.5 D

2.5 cm

feye = 2.6 cm

Lecture 27, ACT 2• Hildegard’s retina is 2.5 cm behind

the lens, which has a minimum focal length of 2.6 cm.

1. What does the focal length fcl of her contact lens need to be?

(a) 65 cm (c) -0.1 cm(b) -65 cm

2.5 cm

feye = 2.6 cm

1 1 1eye clf f f

= +1 1 1

2.5 cmcl eyef f= −

(2.5 cm ) 2.6 2.5 65 cm2.5 cm 2.6 2.5

eyecl

eye

fff

×= = =

− −

Lecture 27, ACT 2• Hildegard’s retina is 2.5 cm behind

the lens, which has a minimum focal length of 2.6 cm.2. What is the power Pcl of the contact lens?

(c) 1000 D(b) -1.5 D

2.5 cm

feye = 2.6 cm

1 1 1.5 D0.65 mcl

clP

f≡ = =

40 D 38.5 Dcl need eyeP P P≡ − = −

Note: We could have solved for the power directly:

(a) 1.5 D

Angular Magnification• Our sense of the size of an object (in the absence of other

clues) is determined by the size of image on the retina.• This is proportional to the angle subtended by the object:

α1h

Bigger image

out

in

M θθ

≡

• The magnification of an optical system can then be equivalently defined as the ratio of the output angular spread to the input angular spread:

h’1 ~ α1

α2hh’2 ~ α2

Smaller image

Magnifying Glass• Our sense of the size of an object is determined by the size

of image on the retina. – If the object were closer to our eye, it would subtend a larger

angle.– However, we can only focus on an object if it is no closer than

the near-point distance Lnp (~25 cm).– We can use a simple magnifier to create an enlarged virtual

image outside Lnp.

α

Lnp

h

Object at Near Point - can’t get nearer

~fβh

Object just inside Focal Pointof simple magnifier

Positive “f” lens

npLh

≈α

fh

≈β

Define Angular Magnification: fL

M np≈≡αβ

⇒ Choose f < Lnp

Ma gnifierAt normal distances a small object (small letter print) subtends a small angle θ. with a lens or magnifier we can magnify the subtended angle to θ′.

The ratio is call angularmagnification

Which is NOT the same aslateral magnification.

θθ ′

=M

Ma

Telescopes• The purpose of a telescope is to gather light from distant

objects and produce a magnified image.– Refracting telescopes use lenses so that the objects can be viewed

directly.

– Reflecting telescopes use mirrors to create the image» Most astronomical telescopes are reflectors, since the most

important feature for these telescopes is the light gathering ability, and it is easier to make a large mirror than it is to make large lenses.

detector

θ1

θ1

θ2

h’

f1 f2

2 2 1

1 1 2

'/'/

h f fMh f f

θθ

= = =

11

'hf

θ ≈ 22

'hf

θ ≈

Telescope

21

12

//M

ff

fyfy

−=′

−=′

=θθ

Angular magnification

Note, light gathering power depends on objective lensDiameter.

Hubble Space TelescopeThe HST was launched in 1990; it was discovered that a lens had been ground incorrectly, so all images were blurry!A replacement “contact lens”, COSTAR, was installed in 1993.

Aperture of primary mirror: 2.4 m (~8 ft.)Mass of primary mirror: 828 kg (~1800 lbs)

Before COSTAR After COSTAR

(Now, Hubble’s instruments have built-in corrective optics, so

COSTAR is no longer needed.)