Diffraction How do we know light is a wave? Waves undergo diffraction if a wave encounters an object that has an opening of dimensions similar to its, part of the wave will flare out through the opening can be understood using Huygen’s argument true for all waves e.g ripple tank

Water waves flare out when passing through opening of width a a

e.g. sound v= f =v/f = (340m/s)/1000Hz =.34 m a of door ~ 1 m => a~ 3 e.g. light ~ 500 nm = 5 x m => need smaller opening tangent to wavelets a =4

Fresnel Bright Spot Geometrical Optics Wave Optics Shine monochromatic light on a solid sphere. What image is produced behind it?

1. C close to B no diffraction=>geometrical shadow 2. C very far from B => Fraunhoffer diffraction 3. Intermediate case- rays not parallel => Fresnel diffraction Fraunhoffer diffraction is the easiest to handle

Single Slit Diffraction

z Assume screen is far enough away that red rays are parallel Path difference between neighbouring rays is z sin  Total electric field due to r 1 and r 2 is E(r 1,t)=E m [sin(kr 1 -  t) + sin(kr 1 -  t+k z sin  )]

Single Slit E(r 1,t)=E m [sin(kr 1 -  t) + sin(kr 1 -  t+k z sin  )] where range on z is 0  z  a Phase difference between top and bottom

Single Slit Amplitude =[2(E m /ksin  )sin(kasin  /2)] =[2(E m a /  )sin(  /2)] = (E m a)sin(  /2)/(  /2) I = I 0 (sin(  /2)/(  /2)) 2 where I 0 = (E m a) 2 is the maximum intensity note: lim sin(x)/x => 1 x => 0 intensity is maximum at  = 0

Single Slit I = I 0 (sin(  /2)/(  /2)) 2 = 0 when  /2 = m  =  asin  / asin  = m for a dark fringe ( m  0) note : m=0 is a maximum! where are the other maxima? maximize sin(x)/x with respect to x (d/dx) [sin(x)/x ] = cos(x)/x - sin(x)/x 2 = 0 or x = tan(x) => x =0 is a solution plot x and tan(x ) versus x and look for intersections

x ~ (m+ 1/2) , m=1,2,3 x  22 33 tan(x)

sin(x)/x [sin(x)/x] 2 