GEOMETRICAL OPTICSANDOPTICAL INSTRUMENTS

Ahmad Alfin A. (x.2/01)

Putri Permata T. (x.2/13)

Shaffan Ula P. (x.2/18)

Yoshinta P. (x.2/26)

Ligita F.A. (x.2/29)

SMAN 5 SURABAYA

INTRODUCTION

Optics is branch of physics concerns the propagation and behavior of light. In general, light is part of the electromagnetic spectrum that extends from X rays to microwaves and includes the radiation that produces the sensation of vision. The study of optics is divided into geometrical optics and physical optics.

Geometrical optics study the application of laws of reflection and refraction of light in the design of lenses, mirrors and other optical instruments.

We’ll discuss about how images are formed by lenses. We’ll also discuss about magnifiers and the human eye.

LENSES

Lens in optics is glass or other transparent substance so shaped that it will refract the light from any object, converging or diverging it, and form a real or virtual image of the object.

Lenses are classified by the curvature of the two optical surfaces. Each surface of a lens can be convex, concave, or flat.

Lens axis is imaginary line perpendicular to both surfaces

axis

CONVEX LENS If the lens is convex, a beam of light traveling parallel

to the lens axis and passing through the lens will be converged (or focused) to a spot on the axis, at a certain distance behind the lens (known as the focal length).

Convex lens has positive focal length In this case, the lens is called a positive (+) or

converging lens. Convex lens usually described as a vertical line with sign (+) on it’s top.

PROPERTIES OF CONVEX LENS

+

f2 f1o

+

f2 f1o

+

f2 f1o

When a beam of ray parallel to the axis passes through the lens, it will converge to a point (f1)

Similarly, rays passes f2 refracted from the lens as parallel rays

Rays passes undeflected straight through center of the lens (O)

IMAGING

+

f2 f1o2f2

2f1

Image Properties:1. Real Image2. Inverted3. Smaller than Object

Condition: S>2f2

S

IMAGING

+

f2 f1o2f2

2f1

Image Properties:1. Real Image2. Inverted3. Same Size as Object

Condition: S=2f2

IMAGING

+

f2 f1o2f2

2f1

Image Properties:1. Real Image2. inverted3. Bigger than Object

Condition: f2<S<2f2

IMAGING

+

f2 f1o2f2

2f1

No Image formed, or Image at infinite distance

Condition: S=f2

IMAGING

+

f2 f1o2f2

2f1

Image Properties:1. Virtual Image2. Upright3. Bigger than Object

Condition: S<f2

SIMULATION

CONCAVE LENS

If the lens is concave, a parallel beam of light passing through the lens is diverged (spread); the lens is thus called a negative (-) or diverging lens. The distance from this point to the lens is also known as the focal length.

Concave lens has negative focal length It also usually described as a vertical line with (-)

sign.

PROPERTIES OF CONCAVE LENS

-

f1 f2o

-

f1 f2o

-

f1 f2o

When a beam of ray parallel to the axis passes trough the lens, it will appear diverge after refraction

Rays converging towards f2 will refracted parallel to the axis

Rays passes undeflected straight through center of the lens (O)

IMAGING

_

f1 f2o2f1

2f2

Image Properties:1. Real Image2. Upright3. Smaller than Object

Condition: S>2f2

S<2f2

S=2f2

Image of concave lens always virtual, upright, and smaller than object

SIMULATION

LENS MAKER EQUATION

)11

)(1(1

211

2

RRn

n

f

M2M1 n2 n1

R2R1

M2M1 n2 n1

R2R1

Rules:R1 or R2 (+) for convex surfaceR1 or R2 (-) for concave surfaceR1 or R2 () for plane surface

LENS EQUATION

fss

111

21

1

2'

s

s

h

hM

Equation of Object-Image relationship

Linear and lateral magnification

Rules:

