University:
California State University, Northridge
Course:
PHYS 100B | General Physics II
Academic year:

2013
Views:

382

Pages:

19
Author:

Kieran Caldwell

Chapter 28 Physical Optics: Interference and Diffraction Chapter Outline 28-1 Superposition and Interference 28-2 Young’s Two-Slit Experience 28-4 Diffraction 28-5 Resolution 28-4 Diffraction The water wave demonstrates that the wave can spread out in all direction after it goes through a small gap/slit: Huygens’s Principle for light diffraction. Figure 28-17 Diffraction of Water Waves Similar, when the light goes through narrow slit, the light is spread out in all directions, and a diffraction pattern of bright - dark fringes is formed. Diffraction Pattern on screen Figure 28-18 Single-Slit Diffraction To find the first dark fringe, we divide the slit into two regions. First dark fringe happens when: w  sin   , 2 2 i.e. w sin    Figure 28-19 Locating the First Dark Fringe in Single-Slit Diffraction To find the second dark fringe, we divide the slit into four regions. The second dark fringe happens, when: w  sin   , 4 2 i.e. w sin   2 Figure 28-20 Locating the Second Dark Fringe in Single-Slit Diffraction To find the third dark fringe, we divide the slit into six regions. The third dark fringe happens, when: w  sin   , 6 2 i.e. w sin   3 And so on: divide the slit as even number regions…. Condition for Dark Fringe in Single-Slit Interference W sin   m , M=1, first dark fringe; M=2, second dark fringe; …… m  1,2,3,    (28  12) The central bright fringe of diffraction slit has a angular width:  Central fringe Unit: radian 2  W (28  13) Exercise 28-2 Monochromatic light passes through a slit of width 1.2 x 10-5 mm. If the first dark fringe of the resulting diffraction pattern is at angle =3.25º, what is the wavelength of the light? Solution: since, W sin   m , (1.2 105 mm) sin 3.25  (1)   6.80 107 mm  680nm Example 28-5 Light with a wavelength of 511 nm forms a diffraction pattern after passing through a single slit of width 2.20x10-6m. Find the angle associated with (a) the first and (b) the second dark fringe above the central bright fringe. Solution: (a) First Dark Fringe, m=1 since W sin   m , (2.2 106 m) sin   (1)(511109 m)   13.4 (b) Second Dark Fringe, m=2 since W sin   m , (2.2 106 m) sin   (2)(511 109 m)   27.7 28-5 Resolution Diffraction pattern of a circular aperture It can be shown: The angle of First Dark Fringe for the Diffraction Pattern of a Circular Aperture is (from the bright point center to the first dark fringe) sin   1.22  D (28  14) Rayleigh’s Criterion for two-point sources: If the first dark fringe of one circular diffraction pattern passes through the center of a second diffraction pattern, the two sources will appear to be a single source, and therefore the two objects can not be seen as separate source and can not be resolved. Two objects can be seen as separate source only if their angular separation is great than the following minimum: Rayleight’s Criterion (minimum angular resolution)  min  1.22 Unit: radian  D (28  15) Active Example 28-3 The asteroid Ida is orbited by its own small “moon” called Dactyl. If the separation between these two asteroids is 2.5 km, what is the maximum distance at which the Hubble Space Telescope (aperture diameter = 2.4m) can still resolve them at 550-nm wavelength? Active Example 28-3 Resolving Ida and Dactyl Solution 1) The minimum angular resolution is  min  550  10 9 m  1.22  1.22( )  2.8 10 7 D 2.4m 2) The maximum distance is y 2.5km 6 L   8 . 9  10 km 7 tan  n tan(2.8 10 rad ) radian Summary Condition for Dark Fringe in Single-Slit Interference W sin   m , m  1,2,3,    (28  12) Two object can be seen as separate source only if their angular separation is great than the following minimum: Rayleight’s Criterion  min  1.22 Unit Radian  D (28  15)