Basic Interference and. Classes of of Interferometers

Similar documents
Chapter 16. Fraunhofer Diffraction

Outline. Basics of interference Types of interferometers. Finite impulse response Infinite impulse response Conservation of energy in beam splitters

Lecture 04: HFK Propagation Physical Optics II (Optical Sciences 330) (Updated: Friday, April 29, 2005, 8:05 PM) W.J. Dallas

MAGNETIC FIELD INTRODUCTION

The condition for maximum intensity by the transmitted light in a plane parallel air film is. For an air film, μ = 1. (2-1)

Collaborative ASSIGNMENT Assignment 3: Sources of magnetic fields Solution

Chapter 3 Optical Systems with Annular Pupils

Physics 2B Chapter 22 Notes - Magnetic Field Spring 2018

SAMPLE PAPER I. Time Allowed : 3 hours Maximum Marks : 70

e.g: If A = i 2 j + k then find A. A = Ax 2 + Ay 2 + Az 2 = ( 2) = 6

Fresnel Diffraction. monchromatic light source

Solution Set #3

Fundamentals of Photonics Bahaa E. A. Saleh, Malvin Carl Teich

DOING PHYSICS WITH MATLAB COMPUTATIONAL OPTICS

Class XII - Physics Wave Optics Chapter-wise Problems. Chapter 10

A moving charged particle creates a magnetic field vector at every point in space except at its position.

Waves and Polarization in General

Faraday s Law (continued)

Physics 2212 GH Quiz #2 Solutions Spring 2016

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer

Department of Chemistry Chapter 4 continued

Sources of the Magnetic Field. Moving charges currents Ampere s Law Gauss Law in magnetism Magnetic materials

PHY2054 Exam 1 Formula Sheet

Velocimetry Techniques and Instrumentation

( )( )( ) ( ) + ( ) ( ) ( )

Sources of Magnetic Fields (chap 28)

3. Electromagnetic Waves II

r r q Coulomb s law: F =. Electric field created by a charge q: E = = enclosed Gauss s law (electric flux through a closed surface): E ds σ ε0

Chapter 31 Faraday s Law

KEPLER S LAWS AND PLANETARY ORBITS

Ch 30 - Sources of Magnetic Field! The Biot-Savart Law! = k m. r 2. Example 1! Example 2!

6.4 Period and Frequency for Uniform Circular Motion

Magnetic Fields Due to Currents

PHYS Dynamics of Space Vehicles

Lecture 1a: Satellite Orbits

( ) Make-up Tests. From Last Time. Electric Field Flux. o The Electric Field Flux through a bit of area is

Magnetic Field of a Wire

16.1 Permanent magnets

Chapter 7. Interference

GM r. v = For Newton s third law, the forces in the action/reaction pair always act on different objects

Chapter 8. Accelerated Circular Motion

AY 7A - Fall 2010 Section Worksheet 2 - Solutions Energy and Kepler s Law

Physics 122, Fall October 2012

[Griffiths Ch.1-3] 2008/11/18, 10:10am 12:00am, 1. (6%, 7%, 7%) Suppose the potential at the surface of a hollow hemisphere is specified, as shown

Inverse Square Law and Polarization

Lab #0. Tutorial Exercises on Work and Fields

Magnetic fields (origins) CHAPTER 27 SOURCES OF MAGNETIC FIELD. Permanent magnets. Electric currents. Magnetic field due to a moving charge.


Phys 201A. Homework 5 Solutions

Multipole Radiation. February 29, The electromagnetic field of an isolated, oscillating source

SEE LAST PAGE FOR SOME POTENTIALLY USEFUL FORMULAE AND CONSTANTS

Circular Motion. Mr. Velazquez AP/Honors Physics

Metrology and Sensing

( ) ( )( ) ˆ. Homework #8. Chapter 27 Magnetic Fields II.

Physics 201, Lecture 6

REVIEW Polar Coordinates and Equations

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

ANNUAL EXAMINATION - ANSWER KEY II PUC - PHYSICS

06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )

Double-angle & power-reduction identities. Elementary Functions. Double-angle & power-reduction identities. Double-angle & power-reduction identities

Electromagnetic Waves

TELE4652 Mobile and Satellite Communications

Doppler Radar (Fig. 3.1) A simplified block diagram 10/29-11/11/2013 METR

Φ E = E A E A = p212c22: 1

Physics 4A Chapter 8: Dynamics II Motion in a Plane

Fri Angular Momentum Quiz 10 RE 11.a; HW10: 13*, 21, 30, 39 Mon , (.12) Rotational + Translational RE 11.b Tues.

Solutions. V in = ρ 0. r 2 + a r 2 + b, where a and b are constants. The potential at the center of the atom has to be finite, so a = 0. r 2 + b.