1. S1 has positive value if object located in front of the lens

2. S1 has negative value if object located behind the lens

3. S2 has positive value if image located behind of the lens (real)

4. S2 has negative value if image located in front of the lens (virtual)

5. Convex lens has positive focus (f)

6. Concave lens has negative focus (f)

7. h’ has positive value means image is upright and virtual

8. h’ has negative value means image is inverted and real

2 LENS WITH SAME MAIN AXIS

f1 f1o f2 f2

s1 s’1 s’2s2

d

objectFinal Image

2 LENS WITH SAME MAIN AXIS (EQUATION)

21' ssd

1

1

1

11

''

s

s

h

hM

2

2

2

22

''

s

s

h

hM

21 MMM t

21

21

1

2 '''

ss

ss

h

hM

Image formed by first lens will be object for second lens

Total magnification is equal to multiplication of both magnification

COMPOUND LENS

f1 f22nd image

1st image

f3

3rd image

1 2 3

COMPOUND LENS (EQUATION)

...11111

321

fffff i it

...321 PPPPPi

it

Total focus of compound lens

Total optical power of compound lens

For compound lens d=0

OPTICAL POWER OF THE LENS The focal length f is positive for converging lenses, and

negative for diverging lenses. The inverse of the focal length, 1/f, is the optical power of the lens. If the focal length is in meters, this gives the optical power in dioptres (1/m).

Convex lens has positive optical power because it has positive focus, while Concave lens has negative optical power because it has negative focus.

We often use this quantity to determine eyeglasses power

fP

1

P is Power of lens

f is focal length (m)'

11

'

11

ssnssP

Sn is normal near or far point

S’ is user near or far point

EXAMPLE

Convergent lens An object placed 30 cm in front of convergent lens that has

focal length 15 cm. Find: a) Image location (S2), b) Image Magnification, c) Image properties.

fSS

1

2

1

1

1

15

1

2

1

30

1

S

30

1

15

1

2

1

S

30

1

30

2

2

1

S

30

1

2

1

S

cmS 302

130

301

2

M

M

S

SMa) b) c)

Real Image, S2= (+)

inverted, M= (-)

Same size as object, M=1

Convergent = Convex

EXAMPLE

Divergent lens An object placed 30 cm in front of divergent lens that has

focal length 15 cm. Find: a) Image location (S2), b) Image Magnification, c) Image properties.

fSS

1

2

1

1

1

15

1

2

1

30

1

S

30

1

15

1

2

1

S

30

1

30

2

2

1

S

30

3

2

1

ScmS 102

3

130

101

2

M

M

S

SMa) b) c)

Virtual Image, S2= (-)

Upright, M= (+)

Smaller than object, M<1

Divergent = Concave

EXAMPLE

Optical Power of The Lens

1. Find the optical power of a convergent lens with focal length 50 cm

2. Find the optical power of a divergent lens with focal length 25 cm

1)

2)

mcmf 5.050

dioptref

P 25

10

5.0

11

mcmf 25.025

dioptref

P 425

100

25.0

11

THE EYE

Eyes are one of the optical instruments, which are the most familiar to us . An eye contains a convex lens with a thickness that can be set . It changes the focal length so that objects that are either far away or near can be seen more clearly .

ANATOMY OF THE EYE

EYE OPTICS The image formed by eye is real, inverted, and smaller than

the object. For an object to be seen sharply, the image must be formed

exactly at the retina. The eye adjust to different object distances by changing the focal length of it’s lens.

For normal eye, an object at infinite distance sharply focused when ciliary muscle is relaxed. To form sharp image for closer objects, the tension of ciliary muscle increase so that the curvature of it’s surface decrease. This process called accomodation.

The extremes of over which distinct vision is possible are known as the far point and the near point of the eye. The far point of a normal eye is at infinity. The average near point of normal eye is 25 cm.

