Wave Optics. April 9, 2014 Chapter 34 1

Transcription

1 Wave Optics April 9, 2014 Chapter 34 1

2 Announcements! Remainder of this week: Wave Optics! Next week: Last of biweekly exams, then 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 9, 2014 Chapter 34 2

3 Single-Slit Diffraction! Diffraction of light through a single slit of width a that is comparable to the wavelength of light that is passing through the slit! Approach: Consider suitable individual rays Spherical wavelets emitted from a distribution of points in the slit Light emitted from these points will superimpose and interfere based on the path length difference for each wavelet at each position! At a distant screen we observe an intensity pattern characteristic of diffraction! Bright and dark fringes! For diffraction we will analyze the dark fringes as destructive interference April 9, 2014 Chapter 34 4

4 Single-Slit Diffraction! We assume that the point P on the screen is far enough away that the rays r 1 and r 2 are parallel and make an angle θ with the central axis! Therefore the path length difference for these two rays is sinθ = Δx a / 2 Δx = asinθ 2! The criterion for the first dark fringe is Δx = asinθ = λ 2 2 asinθ = λ! Although we chose one ray originating from the top edge of the slit and one from the middle of the slit, we could have used any two rays that originated a/2 apart inside the slit April 9, 2014 Chapter 34 5

5 Single-Slit Diffraction! We can generalize our equation to the four, eight, light rays.! The result is that the dark fringes from single slit diffraction can be described by asinθ = mλ m = 1,2,3,...! If the screen is placed a large distance from the slits, the angle θ will be small and we can make the approximation sinθ tanθ = y / L ( )! We can express the position of the dark fringes as ay L = mλ y = mλl a ( m = 1,2,3,... ) April 9, 2014 Chapter 34 6

6 Single-Slit Diffraction! The intensity of light passing through a single slit can be calculated but we will not present the derivation here! The intensity I relative to I max if there were no slit is! We can see that this expression for the intensity is zero for sinα = 0 (unless α = 0), which means α = mπ for m = 1, 2,! For α = 0, we use! So I = I max lim α 0 sinα α sinα α = 1 mπ = πa λ 2 where α = πa λ sinθ sinθ asinθ = mλ ( m = 1,2,3,... ) April 9, 2014 Chapter 34 7

7 Single-Slit Diffraction! If the screen is placed a large distance from the slits, θ is small and can be approximated as sinθ tanθ = y/l and α = πay λl! If the screen is L = 2.0 m away from the slit, the slit has a width of a = m, and the wavelength of the incident light is λ = 550 nm, we get the intensity pattern April 9, 2014 Chapter 34 8

8 Diffraction by a Circular Opening! Here is a diffraction pattern for red laser light with wavelength λ = 633 nm passing through a circular opening with diameter d = 0.04 mm projected on a screen located 1.00 m from the opening! The diameter of the innermost dark circle is measured to be 3.9 cm! There clearly is a circular Diffraction pattern! April 9, 2014 Chapter 34 11

9 Resolution 2 circular openings Resolved Barely Resolved Not Resolved! If their interference patterns overlap, two openings appear as one April 9, 2014 Chapter 34 13

10 Diffraction and resolution! Try to observe two distant point objects, such as two stars, whose angular separation is small, through a lens! Diffraction limits the ability of the lens to distinguish these two objects! The criterion for being able to separate two point objects is based on the idea that if the first image is centered on the first diffraction minimum of the second object, the objects are just resolved! This criterion is called Rayleigh s Criterion given by 1.22λ θ R = sin 1 d d is the diameter of the lens or optical element April 9, 2014 Chapter 34 14

11 Image from two distant stars! Light from two distant stars i.e. two point sources! Stars move closer to each other! Size of image is determined by size of lens April 9, 2014 Chapter 34 15

12 Hubble Space Telescope! The diameter of the Hubble Space Telescope is 2.4 m PROBLEM! What is the minimum angular resolution of the Hubble Space Telescope? SOLUTION! Using Rayleigh s Criterion with λ = 550 nm, we get 1.22λ θ R = sin 1 d m 2.4 m = ! This corresponds to the angle subtended by a dime at a distance of 64 km April 9, 2014 Chapter 34 16

13 Resolution of the Human Eye! The diameter of the pupil of the human eye is about 4mm PROBLEM! What is the minimum angular resolution of the human eye? SOLUTION! Using Rayleigh s Criterion with λ = 550 nm, we get θ R λ = = : m 4 sin : 1/100deg d 4 mm! Example: 4k TV has 4000 lines horizontally. At eye resolution, this fills 40 deg just about the field of view! April 9, 2014 Chapter 34 17

14 Double Slit Interference+Diffraction Pattern for single slit diffraction Pattern for double slit interference April 9, 2014 Chapter 34 18

15 Double Slit Interference+Diffraction! Including diffraction effects, the intensity of the interference pattern from double slits is given by I = I max cos 2 β sinα α! If the screen is placed a sufficiently large distance from the slits then we can write α = πay λl 2 and β = πdy λl where α = πa πd sinθ and β = λ λ sinθ! On the next slide is the interference/diffraction pattern observed by green laser light incident on a double slit compared with a calculation with a = mm, d = mm, and λ = 532 nm April 9, 2014 Chapter 34 19

16 Real Life Two Slit Diffraction Pattern Diffraction minima April 9, 2014 Chapter 34 20

17 Diffraction Gratings! Putting many slits together forms a device called a diffraction grating! A portion of a diffraction grating is shown! In this drawing we see coherent light with wavelength λ incident on a series of narrow slits each separated by a distance d! A diffraction pattern is produced on a screen a long distance L away April 9, 2014 Chapter 34 21

18 Diffraction Gratings! We can expand our drawing to enable our analysis of the path length difference for the light from the slits to the screen! The distance d is called the grating spacing! If the grating is W wide, the number N of slits or gratings will be d = 1 n l! Diffraction gratings are often specified in terms the number of slits or rulings per unit length, n l! We can obtain d from the specified n l N = W d April 9, 2014 Chapter 34 22

19 Diffraction Gratings! We can calculate the path length differences for the paths shown! Using an adjacent pair of rays, the path length difference is Δx = d sinθ! To produce constructive interference this path length difference must be an integer multiple of the wavelength so ( ) d sinθ = mλ m = 0,1,2,...! The values of m correspond to different bright lines m = 0: central maximum at θ = 0 m = 1: the first order maximum m = 2: the second order maximum, etc. April 9, 2014 Chapter 34 23

20 Diffraction Gratings! Diffraction gratings are designed to produce large angular separations between the maxima so we do not make the small angle approximation for diffraction gratings! Because diffraction gratings produce widely spaced narrow maxima, they can be used to determine the wavelength of monochromatic light by rearranging mλ θ = sin 1 d m = 0,1,2,... ( )! Monochromatic light incident on a diffraction grating will produce lines on a screen at widely separated angles! The pattern produced by green laser light incident on a diffraction grating with n l = 787 lines/cm is shown below April 9, 2014 Chapter 34 24