Diffraction Physics 202 Professor Lee Carkner Lecture 24

PAL #23 Interference Applications  Wavelength of laser  D =  d =  y = 3 mm = m (between 0 and 1 maxima)  y = m D/d  = yd/D = (0.003)( )/(1.02) =  Is this reasonable?  No, laser is red and red light has a wavelength between ~ nm 

PAL #23 Interference Applications  What color does a soap film (n=1.33) appear to be if it is 500 nm thick?  We need to find the wavelength of the maxima: = (2Ln) / (m + ½) = [(2) (500nm) (1.33)] / (m + ½)  = 2660 nm, 887 nm, 532 nm, 380 nm …   Real soap bubbles change thickness due to turbulence and gravity and so the colors shift

Diffraction  When light passes though a small aperture it spreads out   This flaring produces an image with a bright central maximum and progressively fainter maxima at increasing angles   Geometric optics assume point images, but all real images are blurry

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

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

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  =

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

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

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

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

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   Small angle is better   Short and large d give better resolution (smaller  R )

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 

Next Time  Read:  Final, Monday, Feb 13, 9-11 am  About 2/3 covers optics  About 1/3 covers fluids, SMH and waves and thermo  Four equation sheets given

If 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

If 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

If 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