Today . . .• Multiple Lenses• The Eye

– Corrective Lenses for Myopic & Hypertropic Eyes

• Magnifiers & Microscopes• Telescopes

Text Reference: Chapter 34.2,4

Examples: 34.9,10,12,13,14 and 15

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 = +1.5, f1 = +1

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

f = +1 f = -4

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

Draw Rays !

3'1 s 2

1

'1

1 ssm

31

5.111111

11'1

sfs

8.0'2 s 5

4

2

'2

2 ssm

45

11

41111

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):

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

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

Note the negative object distance for the 2nd lens.

f = +1 f = -4

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

1

421 mmm

21

'1

1 ssm3'

1 s31

5.111111

11'1

sfs

22

'2

2 ssm4'

2 s

41

21

41111

22'2

sfs

Lecture 27, ACT 1• Suppose we interchange the converging and diverging lenses in

the preceding case. – What is the relation of the new magnification m’ to the original

magnification m?

• What is the nature of the final image?1B

(c) m’ > m(a) m’ < m (b) m’ = m

(a) real (b) virtual

1A

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

Lecture 27, ACT 1• Suppose we interchange the converging and diverging lenses in

the preceding case. – What is the relation of the new magnification m’ to the original

magnification m?1A(c) m’ > m(a) m’ < m (b) m’ = m

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

• Since the formula for the magnification is equal to the product of the magnifications of each lens (m = m 1 m 2), you might think that interchanging the lenses does not change the overall magnification.• This argument misses the point that the magnification of a lens is not a property of the lens, but depends also on the object distance!• Consider the ray shown which illustrates that the magnification must be < 1!

Lecture 27, ACT 1• Suppose we interchange the converging and diverging lenses in

the preceding case. – What is the relation of the new magnification m’ to the original

maginification m?1A(c) m’ > m(a) m’ < m (b) m’ = m

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

1B(a) real (b) virtual

• What is the nature of the final image?

• The ray used in part A actually shows that the image is real and inverted.• The equations:

2312

23111111

22'2

sfs12

115.1

141111

11'1

sfs

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 5.2f

cm 3.2f

cm 5.210111

' ssf

5.21

251111

' ssf

Therefore the normal eye acts as a lens with a focal length which 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!

An intuitive way to view eye correctionsNear-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 retina

Does the orientation of the lens matter?• Not according to our simple lens equation

assumes paraxial rays• In reality…

As a general rule, it is better to split the refraction at both interfaces.

Spherical

aberration

www-optics.unine.ch/education/optics_tutorials/plano_convex_lens_aberration.html

Why use two lenses instead of one?• No choice vision correction• Improve aberrationsSpherical Aberration Chromatic Aberration

www-optics.unine.ch/education/optics_tutorials/achromat.html

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?

(a) 0.015 D (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:

Apparent Magnification• Our sense of the size of an object is determined by the size of

image on the retina. – Consequently, the apparent magnification factor of a lens is just

the ratio of the angular size with the lens to the angular size without the lens.

– Use lens to make close up image fall in focus range of eye

Lnp

h

Object at Near Point - can’t get nearer

~fh

Object just inside Focal Pointof simple magnifier

Positive “f” lens

npLh

fh

Define Angular Magnification: fL

M np

Microscope• Larger magnifications than are possible with a single lens can

be obtained by combining lenses.• For example, the compound microscope consists of two

lenses, a short focal length objective lens and an eyepiece:

I1

I2• The object, placed just beyond the focal point of the objective, produces a real, inverted image (I1) at a position which is just inside the focal length of the eyepiece.

~fo~fe

eyepieceobjectiveL• This image then produces a

virtual image (I2) which is seen as magnified as in the previous slide.

The magnification of the objective is:o

e

ffL

ssM

'

0

The magnification of the eyepiece is:e

npe f

LM

The total magnification M = MoMe is:e

npe

fL

ffLM

0

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.» As in the microscope, we have two lenses, an objective and an

eyepiece. » Since the object is at a large distance, the objective lens should

have a long focal length to obtain a large magnification.– 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

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.)

Summary

• Multiple Lenses– The image of one lens is the object for the next lens.

• The Eye– Corrective lenses place image of object at a point where person

is able to view it, when he/she cannot properly focus on the object itself.

• Optical Instruments– Use multiple lenses to build microscopes, telescopes, etc….

Text Reference: Chapter 34.2,4

Examples: 34.9,10,12,13,14 and 15