Optic

CHAPTER VICHAPTER VI

OPTICS AND OPTICS AND ELECTROMAGNETICS WAVE ELECTROMAGNETICS WAVE

¤ REFLECTION OF LIGHT ¤ REFRACTION OF LIGHT¤ OPTICAL INSTRUMENTS ¤ ELECTROMAGNETICS WAVE

The laws of reflection (Snellius Laws):incident rays, the reflection rays and the normal line are on the one flat plane.incident angle (i) = reflection angle (r)

i

1. REFLECTION OF LIGHT

r

N

i

i

r

Exercise 1:Look at the picture of the light strikes mirror I.

Find the reflection angle of the light on mirror II !

600

400 1200

a) b)

I

II

II

I

A. REFLECTON IN THE FLAT MIRROR

Exercise 2:1. A man of 180 cm tall, stand up in front of a flat mirror. The

distance of his eyes to the end of the head is 10 cm.

Find the minimum length of the mirror so the man can see all of his images completely.

1. Reflection in concave mirrors.a. Special light principals of a concave mirror

To draw the images formation of a concave mirror, we can use special light principals as follows :

1. Light rays parallel to the principles axis will be reflected through the focus.2. Light rays passing through the focus, will be reflected parallel to the principles axis. 3. Light rays arriving at the centre of the mirror will be reflected back in the direction from

which they came.

1

2

3

CFO

B. REFLECTION IN SPHERICAL MIRRORS

1. The object between O and F

Prinsip menggambarkan pembentukan bayangan:1. Gunakan minimal 2 sinar istimewa2. sinar datang harus mengenai ujung atas benda3. Letak bayangan diperpotongan langsung sinar-sinar pantul (nyata) atau perpotongan perpanjangan ke belakang sinar-sinar pantul (maya)

FC O

The properties of the images are :Virtual, Upright, and Magnified (larger)

b. Image formation in Concave Mirrors

2. The object between F and C

FC O

The properties of images are :Real, inverted, and magnified (larger)

b. Larger than C

FC O

d. The object on C point

FC O

e. The object on F point

FC O

Kesimpulan:

Cermin cekung dapat menghasilkan 5 paket sifat bayangan;Paket I : Ketika benda terletak antara O – FPaket II : Ketika benda terletak antara F – CPaket III : Ketika benda terletak lebih jauh dari pada CPaket IV: Ketika benda terletak tepat di CPaket V : Ketika benda terletak tepat di F

To draw an image on a convex mirror we use special rays principles as follows :1). Rays directed parallel to the principle axis will reflect as it the rays came from the focus point. 2). Rays directed through focus will reflect parallel to the principle axis.3). Rays along a radius of the mirror will reflect back on it self.

B. Reflection in Convex Mirrors

3

F C

1

2

a. Special Rays in Convex mirrors

The properties of images are : Virtual, Upright, and Diminished (Smaller)

F C

ROOM NUMBER OF OBJECT + ROOM NUMBER OF IMAGE = 5

4 1 2 3O CF

THE SYSTEM NUMBER OF ROOM

123 4FC O

If R.N. Image > R.N. Object → the image is bigger

If R.N. Image < R.N. Object → the image is smaller

The Formula:

's

1

s

1

f

1 +=

's

1

f

1

s

1 −=

s

1

f

1

's

1 −=

f's

f'.ss

−=

fs

f.s's

−=

s's

s'.sf

+=

1.

2.h

'h

s

'sM ==

s.M's −=

s.M's =

If the image are virtual/upright/behind of mirror

If image are real/inverted/front of mirror

R = Radius

f = Focus distance

S = Object distance

S’ = Image distance

h = Object height

h’= Image height

M = Image Magnification

f (+), R(+) → for concave mirror

f (-), R(-) → for concave mirror

s’(-) → if the image is virtual/ upright/behind the mirror

s’(+)→if the image is real/ inverted/front of the mirror

f =1/2 R

Exercise 3:1. The focus of a concave mirror 15 cm. Find the location of

the image, the magnify and image properties of an object which is put in front of the mirror as follows : a. 20 cm b. 10 cm c. 50cm

2. A concave mirror has a radius of 18 cm, what is the distance of the object in front of the mirror if the properties of the image :a. Real and magnified 1/3 timesb. Virtual and magnified 3 times

3. A concave mirror has a focus 30 cm. Find the position of object if the image that formed is upright at 10 cm from the mirror.

