Lensa cermin dan gelombang

Gelombang dan optika

Keywords (kata kunci)

Ligght theory (teori cahaya)

Reflection of light (pemantulan cahaya)

Mirror (cermin) Mirror equation

(persamaan cermin) Image formation

(pembentukan bayangan)

Refraktion of light (pembiasan cahaya)

Refractive index (indeks bias)

Lens (lensa) Lens equation

(persamaan lensa) Optical instruments

(alat-alat optik) Electromagnetic waves

spectrum (spektrum gelombang elektromagnetik)

A. Nature of Light (sifat dasar cahaya)Sifat cahaya ada dua, yaitu:

1. Cahaya sebagai gelombang (waves)2. Cahaya sebagai partikel (particles)

1. Emission of Light (Pancaran Cahaya)

Elektron

Cahaya dipancarkanLight is emited

Inti

Excited stateKeadaan tereksitasi

Lower energy levelTingkat energiLebih rendah

Lowest energy levelTingkat terendah

+

2. Electromagnetic Waves (gelombang Elektromagnetik)Cahaya polikromatik (polychromatic light) adalah cahaya yang terdiri dari berbagai panjang gelombang dan frekuensi.contoh : cahaya matahari (sunshine)cahaya monokramatik (monochromatic light) adalah cahaya yang hanya terdiri dari satu panjang gelombang dan frekuensi.contoh : laser

Hubungan panjang gelombang dan frekuensi gelombang elektromagnetik (EMG), dirumuskan:

fcv

Dengan: v = c = light speed (laju cahaya)

= 3 x 108 m/s = wavelength/panjang gelombang

(m) f = frequency/frekuensi (Hz)

3. Photon (foton)adalah paket-paket energi cahaya atau energi yang dibangkitkan oleh gerakan muatan-muatan listrik (radiasi elektromagnetik)Foton merupakan partikel-partikel yang tidak bermuatan listrik dan tidak bermassa,tetapi mempunyai energi dan momentum.

Besarnya energi sebuah foton dirumuskan:

Dengan : E = photon energy

(J) h = Planck’s

constant = 6,63 x 10-34 Js f = frequency (Hz) 1 eV = 1,6 x 10-19J

hfE

Contoh soal

Calculate the amount of photon emitted by a 100 watt lamp in 2 second, if the light that radiated by the lamp has wavelength of 600 nm!

Diket : P = 100 watt t = 2 s = 600 nm = 600 x 10-9m c = 3 x 108 m/s h = 6,63 x 10-34 Js

Ditanya: n

b. Opaque Subtances (bahan tak tembus cahaya)

light raySinar cahaya

MirrorCermin

4. Interaction of Light with substances(interaksi cahaya dengan bahan)

a. Transparent Subtstances (bahan tembus cahaya)

lens

c. Translucent Substances (bahan buram)- meneruskan- memantulkan- menyerap- menghamburkancontoh : air keruh

5. Interference, Diffraction, and Polarization (interferensi, difraksi, dan polarisasi)

a. Interference (Interferensi)adalah sebuah peristiwa yang terjadi ketika dua buah gelombang bertemu pada saat bergerak dalam medium yang sama.Interferensi gelombang ada 2 yaitu: interferensi konstruktif dan destruktif

b. Difraction (difraksi)Pembelokan atau penyebaran gelombang cahaya ketika cahaya tersebut dilewatkan melalui celah sempit.contoh : difraksi sinar – x oleh kisi kristal padat

c. Polarization proses pengubahan cahaya tak terpolarisasi menjadi cahaya terpolarisasi.Proses polarisasi:- transmisi- pemantulan- pembiasan- hamburan menggunakan polaroid filter.

6. The development of Theories of Light (Perkembangan Teori-teori Cahaya)

a. Impuls Theory of Light (teori impuls cahaya)

b. Corpuscular Theory (teori Korpuskuler)

c. Waves Theory (teory gelombang)d. Theory of Electromagnetic Waves

(teori gelombang elektromagnetik)e. Quantum Theory (teori kuantum)

B. Reflection of Light (Pemantulan Cahaya)

1. Stremam of Ligth (Berkas cahaya)

Source of lightSumber cahaya

Waves frontMuka gelombang

Rays/sinar

Kinds of stream of ligth (jenis-jenis berkas cahaya)

Parallel/sejajar

DivergingMenyebar

Converging mengumpul

2. Types of Light Reflection (jenis-jenis pemantulan cahaya)

Specular Reflection(smooth surfaces)

Diffuse Reflection(rough surfaces

3. The Law of Light Reflection (Hukum Pemantulan cahaya)

I R

i R

N

I = incident ray sinar datang

R = reflected ray sinar pantul

The Law of light reflection:a. Incident ray, reflected ray, and the

normal line cut at one point and lie on one straight plane.

b. The angle of incidence (i) is equal to the angle reflection (R)

I = R

4. Reflection of Light on Plane Mirrors (Pemantulan pada Cermin Datar)

a. The Characteristics of Image on Plane Mirrors (Sifat-sifat Bayangan pada Cermin Datar)

1) Cannot catched by screen (virtual image) (bayangan maya)

2) Upright and face invertedly to the object (tegak dan menghadap berlawanan arah terhadap bendanya)

3) The image is equal in size as the object (bayangan sama besar dengan bendanya)

4) The image distance to the mirror is equal to the object distance to the mirror (jarak bayangan ke cermin sama dengan jarak benda ke cermin)

S S’

b. Drawing Image Formation in Plane Mirrors with Ray Diagram (melukis pembentukan bayangan pada cermin datar dengan diagram sinar)

c. The Sum of Image on Plane Mirror (jumlah bayangan pada cermin datar)

mn

0360

Contoh

Two plane mirrors form an angle of 90o of each. If an object is placed between both mirrors, determine the sum of image formed!