NEARSIGHTEDNESS (MYOPIA)

Someone who can see objects which has distance 25cm clearly, but can't see objects with far distance clearly is called nearsightedness. It's also called far-blurred . This is caused by the eyepiece can't flat in proper order so the farthest image object doesn't reach the retina . It can be cured by using concave eyeglasses . Before the light enters the eye, it is refracted by concave lens, so the reflection precise in the retina .

IMAGING

_Parallel light

FARSIGHTEDNESS (HYPERMETROPIA)

Farsightedness, person can't see clearly close to the object, but can see clearly far to the object . This is caused by the eyepiece can't curve in proper order so the closest reflection object is formed behind the retina . It can be cured by using convex eyeglasses. The function of convex lens is to circumcise the light before it enters the eyes. This is intended to form a reflection precise in the retina.

IMAGING

+Near Object

PRESBYOPIA

An old man, usually can't see far object or near object. This is caused by the near point . Is longer than the near point of normal eyes and the far point is shorter than the far point of normal eyes . Presbyopia can be cured by bifocals lens. The negative lens is for looking at the far objects . While the positive lens is for reading or looking at the near object.

ASTIGMATISM

Is cause by the misdistribution of the curvedness of cornea. it affects the light that enters the eye don't refract evenly . it's caused why astigmatic person can't differentiated the horizontal and vertical lines . it can be cured by using cylindrical eyeglasses .

LASER CORRECTION

EXAMPLE

mf

cmf

f

f

ssf

1

100

100

11

100

111

'

111

A man suffers myopia has far point 100 cm. He wants to see far objects. Find the focal length and optical power of the lens he should use.Hints: place the object in the man’s far point

dioptreP

P

fP

11

1

1

ANGULAR SIZE When you see a ship moving toward a harbor you will realize

that at first the ship looks very small, but when it’s getting closer it will looks bigger. We can explain this occurrence with angular size concept.

3 2 1

h

A B C

A’ B’ C’

• If the angular magnitude () of an object bigger, the image formed will also looks bigger in the retina.

3> 2 > 1 then C’>B’>A’

ANGULAR MAGNIFICATION Angular magnification defined as ratio between

angular size of image of an object () and angular size of the object ().

aM

• Memorize well…• With this concept we can derive equations of magnifying glass and

microscope.

MAGNIFYING GLASS A magnifying glass is a large convex lens commonly used to

examine small objects. The image is virtual because it is only perceived by the viewer’s brain, and cannot be captured on a screen.

Angular size if we see objects using magnifying glass is bigger than if we see objects directly.

Magnifying glass has angular magnification, there are 3 conditions of angular magnification on a magnifying glass:

a.Eye accommodate at x distanceb.Eye at maximum accommodationc. Eye with no accommodation

EYE ACCOMMODATE AT CERTAIN “X” DISTANCE

h’ h

S

S’ = -x

h

Sn

na

na

Sfx

fxM

S

SM

..

x

S

f

SM nn

a

EYE AT MAXIMUM ACCOMMODATION When eye fully accommodate, image formed from

magnifying glass must placed in eye’s near point, so S’=-Sn and x=Sn, subtitute Sn to x in our previous equation

1

f

SM

S

S

f

SM

x

S

f

SM

na

n

nna

nna

h’ h

S

S’ = -x

h

Sn=near point

EYE WITH NO ACCOMMODATION Image formed from magnifying glass must be placed in

eye’s far point, far point = , to obtain image at infinite distance we must place the object in the focal point.

f

SM

f

SM

S

f

SM

x

S

f

SM

na

na

nna

nna

0

h’

f

EXAMPLE

cmf

mf

f

Pf

5

05.020

1

1

A magnifying glass with optical power 20 dioptre used by a man with near point 30 cm and far point 200 cm. Find the location of the image and magnification of the magnifying glass when man’s eye at maximum accommodation.

cms

s

s

s

sfs

ssf

7

3030

7130

1

30

6130

1

5

11

'

111

'

111

a.

b.

75

355

5

5

30

1

a

a

na

M

M

f

SM