Double Slit Interference
Intensity of Double Slit
E= E1 + E2
I= E2 = E12 + E22 + 2 E1 E2
= I1 + I2 + “interference” <== vanishes if incoherent
Refraction
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In general v = f and  changes if v does
in vacuum c = f
in a medium c/n = nf
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= /n2
lower wave has 1= /n1
Refraction
• Wave 2 has N2 = L/2 = (L/)n2
wavelengths in block
• Wave 1 has N1 = L/1 = (L/)n1
wavelengths in block
• hence N2 -N1 = (L/)(n2 -n1)
• phase change of wave 1 is k1x-t
(2/1)L - t
• phase change of wave 2 is k2x-t
(2/2)L - t
• phase difference =(2L/)(n2 -n1)
= 2(N2 -N1 )
Refraction
• Emerging waves are out of phase
• interfere constructively if phase difference
is 2 x integer
• (2L/)(n2 -n1) = 2m
• hence L = m /(n2 -n1)
Problem
• Which pulse travels through the plastic in
less time?
Solution
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t=d/v
pulse 2: t=t1+t2+t3+t4
v1=c/1.55, v2=c/1.70, v3=c/1.60, v4=c/1.45
t=(L/c)( 1.55+1.70+1.60+1.45)=6.30(L/c)
pulse 1: t= t1+t2+t3
v1=c/1.59, v2=c/1.65, v3=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
Why are there
different colours?
Why does top
portion of film
appear dark?
• 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
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nwater/)(2t)
• if  = 2m, then constructive interference
• if  =  (2m-1), then destructive interference
•  = (2nwater/)(2t) =  (2m-1), i.e. t = (2m-1)(/4nwater)
• non-reflecting glass uses this principle
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.
• 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
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d19 = (19 – 1/2)  /2 = 5457 nm; d20 = 5753 nm
• hence
5.46 µm < d < 5.75 µm
Newton’s Rings
Light reflected from
curved lens interferes
with lift reflected from
plate: bright ring
= + (2/)(2d)=2m
2d=(m-1/2)  max
r
1
2dR  (m  ) R m  1, 2,3,...
2
d  R  R2  r 2