Wave Optics. April 11, 2014 Chapter 34 1

Transcription

1 Wave Optics April 11, 2014 Chapter 34 1

2 Announcements! Exam tomorrow! We/Thu: Relativity! Last week: Review of entire course, no exam! Final exam Wednesday, April 30, 8-10 PM Location: WH B115 (Wells Hall) Comprehensive, covers material from the entire semester me if you have an exam conflict a) at 8pm, b) on Wednesday Make-up exam: To be announced April 11, 2014 Chapter 34 2

3 Geometrical Optics! Law of reflection: θ i =θ r! The focal length, f, of a spherical mirror with radius R is f = R 2! We define all distances on the same side of the mirror as the object to be positive and all distances on the far side of the mirror from the object to be negative! In the case of real images, d i is positive! In cases where the image is virtual, we define d i to be negative! If the image is upright, then h i is positive and if the image is inverted, h i is negative Physics for Scientists & Engineers 2 3

4 Spherical Mirrors! Mirror equation: or! Magnification:! Image tpye and location = d d f o h i i m = = h o d o i d d i = d d f o o f Case Type Direction Magnification d o < f Virtual Upright Enlarged d = f Real Upright Image at infinity f < d o < 2 f Real Inverted Enlarged d o = 2 f Real Inverted Same size d o > 2 f Real Inverted Reduced! Concave mirrors: f positive! Convex mirrors: f negative For a convex mirror, the image is always virtual, upright, and reduced Physics for Scientists & Engineers 2 4

5 Refraction! The index of refraction, n, is given by n = c v Law of refraction - Snell s Law: n sinθ = n sinθ ! Special case: total internal reflection (n 2 < n 1 ) When θ 2 reaches 90, total internal reflection takes place instead of refraction n2 sin θ c = ( n2 n1) n 1 Physics for Scientists & Engineers 2 5

6 ! Lens Maker s equation Lenses = ( n 1) f R R 1 2! Lens equation 1 d o + 1 d i = 1 f h d i i m = = h o d o! Conventions for distances and heights We define The focal length f of a converging lens is positive The focal length of a diverging lens is negative We define the object distance d o to be positive If the image is on the opposite side of the lens from the object, the image distance d i is positive and the image is real If the image is on the same side of the lens as the object, the image distance d i is negative and the image is virtual If the image is upright, then h i is positive and if the image is inverted, h i is negative Physics for Scientists & Engineers 2 6

7 Lenses! For a converging lens, we find that for d o > f we always get a real, inverted image formed on the opposite side of the lens! For a converging lens and d o < f, we always get a virtual, upright, and enlarged image on the same side of the lens as the object! The special cases for d o > f for a converging lens are Case Type Direction Magnification f < d o < 2 f Real Inverted Enlarged d o = 2 f Real Inverted Same size d o > 2 f Real Inverted Reduced! For diverging lenses, we always get an image that is virtual, upright, and reduced in size! Power of a lens: 1 m D = f Physics for Scientists & Engineers 2 7

8 Optical Instruments! Magnifying glass: Assuming a typical value for the near point of 25 cm we can write the angular magnification as d near 0.25 m m θ f f! Magnification of a microscope m 0.25L = f f o e! Near-sightedness is corrected with diverging lens! Far-sightedness is corrected with converging lens Physics for Scientists & Engineers 2 8

9 Interference! Two coherent waves: The criterion for interference is given by a path difference Δx! Constructive interference:! Destructive interference: ( 0, 1, 2,... ) Δ x= mλ m= m=± m=± 1 Δ x= m+ λ m= m=± m=± 2 ( 0, 1, 2,... ) Physics for Scientists & Engineers 2 9

10 Double Slit! Coherent light with wavelength λ incident on two narrow slits separated by a distance d produces an interference pattern on a screen located a large distance L from the slits! The position of the bright fringes from the center line is given by y mλl = d! The position of the dark fringes from the center line is given by y 1 m+ λl 2 = d ( ) dsin θ = mλ m= 0, m=± 1, m=± 2,... 1 dsin θ = m+ λ m= 0, m=± 1, m=± 2,... 2 ( )! Intensity: I π dy λl 2 = 4Imax cos Physics for Scientists & Engineers 2 10