5. Reflection in Curved Mirrors (Pemantulan pada cermin Lengkung)

a. The Anatomy of Concave and Convex Mirror (anatomi cermin cekung dan cermin cembung)

OFM

OF M

R RConcave mirror Convex mirror

b. Reflection in Concave Mirrors (Pemantulan Pada Cermin Cekung)

1) Special Rays in Convave Mirrors (sinar-sinar istimewa pada cermin cekung)

a) The incident ray parallel to the principal axis will be reflected passing through the focal point.

OFM

+

b) The incident ray passing through the focal point will be reflected parallel to the principal axis.

OFM

+

c) The incident ray passing through the mirror’s center of curvature will be reflected again through the same point.

OFM

+

2) Drawing Image Formation in Concave Mirrors with Ray Diagrams (melukis pembentukan bayangan pada cermin cekung dengan diagram sinar)

3) Spherical Aberration (Aberasi Sferis)

FM

c. Reflection in Convex Mirror (Pemantulan pada Cermin Cembung)

F M

1) Special Rays in convex Mirrors (Sinar-sinar Istimewa pada Cermin Cembung)

a) The incident ray parallel to the principal axis will be reflected as if it comes from the focal point.

F MO

2) The incident ray that seems to wards the focal point will be reflected parallel to the principal axis.

F M

c) The incident ray that seems to wards the mirror’s center of curvature will be reflected as if it comes from that point.

F M

2) Drawing Image Formation in Convex Mirrors with Ray Diagram (Melukis Pembentukkan Bayangan pada Cermin Cembung dengan Diagram Sinar)

d. Esbach’s Theorem (Dalil Esbach)

OFM

III II I IV

+

CONCAVE MIRRORS

O F M

IV I II III

-

CONVEX MIRRORS

The image characteristics of concave and convex mirrors can be determined based on Esbach’s theorem according to the rules as follows:

1. R + R’ = 5

2. All images in front of the mirrors are real and inverted.

3. All images behind the mirrors are virtual and upright.

4. R’ R (then the image is magnified)

5. R’ R (then the image is reduced)

e. The Curved Mirror Equation (Persamaan Cermin Lengkung)

'

112

'

111

ssR

atau

ssf

Where: f = mirror focal length (panjang

fokus cermin) S = object distance to the mirror

(jarak benda ke cermin) S’= image distance to the mirror

(jarak bayangan ke cermin) R = mirror’s radius of vurvature

(jari-jari cermin) = 2f

Note (catatan) In the curved mirror equation, there

are rules of mark, those are: f and R is positive (+) for concave

mirrors f and R is negative (-) for convex

mirrors S is positive (+) if the object is in

front of the mirror and s is negative (-) if the object lies behind the mirror.

S’ is positive (+) if the image lies is in front of the mirror and s’ is negative (-) if the image lies behind the mirror.

Linear magnification is defined as the ratio of image height (h’) with object height (h), this magnification is formulated by the following equation.

s

s

h

hM

''

Where: M = linear magnification

(perbesaran linier) h’ = image height (tinggi

bayangan) h = object height (tinggi benda)

Sample Problem

1. A convex mirror has focal length of 20 cm. If an object lies 10 cm in front of the mirror, determine:

a. Image distance to the mirrorb. Image linier magnification.

2. An object of 2 cm in height stands upright in front of a concave mirror which has the focal length 10 cm. If the object distance to the mirror 15 cm, ditermine:

a. Image magnification b. Image height

f. Problem Solving of Two Mirrors which Face Each Other (Penyelesaian Masalah Dua Buah Cermin yang saling Berhadapan)Secara matematis jarak antar cermin dirumuskan:

2'1 ssd

Where: d = distance between mirror (jarak

antar cermin) s1’ = first image distance to the

first mirror (jarak bayangan pertama ke cermin pertama)

S2 = first image distance to the second mirror (jarak bayangan pertama ke

cermin ke dua)

The final image resulthan from the curved mirror system that face each other has the total magnification as follows.

2

'2

1

'1

21 s

sx

s

sxMMM tot

Refraction of Light(Pembiasan Cahaya)

1. The Definition of Light Refraction (Pengertian Pembiasan Cahaya)Pembiasan cahaya adalah: peristiwa pembelokan arah cahaya ketika meliwati bidang batas diantara dua medium yang berbeda.