The geometric construction of Ewald sphere and Bragg condition:

Phys101 Lectures 30, 31. Wave Motion

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

Review for 2 nd Midterm

1) Consider an object of a parabolic shape with rotational symmetry z

Physics 107 TUTORIAL ASSIGNMENT #8

Phys-272 Lecture 17. Motional Electromotive Force (emf) Induced Electric Fields Displacement Currents Maxwell s Equations

Physics: Dr. F. Wilhelm E:\Excel files\130\m3a Sp06 130a solved.doc page 1 of 9

OSCILLATIONS AND GRAVITATION

Circular motion. Objectives. Physics terms. Assessment. Equations 5/22/14. Describe the accelerated motion of objects moving in circles.

, and the curve BC is symmetrical. Find also the horizontal force in x-direction on one side of the body. h C

Chapter 13: Gravitation

Mechanics Physics 151

Outline. Classes of polarizing devices Polarization states. Eigen-polarization of crystals. Momentum matching at boundaries Polarization calculations

Electromagnetism Physics 15b

Calculate the electric potential at B d2=4 m Calculate the electric potential at A d1=3 m 3 m 3 m

Extra notes for circular motion: Circular motion : v keeps changing, maybe both speed and

Course Updates. Reminders: 1) Assignment #8 will be able to do after today. 2) Finish Chapter 28 today. 3) Quiz next Friday

Physics 11 Chapter 3: Vectors and Motion in Two Dimensions. Problem Solving

2/26/2014. Magnetism. Chapter 20 Topics. Magnets and Magnetic Fields. Magnets and Magnetic Fields. Magnets and Magnetic Fields

Physics 506 Winter 2006 Homework Assignment #9 Solutions

11) A thin, uniform rod of mass M is supported by two vertical strings, as shown below.

Lecture 8 - Gauss s Law

We will consider here a DC circuit, made up two conductors ( go and return, or + and - ), with infinitely long, straight conductors.

cos kd kd 2 cosθ = π 2 ± nπ d λ cosθ = 1 2 ± n N db

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

Magnetic Field. Conference 6. Physics 102 General Physics II

B. Spherical Wave Propagation

How Electric Currents Interact with Magnetic Fields

Flare Calculation on EUV Optics

PHYSICS PRACTICE PROBLEMS

FOLDS (I) A Flexure (deformation-induced curvature) in rock (esp. layered) B All kinds of rocks can be folded, even granites

ME 425: Aerodynamics

Transcription:

Basic Intefeence and Classes of Intefeometes Basic Intefeence Two plane waves Two spheical waves Plane wave and and spheical wave Classes of of Intefeometes Division of of wavefont Division of of amplitude Optics 505 - James C. Wyant Page 1 of 26 Optical Detectos Respond to Squae of Electic Field E = E 1 + E 2 = E 1ˆ a 1 e i ( ω 1t +α 1 ) + E 2 a ˆ 2 e i ( ω 2t+α 2 ) I = Constant E 1 + E 2 2 = Constant E 2 1 + E 2 2 + 2E 1 E 2 ( a ˆ 1 a ˆ 2 )cos[(ω 1 ω 2 )t + α 1 α 2 ] [ ] I = I 1 + I 2 + 2 I 1 I 2 ( a ˆ 1 a ˆ 2 )cos[(ω 1 ω 2 )t + α 1 α 2 ] Iadiance at each point vaies cosinusoidally with time at the diffeence fequency Optics 505 - James C. Wyant Page 2 of 26 Page 1

Intefeence Finges I = I 1 + I 2 + 2 I 1 I 2 ( a ˆ 1 a ˆ 2 )cos[(ω 1 ω 2 )t + α 1 α 2 ] Let ω 1 = ω 2 I = I 1 + I 2 + 2 I 1 I 2 ( a ˆ 1 a ˆ 2 )cos( α 1 α 2 ) Bight intefeence finge α 1 α 2 = 2πm Dak intefeence finge α 1 α 2 = 2π m + 1 2 Optics 505 - James C. Wyant Page 3 of 26 Intefeence of Two Plane Waves k1 k 2 θ 1 θ2 y α = α 1 α 2 = k 1 k 2 + φ 1 φ 2 x E 1 = E 1 e i ( k 1 ωt+φ 1 )ˆ a 1 E 2 = E 2 e i ( k 2 ωt+φ 2 )ˆ a 2 k 1 = k( cosθ 1 i ˆ + sin θ 1 ˆ j ) k 2 = k cosθ 2 i ˆ + sin θ ˆ 2 j = x i ˆ + y ˆ j, k = 2π λ ( ) Let φ 1 = φ 2 Bight finge α = 2π m = kxcosθ ( 1 cosθ 2 )+ y( sinθ 1 sinθ 2 ) [ ] Dak finge ( ) α = 2π m + 1 2 Optics 505 - James C. Wyant Page 4 of 26 Page 2