11 Single Slit! If coherent light with wavelength λ is incident on a single slit of width a, a single-slit diffraction pattern will be produced! The angles of the minima are given by ( ) asin θ = mλ m= 1,2,3,... m is the order of each minimum! On a screen a large distance L away from the slit, the position on the screen, measured from the center line, is given by mλl y= = a ( m 1,2,3,... )! Intensity: I sinα = Imax where α = α 2 πa sinθ λ Physics for Scientists & Engineers 2 12

12 Thin Film Interference! Reflection at the boundary of two media:! For reflected light If n 1 < n 2, the phase of the reflected wave will be changed by half a wavelength If n 1 > n 2 then there will be no phase change! Special cases: The criterion for constructive interference of light incident on a thin, optically clear medium in air such as a soap bubble is 1 λair m+ = 2 t m= 0, m=± 1, m=± 2,... 2 n ( )! The minimum thickness t min that will produce constructive interference corresponds to t min = λ air 4n Physics for Scientists & Engineers 2 14

13 Thin Film Interference! We get the same answer for the destructive interference of light passing from air to two clear optical media such that n air < n 1 < n 2 such as the coating on a camera lens 1 λair m+ = 2 t m= 0, m=± 1, m=± 2,... 2 n ( ) t min λair = 4n Physics for Scientists & Engineers 2 15

14 Interferometer! An interferometer is a device designed to measure lengths or changes in length using interference of light! An interferometer can measure lengths or changes in lengths to a fraction of the wavelength of light using interference fringes! Here we will describe an interferometer similar to one constructed by Albert Michelson and Edward Morley at the Case Institute in Cleveland, Ohio in 1887! Our interferometer is simpler than the one used by Michelson and Morley but is based on the same physical principles April 11, 2014 Chapter 34 16

15 Interferometer! A photograph and drawing of a commercial interferometer used in physics labs are shown below April 11, 2014 Chapter 34 17

16 Interferometer! Transmitted light is totally reflected from m 2 back to m 1! Reflected light is totally reflected from m 3 back toward m 1! Part of the light from m 2 is reflected by m 1 toward the viewing screen and part of the light is transmitted and not considered! Part of the light from m 3 is transmitted through m 1 and part is reflected and not considered! The light from mirrors m 2 and m 3 will interfere based on the path length difference between the two ways of arriving at the viewing screen April 11, 2014 Chapter 34 18

17 Interferometer! Both paths undergo two reflections, each resulting in a phase change of half a wavelength! The condition for constructive interference is Δx = mλ m = 0,m = ±1,...! The two different paths will have a path length difference Δx = 2x 2 2x 3 ( ) ( ) Δx = 2 x 2 x 3! The viewing screen will display concentric circles or vertical fringes corresponding to constructive and destructive interference depending on the type of de-focusing lens April 11, 2014 Chapter 34 19

18 Interferometer! If the movable mirror m 2 is moved a distance of λ/2, the fringes will shift by one fringe! Thus this type of interferometer can be used to measure changes in distance on the order of a fraction of a wavelength of light, depending on how well one can measure the shift of the interference fringes April 11, 2014 Chapter 34 20

19 LIGO! The LIGO experiment uses a Michelson interferometer to detect small movements of one mirror caused by gravitational waves. Gravitational waves have never been seen before, but LIGO hopes to be the first! Each arm of the interferometer is 4 km long! The sensitivity is ΔL/L~10-20 April 11, 2014 Chapter 34 21

20 Diffraction Gratings! To produce constructive interference the path length difference must be an integer multiple of the wavelength so d sinθ = mλ m = 0,1,2,...! The grating spacing (distance between adjacent slits) is d! If the grating is W wide, the number N of slits or gratings will be and the line density n l is N = W d ( )! The angle to the m th order maximum is mλ θ = sin 1 d m = 0,1,2,... ( ) n l = 1 d April 11, 2014 Chapter 34 22