4. Where should an object to be part in front of a concave mirror with radius at 12 cm so the image will be magnified 2 times to the object

Exercise 4:

1. The focus of a convex mirror is 12 cm. An object of 3 cm tall is placed 15 cm in front of the mirror.Find :a. Image positionb. The magnificationc. Image height

2. An object is placed 15 cm in front of a mirror. The mirror formed virtual image and 1/5 times the size of the object. Find the radius of the mirror.

3. A car mirror has a radius of 40 cm. Find the position of the object if the image formed at 10 cm behind the mirror.

2. REFRACTION OF LIGHT

The Refraction Laws of light (Snellius Laws);

1. When rays strike from a less dense medium to denser medium, it will refracted a rays close to the normal line.

2. When the rays strike from denser medium to less dense medium, it will refracted away from the normal line .

3. When the rays come perpendicular to the surface, it won’t refracted

4. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant , and called index of refraction.

i

r’

N

water

air

i

i = 0

i = 0

1 2 3

r’

A. Index of Refraction1. The absolute index of refraction

“ the ratio between light speed in the vacuum (c) and the speed of light in that medium (v)”

n=The absolute Index of refraction of medium

v= Speed of light in medium c= Speed of light in vacuum/air = 3 x 108 m/s

2. The relative Index of refraction

v

cn =

)r(Sin

)i(Sin

v

v

n

nn

2

1

2

1

1

221 =

λλ===

i

r

n1

n2

λ1

v2λ2

v1

• n21 = The relative Index of refraction from the medium 2 to medium 1

• n1, n2 = The absolute Index of refraction of medium 1, medium 2

• λ1 , λ2= The wave length of light in medium 1, medium 2 (m)

• v1, v2 = Speed of light in medium 1, medium 2 (m/s)

1. If the absolute index of refraction of air, water , glass and diamond each is 1, 4/3, 3/2 and 5/2. Find :a. The relative index of refraction from water to glass b. The relative index of refraction from diamond to glass. c. Speed of light in water, glass and diamond (c = 3 x 108m/s)d. The wave length green light in water, glass and diamond (if the wave length green light in air is 5000Ao) e. The green light frequency f. If a light ray enters from air into water at an angle 45 o. Find an angle of refraction in water.

Problems:

2. A path of yellow rays with a wave length of 550 nm strikes into water ( n water= 4/3 ).

Find: a. The speed of the light in water. b. The wave length and frequency of the light in water

Total internal RefractionThe conditions to Total internal Refraction are ;

1. The rays should come from a denser medium (higher refractive index) to less dense medium (lower refractive index ).

2. The angle of incident is bigger then the critical angle (ic)

Less dense medium ( n2 )

Denser medium

( n1 )

ic

r= 90

Lamp

Total light refraction

• The critical angle/sudut batas ( ic ): The incidence angle to produce the refraction angle of 90o.

• Based on Snellius Law the critical angle can be found:

n2 < n1

=

=

=

=

−

1

21c

1

2c

o2c1

21

n

nsini

n

nisin

90sinnisinn

rsinn i sinn

n1 : The refraction index of medium 1

n2 : The refraction index of medium 2

ic : The critical angle

1. If the absolute index of refraction of water and glass are 4/3 and 3/2 respectively. Find the critical angle of water and glass.