Pada Pembiasan cahaya terjadi: Perubahan arah Perubahan kecepatan Perubahan panjang gelombang Frekuensi dan fase gelombang tetap

2. The Law of Refraction (Snell’s Law)1. Snell’s I law:

“The incident ray, refracted ray and normal line all lie on one plane”

2. Snell’s II law:“If the incident ray travels from a less dense to a denser medium, then it bends (refracts) towards the normal line, and if the incident ray travels from a denser to a less dense medium then it bends (refracts) away from the normal line.

n2 > n1

n2

n1

n2 < n1

n1

n2

Secara matematis dirumuskan:

2211 sinsin nn

Where:n1 = refractive index of medium 1

n2 = refractive index of medium 2

Θ1 = angle of incidence

Θ2 = angle of refraction

3. Refractive Index (Indeks Bias)1. Absolute refractive index

v

cn

Where:n = absolute refractive indexc = light speed in air = 3 x 108 m/sv = light speed in medium (m/s)

2. Relative Refractive Index

2

112 n

nn

Generally, for two medium, the Snell’s law equation is:

212

1

1

2

2211

sin

sin

sinsin

nn

n

or

nn

When light travels a certain medium to another medium and is refracted, then it has different speed in the two medium. Therefore, holds the following eqution.

212

1

1

2

2

1 nn

n

v

v

Sample Problem

A stream of light travels from air to a glass with the angle of incidence 60o, if nair = 1 and nglass = 3, determine the angle of light refraction!

The speed of light in air 3 x 108 m/s and its frequency 6 x 1014 Hz, determine:

a. Light speed in water (n = 1,33)b. The change of wavelength in water and in

air

Scientific Activity(Kegiatan Ilmiah)

Refraction of Light in Planparallel Glass( Pembiasan cahaya pada kaca Planparalel)

Refraction of Light in Prism (Pembiasan Cahaya pada Prisma)

Total Reflection (PemantulanTotal/Sempurna)

Total reflection can occur if the following two conditions are complied, those are, light travels from a denser to a less dense medium and the light angle of incidence is larger than the critical angle.

1

2

21

sin

90sinsin

n

n

nn

k

ok

Where : n1 = refractive index of medium 1 (denser

medium) n2 = refractive index of medium 2 (less dense

medium) k = critical angle

Reflection of Light in Planparallel Glass

(Pembiasan pada Kaca Planparallel)N1 N2

d

n1

n2 n1

n1

1

2

t

1’

2’

The magnitude of light displacement complies the following equation:

2

21

cos

sin

d

t

Where: t = displacement of light d = planparallel glass thickness 1= angle of incidence

2= angle of refraction

Refraction of Light in Prism(Pembiasan Cahaya pada

Prisma)

N1

N2

1 2 34

D

Based on the figure above, then the refraction in prism the following equations:

41

32

D

and

Where: = angle of refrator1 = first angle of incidence2 = first angle Of refractionD = deviation angle3 = second angle of Incidence4 = second angle of refraction

If 1 = 4, then:

12mD

Dm = angle of minimum deviation

Because at the moment of minimum deviation 1 = 4, then 2= 3, so that 1= ½ ( + Dm), and = 22 = 23

Then, Snell’s law equation: 2

121 sinsin pmm nDn

Where :nm = refractive index of medium

np = refractive index of pris

Specifically for ≤ 150, then holds the following equation :

1

m

pm n

nD

Sample problem

The ray of light shown in Figure 1 is incident upon a 600-600-600 glass prism, n = 1,5

1=450

2 1’

2’

600

600 600

P Q

a. Using Snell’s law of refraction, determine the angle 2, the nearest degree.

b. Using elementary geometri, determine the value of 1’

c. Determine 2’

A light hits one surface of a thick glass by angle of incidence 600. If the refraction index of glass 1,5, then calculate the angle formed by the light coming out from the glass to the normal line!

Refraction of Light in Curved Plane

(Pembiasan Cahaya pada Bidang Lengkung)

Light refraction in curved plane

s

S’

n1 n2

Mathematically, the image formation in transparent curved plane complies the following equation:

R

nn

s

n

s

n 12,21

Where:n1 = refractive index of medium 1n2 = refractive index of medium 2S = object distance to the curved plane surfaceS’ = image distance to the curved plane surfaceR = radius of curvature

While the magnification of image formed can be ditermined by the following equation:

2

1''

n

nxs

s

h

hM

Where:M = image magnificationh’ = image heighth = object height

The value of R, s and s’ from the above equtions comply the following rules: R positive (+) if the surface of plane is

convex and R negative (-) if the surface of plane is concave.

S positive (+) for real object and s negatif (-) for virtual object.

S’ positive (+) for real image and s’ negatif (-) for virtual image.

Object focal points in curved plane

F1

f1

n1 n2

S’ =

Based on the figure above, for s = f1, then s’= ~, therefore the object focal length (f1) can be determined as follows:

12

11

1

12

1

121

1221

,

nn

Rnf

thenfs

because

nn

Rns

R

nn

s

nR

nnn

s

n

Where:f1 = object focal

length

Image focal point in curved plane

n1 n2

F2

S = ~

12

22 nn

Rnf

Where:

f2 = image focallength