Finge Spacing I = I 1 + I 2 + 2 I 1 I 2 ( a ˆ 1 a ˆ 2 )cos( α 1 α 2 ) Bight finge α = α 1 α 2 = 2πm = kxcosθ ( 1 cosθ 2 )+ y( sinθ 1 sinθ 2 ) Staight equi-spaced finges Look in x=0 plane Finge spacing y = [ ] λ sinθ 1 sinθ 2 Optics 505 - James C. Wyant Page 5 of 26 Finge Visibility I I 1 + I 2 + 2( a ˆ 1 a ˆ 2 ) I 1 I 2 AC I 1 + I 2 I 1 + I 2 2( a ˆ 1 a ˆ 2 ) I 1 I 2 DC y Finge Visibility = V = I max I min I max + I min if a ˆ 1 a ˆ 2 = 1 V = 2 I 1I 2 = I 1 + I AC 2 DC Optics 505 - James C. Wyant Page 6 of 26 Page 3

Finge Spatial Fequency (1) φ φ (2) /2 /2 φ λ=633 nm (1) ν s (l / mm) 2sin ( φ /2) ν s = λ ν s = sin φ λ (2) φ (Degees ) Optics 505 - James C. Wyant Page 7 of 26 Moié Patten - Two Plane Waves Optics 505 - James C. Wyant Page 8 of 26 Page 4

Effect of Polaization Diection E p y E s θ 2 E p θ 1 x E s Dependence of a ˆ 1 a ˆ 2 on angle fo s and p polaization s polaization: a ˆ 1 a ˆ 2 = 1 fo all angles p polaization: a ˆ 1 a ˆ 2 depends upon angle Optics 505 - James C. Wyant Page 9 of 26 Intefeence of Two Spheical Waves S 1 1 P( ) S 2 2 E 1 = a ˆ B 1 1 e ik [ 1 ωt+φ 1 ] 1 E 2 = a ˆ B 2 2 e ik [ 2 ωt+φ 2 ] 2 Optics 505 - James C. Wyant Page 10 of 26 Page 5

Two Spheical Waves - Finge Shape E 1 = a ˆ B 1 1 e ik [ 1 ωt+φ 1 ] 1 E 2 = a ˆ B 2 2 e ik [ 2 ωt+φ 2 ] 2 I 1 = I 2 = B 1 Constant 1 B 2 Constant 2 α = 2π { λ 1 2 }+ φ 1 φ 2 = Constant fo given finge Hypebolic Finges Optics 505 - James C. Wyant Page 11 of 26 Moié Patten - Spheical Waves Optics 505 - James C. Wyant Page 12 of 26 Page 6

Spheical Waves - Special Case #1 1 2c 2 x o if x o >> 2c then mλ = 2cy x o Same esult as fo two plane waves Optics 505 - James C. Wyant Page 13 of 26 Moié Patten - Staight Line Finges Optics 505 - James C. Wyant Page 14 of 26 Page 7

Spheical Waves - Special Case #2 y 2c Y o x Fo bight finge mλ = 1 2 ( = 2c x2 + z 2 ) c Y o 2 Finges ae concentic cicles = 2c mλ c Y o spatial fequency = 1 = 2c y o2 λ Optics 505 - James C. Wyant Page 15 of 26 Concentic Cicula Finges Optics 505 - James C. Wyant Page 16 of 26 Page 8

Intefeence of Plane Wave and Spheical Wave x o y x If x o >> y and z then θ α = 2π ysin θ x λ o 1 + y2 + z 2 2x2 o If θ = 0 bight finge when y2 + z 2 2x o = (mλ + x o ) = m λ Cicula finges of adius = 2x o m λ Optics 505 - James C. Wyant Page 17 of 26 Two Basic Classes of Intefeometes Division of Wavefont Division of Amplitude Optics 505 - James C. Wyant Page 18 of 26 Page 9

Division of Wavefont (Young s Two Pinholes) Souce Two Pinholes Intefeence of two spheical waves Optics 505 - James C. Wyant Page 19 of 26 Division of Wavefont (Lloyd s Mio) S 1 S 2 Mio Intefeence of two spheical waves Optics 505 - James C. Wyant Page 20 of 26 Page 10

Division of Wavefont (Fesnel smios) S M 1 S 1 M 2 S 2 Optics 505 - James C. Wyant Page 21 of 26 Division of Wavefont (Fesnel s Bipism) S 1 S S 2 Optics 505 - James C. Wyant Page 22 of 26 Page 11

Division of Amplitude (Beamsplitte) Beamsplitte Optics 505 - James C. Wyant Page 23 of 26 Division of Amplitude (Diffaction) θ d sin θ = mλ Diffaction Gating Optics 505 - James C. Wyant Page 24 of 26 Page 12

Division of Amplitude and Division of Wavefont Polaization Lateal Displacement Angula Displacement E O O E OA OA Savat Plate Wollaston Pism Optics 505 - James C. Wyant Page 25 of 26 Division of Amplitude and Division of Wavefont Plane Paallel Plate Optics 505 - James C. Wyant Page 26 of 26 Page 13