2. Will the perfect reflection happen if light come from water to air with incident angle:a. 35o

b. 60o

Refraction on a Parallel shift(Dua bidang batas sejajar )

t =d Sin (i – r)

Cos r

t = the displacement of light

d = the thickness of planparallel glass

i = angle of incident

r = angle of refraction

t

d

i

r

Planparallel glass

1. Rays enter the glass parallel with incident angle of 600. If the glass thickness 2 cm and refraction index of glass 3/2, count the displacement rays?

Example problems:

REFRACTION ON A SPHERICAL MEDIUM

R

nn

S

n

S

n 1221

'

−=+

The position of the image formed by refraction on a plane with two different refractive medium can be found with the formula:

EXERCISE 2.1. A ball aquarium with a radius of 50 cm filled with water ( nwater=

4/3 ). A gold fish is on 40 cm in front of the aquarium wall. Alex is 30 cm out side of the aquarium wall.

Find: a. the distance image of the fish from the aquarium wall seen by Alex. b. the distance image from the aquarium wall of Alex seen by fish.c. the distance of the fish from Alex seen by Alex.

2.There is an object D at the bottom of the pot shown in the figure. An observer looks in the direction of the normal. How far away does the observer see the object ( n water= 4/3 ).

1 m

D

Jika medium tempat benda indeks biasnya lebih besar dibandingkan medium tempat pengamat, maka benda

seolah-olah lebih dekat, dan sebaliknya

Seekor burung terbang 1 m diatas permukaan air , diamati oleh orang yang menyelam dalam air

LENSESThe definition : a transparent material which is limited by spherical surfaces or

spherical and plane surfaces.

The kind of lenses: converging lenses (+ ) and diverging lenses ( - )

F(virtual)( - )

F(real) ( + )

The Properties of covex lenses :

1.To converge rays

2.The focus of the lenses is real ( + )

3. On the centre part is thicker

The Properties of concave lenses :

• To diverge rays

• The focus the lenses is virtual ( - )

• On the centre part is thinner

PRINCIPAL RAYS FOR LENSESa. Pricipal rays for Converging Lenses

F22.F2 F1 2.F1

12

3

1. Rays parallel to the principal axis pass trough the principal focus after trough the lens.

NOTE :

2. Rays coming from the principal focus emerge parallel to the principal axis after refraction trough the lens.

3. Rays passing trough the optical center are not deviated.

O

( + )

b. Principal Rays for Diverging Lenses

NOTE :• Rays parallel to the principal axis are refracted and the

extension of the refracted rays passes trough the focus.

• If rays are incident on the lens in such a way that their extension pass trough the focus, they are refracted parallel to the principal ( optical ) axis.

• Rays passing trough the optical center are not deviated.

F1 2F1F22F1

1

2

3

THE LENSE FORMULA

1. S'

1

S

1

f

1 +=

h

'h

s

'sM ==

P = R1: (+) R1: ( + )

R2: ( - )R2: ( + )

R1: ( + )

R2: ~

+

−=

21m

L

R

1

R

11

n

n

f

12.

3.

4.

)m()cm( f

1

f

100P ==

NOTES :

f (+) for convex lens, f(-) for concave lens

R ( + ) for the convex surface

R (-) for the concave surface

R ( ~ ) for the flat surface

nm : the refractive index of the medium surrounding area of the lens

nL : the refractive index of the lens materials.

R : the radius of the spherical surface of the lens

s’(-) if the image is virtual/ upright/front of the lens

s’(+) if the image is real/ inverted/behind the lens

Lensa konkaf konveks, jari-jari permukaan cembungnya 10 cm dan permukaan cekungnya 20 cm, indeks bias lensa 1,5. Sebuah benda setinggi 5 cm diletakkan sejauh 10 cm di depan lensa. Hitung:a. jarak bayangan ke lensab. tinggi bayangan c. sifat-sifat bayangand. kekuatan lensa jika dimasukkan ke dalam air (n= 4/3)

EXERCISE 3:1. A biconvex Lens ( n = 3/2 ) has same surface radius of 10 cm. Find : a. Lens focus length in air. b. Lens focus length in water (n water = 4/3 ).2. A Concave Convex ( n: 1,5 ) has concave radius of 20 cm

and convex radius of 30 cm.Find : a. Lens focus length in air b. Lens focus length in sulphuric acid ( n = 1,8 )

c. what kind of lens from (b)3. A biconvex lens ( n = 3/2 ) in a liquids has a focus of 50 cm

and 10 cm in the air. If the radius of the surface is the same, R.

