Diffraction Physics 202 Professor Lee Carkner Lecture 26

PAL #25 Interference Applications  Wavelength of laser  D =  d = 0.25 mm = m  y = 1.5 cm = m (between any 2 maxima)   = yd/D = (0.015)(2.5X10 -4 )/(5.5) = 682 nm  Is this reasonable? 

Diffraction  When light passes though a small aperture it spreads out   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

Diffraction and Optics  Real point sources of light always experience diffraction and so we never see true point images   Diffraction limits even the best optics   Degree of diffraction (and blurriness) depends on aperture size and wavelength

Diffraction and Interference   Light rays from different parts of the same aperture can also produce interference   Instead of two rays from two slits, we have a continuum of rays emerging from one slit

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

Single Slit Diffraction

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

Diffraction Patterns  Since a sin  = for m=1, we can say for the more general case: a sin  = m  (min)   The maxima are located halfway between the minima   Since waves from the top and bottom half cancel   Smaller slit means more flaring

Rays from Slit

Intensity   Intensity of maxima decrease with increasing    We will use  which is half the phase difference between the top and bottom of the slit   = ½  = (  a/ ) sin  I = I m (sin  /  ) 2 

Diffraction and Interference

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 = 1.5, 2.5, 3.5 … maxima

Diffraction and Circular Apertures   The location of the minima depend on the wavelength and the diameter instead of slit width:  For m = 1   The minima and maxima appear as concentric circles

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 

Resolution and Circular Aperture

Rayleigh’s Criterion   This will be true if their angular separation satisfies the diffraction formula   R = 1.22 /d  This is called Rayleigh’s criterion 