1. Diffraction Physics 202 Professor Lee Carkner Lecture 26

2. PAL #25 Interference Applications  Wavelength of laser  D = 5.5 m  d = 0.25 mm = 0.00025 m  y = 1.5 cm = 0.015 m (between any 2 maxima)  y = m D/d  = yd/D = (0.015)(2.5X10 -4 )/(5.5) = 682 nm  Is this reasonable?  Yes, laser is red and red light has a wavelength between ~ 600- 700 nm

3. If twice the thickness of the middle layer is ½ wavelength, what kind of interference would you see? a)Constructive b)Destructive c)None n=1 n=1.5n=1.3

4. If twice the thickness of the middle layer is ½ wavelength, what kind of interference would you see? a)Constructive b)Destructive c)None n=1.3 n=1.5n=1.1

5. If twice the thickness of the middle layer is ½ wavelength, what kind of interference would you see? a)Constructive b)Destructive c)None n=1.3 n=1n=1.1

6. Diffraction   A plane wave becomes a semicircular wavefront  This effect also occurs when light passes by an obstacle   The pattern consists of minima and maxima of decreasing intensity as you move away from the center

7. Diffraction and Optics   Geometric optics assume point images, but all real images are blurry   Microscopes and telescopes are limited in how small or distant an object they can resolve by diffraction  Degree of diffraction (and blurriness) depends on aperture size and wavelength

8. Diffraction and Interference  Young’s experiment is an example of light rays from two different apertures producing interference   This is called single slit diffraction  Instead of two rays from two slits, we have a continuum of rays emerging from one slit

9. Path Length Difference  Minima (dark fringes) should occur at the point where half of the rays are out of phase with the other half   If we assume that the distance to the screen (D) is much larger than the slit width (a) then the path difference is   where d is the distance between the origin points of the two rays   We will pair up the rays, and find the path length for which each pair cancels out

10. Location of the Minima  Where is the first minima?  Since:  L /d = sin    How far apart can a pair of rays get?   For the first minima  L must equal /2: (a/2) sin  = /2 a sin  =

11. Diffraction Patterns  a sin  = m  (min)  Where  is the location of the minima corresponding to order m   Note that this relationship is the reverse of that for double slit interference [d sin  = (m+½)  : min]  Since waves from the top and bottom half cancel

12. Intensity   Intensity of maxima decrease with increasing   The intensity is proportional to the value of E 2, which in turn depends on the phase difference   = ½  = (  a/ ) sin  I = I m [(sin  /  ] 2  where I m is the maximum intensity of the pattern

13. Intensity Variations  The intensity falls off rapidly with linear distance y   Remember tan  = y/D   The narrower the slit the broader the maximum  Remember:   m = 1,2,3 … minima  m = 1.5, 2.5, 3.5 … maxima

14. Diffraction and Circular Apertures   The location of the minima depend on the wavelength and the diameter instead of slit width: sin  = 1.22 /d  For m = 1   The minima and maxima appear as concentric circles

15. Rayleigh’s Criterion  We will consider two near-by point sources to be resolvable if the central maximum of one lies on the first minimum of the other   For small angles:  R = 1.22 /d  This is called Rayleigh’s criterion   Short and large d give better resolution (smaller  R )

17. Resolution  Since virtually all imaging devices have apertures, virtually all images are blurry   If you view two point sources that are very close together, you may not be able to distinguish them 

18. Next Time  Read: 36.7-36.9  Homework  Final, Monday, 9-11 am  2/3 covers optics  1/3 covers fluids, SMH and waves (test 1)  1/3 covers sound and thermo (test 2)  Three equation sheets given