Find : a. the radius of the spherical surface. b. the refractive Index of the liquids.

4. A Plan Concave (n = 3/2 ) has a curvature radius of 30 cm . Find : a. the focus length in air. b. the focus length in water. (n water = 4/3 )

1

Object room number and Image room number

CONVEX Lenses

CONCAVE Lenses

IIIIIIIV

IVI II III

NOTE :

*) I, II, III, IV : OBJECT ROOM number

*) 1, 2, 3, 4 : IMAGE ROOM number

*) Object room Number + image room number = 5

4 11 12 13

14111213

EXERCISE 5.1

4

3

2

5

Draw the image formation of an object asin the figure below.

EXERCISE 6.1) An object is 5 cm in front of a lens. The image is 10 cm in front of a

lens. Find the focus lens and kind of the lens.

2) A magnified and inverted image is located at distance of 30 cm from convex lens with a focus of 10 cm.

Find : a. the object distance.

b. image magnification

3) An object high of 12 cm is placed

at distance of 30 cm from concave lens which has a focus 20 cm.

Find : a. the image distance

b. the image high.

4) An object is in front of a lens, will form an image which is upright.

If a focus lens 24 cm.

Find : a. the distance object and image if image magnification is 0,5 X

b. the distance object and image if image magnification is 4 X

III. OPTICAL INSTRUMENTS

*Seeing processes : The optical axis of the lens is directed to the object seen (third), the image made up by the lens is in the second region real, will be smaller, inverted and exactly in the yellow spot of the retina.*The lens of the eye is a convex lens (positive).

*Maximum Accommodation: the conditions of the eye lens becomes spherical (the focus = small), the eyes can see the shortest distance ( near point = punctum procximum = PP, for an adult = 25 cm ).*Minimum Accommodation : the conditions of the eye lens becomes less spherical the eyes is relax, seeing the far point ( punctum remotum : PR ), the focus is bigger ).

1. EYE

The Sight Range For Normal eye and Eye Defect

• Normal Eye

• Myopic Eye (Young Eye / near sighted)

• Old Eye (Presbiopic)

• Hyperopic Eye (far sighted)

25 cm ∞PP

PP

PP

PP

PR

PR

PR

PR

2. Spectacles / Glasses To make a better sight, glasses are needed. The Power of the glasses can be found with the formula below;

A. For the myopic/near sight (need the negative lens glasses)• If he wants to have the range sight at certain far point (x)

)(

1

mfP = we count focus (f) from the

formula:

s’1

s1

f1 +=

Where:

S= x (the far point after using a glasses)

S’= -PR (the far point before using a glasses)

• If he wants to have a sight range likes the normal eye

) (

-1

mPRP =

)(-100cmPR

P =or

B. The hyperopic/far sight (need the positive lens glasses)

• If he wants to have a sight range at a certain near point (x)

)(

1

mfP =

we count focus (f) from the formula:

s’1

s1

f1 +=

Where:

S= x (the near point after using a glasses)

S’= -PP (the near point before using a glasses)

EXERCISE 7.1. Ani has a 2D minus glasses. What is Anis’s farthest distance

without glasses ? 2. The Joyo grandpa when read the news paper, the paper

should be placed 50 cm in front of the his eyes. What is the kind and the size of the glasses to help him so he can read the news paper as a common people (the Joyo grandpa ).

