 # Double Slit Interference. Intensity of Double Slit E= E 1 + E 2 I= E 2 = E 1 2 + E 2 2 + 2 E 1 E 2 = I 1 + I 2 + “interference” <== vanishes if incoherent.

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Double Slit Interference

Intensity of Double Slit E= E 1 + E 2 I= E 2 = E 1 2 + E 2 2 + 2 E 1 E 2 = I 1 + I 2 + “interference” <== vanishes if incoherent

Refraction In general v = f and changes if v does in vacuum c = f in a medium c/n = n f hence n = /n which is less than consider two light waves which are in phase in air (n=1) and each passes through a thickness L of different material upper wave has 2 = /n 2 lower wave has 1 = /n 1

Refraction Wave 2 has N 2 = L/ 2 = (L/ )n 2 wavelengths in block Wave 1 has N 1 = L/ 1 = (L/ )n 1 wavelengths in block hence N 2 -N 1 = (L/ )(n 2 -n 1 ) phase change of wave 1 is k 1 x-  t (2  / 1 )L -  t phase change of wave 2 is k 2 x-  t (2  / 2 )L -  t phase difference =(2  L/ )(n 2 -n 1 ) = 2  (N 2 -N 1 )

Refraction Emerging waves are out of phase interfere constructively if phase difference is 2  x integer (2  L/ )(n 2 -n 1 ) = 2  m hence L = m /(n 2 -n 1 )

Problem Which pulse travels through the plastic in less time?

Solution t=d/v pulse 2: t=t 1 +t 2 +t 3 +t 4 v 1 =c/1.55, v 2 =c/1.70, v 3 =c/1.60, v 4 =c/1.45 t=(L/c)( 1.55+1.70+1.60+1.45)=6.30(L/c) pulse 1: t= t 1 +t 2 +t 3 v 1 =c/1.59, v 2 =c/1.65, v 3 =c/1.50 t=(L/c)(2 x 1.59 + 1.65 +1.50)=6.33(L/c) pulse 2 takes least time

Phase Change due to Reflection Soap films, oil slicks show interference effects of light reflected from the top and bottom surfaces When a wave moves from one medium to another there is a phase shift of  if it moves more slowly in the second medium and zero if it moves more quickly Why does top portion of film appear dark? Why are there different colours?

Fixed End Phase change of 

Free End No phase change

What is phase difference between rays 1 and 2 ? ray 2 travels further => phase difference due to path difference phase difference due to extra thickness is (2  / `)(2t) but ` is the wavelength in the water medium! ` = /n ray 1 is reflected from a medium with slower speed ray 2 is reflected from a medium with higher speed extra phase difference of  due to reflection of ray 1 total phase difference  =  + (2  n/ )(2t)

Both rays reflected from media in which wave moves more slowly phase difference only due to path difference  = (2  n water / )(2t) if  = 2  m, then constructive interference if  =  (2m-1), then destructive interference  = (2  n water / )(2t) =  (2m-1), i.e. t = (2m-1)( /4n water ) non-reflecting glass uses this principle

1. Find d for the 19th and 20th bright fringe: path difference? black ray travels extra distance 2d in air =>phase diff = (2  / )(2d) note: n=1! black ray has extra phase difference of  due to reflection bright fringe when  = (2  / )(2d)+  =2  m => d=(m-1/2)( /2) 2. Give the limits on d d 19 = (19 – 1/2) /2 = 5457 nm; d 20 = 5753 nm hence 5.46 µm < d < 5.75 µm The diameters of fine wires can be accurately measured using interference patterns. Two optically flat pieces of glass of length L are arranged with the wire between them as shown above. The setup is illuminated by monochromatic light, and the resulting interference fringes are detected. Suppose L = 20 cm and yellow sodium light (  590 nm) is used for illumination. If 19 bright fringes are seen along this 20-cm distance, what are the limits on the diameter of the wire? Hint: The nineteenth fringe might not be right at the end, but you do not see a twentieth fringe at all.

Newton’s Rings Light reflected from curved lens interferes with lift reflected from plate: bright ring  =  + (2  / )(2d)=2m  2d=(m-1/2) max

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