21 Dispersion! In addition, diffraction gratings can be used to separate out different wavelength light from a spectrum of wavelengths! Sunlight will be dispersed into multiple sets of rainbow-like colors as a function of λ! If the light is composed of several discrete wavelengths, the light will be separated into sets of a few lines corresponding to those wavelengths! A diffraction grating can be quantified by its dispersion! The dispersion describes the ability of a diffraction grating to spread apart the various wavelengths in a given order! Dispersion is defined as D = Δθ Δλ! Δθ is the angular separation between two lines with wavelength difference Δλ April 11, 2014 Chapter 34 23

22 Dispersion! We can get an expression for the dispersion by differentiating with respect to λ dθ dλ = d dλ sin 1 mλ d! We can use the relation dsinθ = mλ to get dθ dλ = 1 1 sin 2 θ m d = = m d cosθ! Taking intervals of θ and λ that are not too large gives 1 1 mλ d 2 m d = m d 2 ( mλ) 2 D = Δθ Δλ = m d cosθ ( m = 1,2,3,... ) April 11, 2014 Chapter 34 24

23 Resolving Power! The resolving power R of a diffraction grating describes the ability of the diffraction grating to resolve closely spaced maxima, which depends on the width of each maximum! We define the power R of a diffraction grating to resolve two wavelengths, λ 1 and λ 2, as R = λ ave Δλ where λ = λ ave ( + λ 1 2 )/ 2 and Δλ = λ 2 λ 1! Thus to discuss the resolving power, we need an expression for the width of each maximum! The width of each maximum is defined by the position of the first minimum on each side of the maximum! We can then define the half-width θ hw of the maximum as the angle between the maximum and the first minimum April 11, 2014 Chapter 34 25

24 Resolving Power! We base our argument our analysis of single slit diffraction using the whole grating as the single slit as shown! The angle of the first minimum for single slit diffraction can be obtained where we substitute Nd for the slit width a Nd sinθ hw = λ! One can show that the width of the maxima θ hw is λ θ hw = θ is the angle corresponding to the maximum intensity for that order! The resolving power R is R = λ Δλ = Nm Nd cosθ λ ( λ + ( λ + Δλ) )/ 2 April 11, 2014 Chapter 34 26

25 X-Ray Diffraction! Wilhelm Röntgen discovered X-rays in the late 1800 s! These experiments suggested that X-rays were electromagnetic waves with a wavelength of about m! At about the same time, the study of crystalline solids suggested that the atoms of those solids were arranged in a regular repeating pattern with a spacing of about m! Putting these two ideas together, Max von Laue proposed in the early 1900 s that a crystal could serve as a three dimensional diffraction grating for X-rays! Von Laue and Friedrich Knipping did the first X-ray diffraction experiment that showed diffraction of X-rays by a crystal in 1912! Soon after Sir William Bragg and his son William Bragg derived Bragg s law and carried out a series of experiments involving X-ray diffraction from crystals April 11, 2014 Chapter 34 27

26 Crystal structure! Let s assume that we have a cubic crystal as shown! Each atom in the lattice is a distance a away from the next atom in all three directions! We can imagine various planes of atoms in this crystal that can serve as diffraction gratings April 11, 2014 Chapter 34 28

27 Planes in a crystal! For example, the horizontal planes are composed of atoms spaced a distance a apart with the planes themselves being spaced a distance a from each other! We can imagine X-rays incident on these planes and that the rows of atoms in the crystalline lattice can act like a diffraction grating! The X-rays can be thought of as scattering from the atoms April 11, 2014 Chapter 34 29

28 Diffraction on a Crystal! Interference effects are caused by path length differences! Considering adjacent planes, we can see below that the path length difference for the scattered X-rays from the two planes is Δx = Δx 1 + Δx 2 = 2asinθ! The criterion for constructive interference is given by Bragg s scattering is ( ) 2asinθ = mλ m = 0,1,2,...! This equation is known as Bragg s Law April 11, 2014 Chapter 34 30