).(

14

cmPPP −=

• If he wants to have the range sight likes the normal eye

).(

1004

cmPPP −=

C. The Old Eye/near and far sight (need the double lenses = the positive and negative lens glasses)

3. Magnifying Glass/Loupe1. Consist of 1 positive (+)lens/Converging lens.2. The object placed in region I (O - F)3. The image in region IV are virtual, upright, and

magnified/larger. The conditions of the eyes when using optical

instruments .1. Maximum Accommodation, so:

Location of objects : S = range O - FLocation of image : S’= - PP = -Sn = - 25 cm

M =PP

f+ 1

2. Minimum Accommodation ; so Location of objects :S = fLocation of image :S’ =

3. Accommodated on a certain distance (X); Location of objects :S = range O - FLocation of image :S’= - X

~

M =PP

f

=PP

f+

X

PPM

The rays diagram on Magnifier glass.

a. Max Accommodation

F12 F2 2 F1

(+)

h

h’

SF2

S’ = -PP

b. Accommodation on the distance X

F12 F2 2 F1

(+)

h

h’

SF2

S’ = - X

c. Min Accommodation

F2

2 F1F1

2 F2

S’ = ~

S

EXERCISE 1.1. A magnifier glass has the power of 25 D used by people of

normal eyes Find the angular magnification if the eyes see the object;

a. a min accommodation (tanpa akomodasi ) b. accommodation on the distance of 1 m. (berakomodasi pada

jarak 1 m).

c. max accommodation (berakomodasi max).

d. Accommodation on the distance of 1 m but the distance the eye to magnifier glass is 10 cm.

1. A Magnifier glass has a lens with a power of 20 D. An observer with a near point of 30 cm use the loop. Find the object distance and the magnification of the if :

a. Max. accommodation. b. Accommodation on the distance of 20 cm.

c. Min accommodation

4. MicroscopeMicroscope is an optical instrument to help the eyes to see small object in order to see bigger and clear

The device consists of 2 positive lenses as fallows : Objective Lens : close to object . Ocular Lens/Eyepiece Lens : close to eyes, this

lens has the same function as magnifier glass.

fob < foc

The image that formed by objective lens is: real, inverted, and larger

The final image that formed by ocular lens is: virtual, inverted, and larger

The ray diagram of Microscope

a.Min Accommodation

Fok

F ok

Oc. L. ( + ) Ob. L.(+)

Sok

S’ob

Sob

d

b. Max Accommodation

Fob Fok

Oc. L. ( + )Ob. L ( + )

S’ok= -PP

SokSob

S’ob

Fok

d

and than, Mok depends conditions of the eyes when using the microscope; .

1. For Maximum Accommodation:Location of the final image: Sok’= - PP =-Sn = - 25 cm

Mok=PP

fok

+ 1

Microscope Magnification

Mtotal = Mob x Mok

Where: Mob =S’

ob

S ob

d = s’ob + sokobtain

2. For Minimum AccommodationLocation of the objects for ocular lens :Sok = fok

Location of the final image :Sok’ =

3. For Accommodation at certain distance (X); Location of the final image :Sok’= - X

~

Mok =PP

fok

=PP

fok

+ X

PPMok

d = s’ob + fok

d = s’ob + sok

obtain

obtain

d = the microscope length(the distance of the objective lens to the ocular lens)

Exercise 2 .1. A microscope has an objective and eyepiece lens earli of which have a

focus of 3 cm and 5 cm in front 0f the objective lens with min accomodation. If the distance between the two lenses is 14 cm. Find :

a. the magnification, of the objective lens. b. the magnification of the eyepiece

c. the total magnification

2. A microscope has an objective and eyepiece each of which has focus of 4/3 cm and 2,5 cm. The distance between objective and eyepiece lens is 69/11 cm. Find the total magnification ( max accommodation).

3. A microscope has an objective and eyepiece lens each of which has a power of 13 1/3 D and 412/3 D and the length of microscope is 23 cm. Find the magnification of the microscope to max accommodation and min accommodation.

5. TELESCOPE / BINOCULARS

1. Refractive telescope / Binoculars stars.Refractive telescope is used to watch the stars.

Telescope is an optical instruments used to see a object infinitely in order to visible near and clear usually the sight of distant object is done in a long period of fine, so it is done by minimum accommodation .

~

•This instruments consists of two convex lens ( + ). Objective Lens : is the lens close to object. The eye piece (ocular lens ) : is close to eyes

•The object : is infinitively. ( Sob= )•The image : is in the focus of the objective lens ( S’ob = fob ) • fob > fok

•The properties of the final image are : - virtual - larger

- inverted

M =f ob

S ok

d = f ob + S ok

The telescope length :

Fob Fok

Fok

a. Max Accommodation (s’ok= - PP)

S’ok

SokS’ob= fob

L. Okuler ( + )L. Obyektive(+)

The magnification;

The ray diagram of Binoculars stars

M =f ok

f ob

d = f ob + f ok

The telescope length :The magnification;

d

b. Min Accommodation (sok = fok)

Fob

F ok

Fok

Sok= fokS’ob= fob

L. Obyektive(+) L. Ocular ( + )

2. Binoculars earth / Ground telescopes• Consists of 3 lenses ( + ) : objective lense, thin lense and eyepiece lense. • The function of the thin lense is to reverse the object on the eyepiece lense.

OB (+) OK (+)I (+)

Fok

F ob, 2FP F ok

d = f ob + 4 f P + s okM =

s ok

f ob

2FP

S’ob 4 fp

sok

The conditions of the eyes to use Binocular earth: a. Maximum Accommodation (s’ok = -PP)

OB ( + ) OK ( + )

Inverter ( + )

F ok , F P

F ob, FP F ok

d = f ob + 4 f P + f okM =

f ok

f ob

b. Minimum Accommodation (sok = fok)

EXERCISE 3

1. A telescope has an objective lens with the power of 25/18 D and eyepiece lens of 25 D. The telescope is used for watch an object in the sky.Find the length and the magnification of the telescope if :a. The eye is relaxedb. Max accommodation.

2. A telescope when the eye is relaxed has a magnifikation of 50 x . If the eyespiece lens has a focus of 50 mm. Find the length of the telescope if :a. The eye is relaxedb. Max accommodation.

Gallilean binoculars• Consists of one convex lens as an objective lense one concave lense

( + ) as an eye piece lens ( - ).• The objective lens ( + ) :

– The object is at infinitive ( S = )– The image is at the focus point ( S’ = fouler )

• The eye piece lens ( - ) :– Min accommodation.....

» the object is at focus ( Sok= fok )» The image is at far point ( Sok = )

– Max accommodation ....» the object is at the first region» the image is at the near point ( S’ok = 25 cm )

• The final properties of the image are :– Virtual– inverted– larger

The magnification Gallilean bioculars

okular

objak S

FM =max

fobyfok

Exercise 4.

1. A binocular has an objective lense, thin lense and eye piece lense each of them has focus of 20 cm and 50 cm. The device is used to watch a distant object with the eye relaxed .

Find : a. The length of the ground telescope

b. The angular magnification of the telescope.

2. A Gallilean binocular consists of objective lense with a focus 21 cm and eye piece concave lense with a focus of 6 cm.

Find : a. The length of the binocular

b. The magnification telescope.

3. After the docter checked his eyes, Agus advise to change his glass from – 2 D to 2,5 D. What is the shiff of the far point of his eyes ?

REFRACTION ON A PRISM(Two sided crossing / Non Parallel sided Transparant medium)

D = (i1+r’) –

If the is quite big more than 15o (B>15o)

If the is smaller than 15 o (B < 15o):

The formula is;

( ) ( )ββδ2

1

2

1min Sin

n

nSin

kel

p=+

N1

N2

β

i1 r2

Sinar keluar menyimpang dari arah sinar masuk prisma : sudut DEVIASI

β

βδ

−= 1min

kel

p

n

n

The Deviation angle can be found from the formula :

D

np

nkel

β

β

i1