GR0177 #69



Alternate Solutions 
atomike238 20110310 12:50:58  Assuming the boundaries are vacuumoil, oilglass, then there will be a phase shift at the vacuumoil boundary since < and another phase shift at the oilglass boundary because < .
That tell us that constructive interference happens for integer wavelengths (m).
Light crosses a distance d in t=d/c. but since its speed is lower in oil, . Then t=d/v=nd/c. Its as if light had to travel nd instead of d.
Therefore the condition for constructive interference becomes
The next minimum thickness other than zero is when m=1.
So = 200nm   ee7klt 20051111 04:25:27  Hi,
I think E123 has got it right and just wanted to add my 2 cents to this to see if this reasoning is right.
I think the quantity of interest is the optical path length (OPL) where since this is really the distance that the light 'sees'?
Thus for first order constructive interefence at the air/oil boundary, we require giving .
 

Comments 
kaic 20131015 23:14:18  I don't know if I'm completely misunderstanding this problem or what, but it seems to me that when the wave which reflects off the glass reemerges, it should be 180 degrees out of phase with the wave which reflected off the oil.
We have a wave incident on a surface of higher index of refraction than air. Part of it reflects with a phase change of 180 degree, part of it is transmitted, also with a phase change of 180 degrees.
The portion which is transmitted then reflects off of the glass, again undergoing a 180 degree phase change, making it back in phase with the initial incoming wave, and now 180 degrees or out of phase with the first reflected wave.
In which case shouldn't we want the path length difference to be equal to half a wavelength (in medium) so that overall the second reflected wave is 180 degrees out of phase with the incident wave, and in phase with the first reflected wave?
for odd
kaic 20131015 23:19:18 
Apparently I've been working under the false assumption that the transmitted wave also undergoes a phase change, which is apparently not true.

  Jovensky 20130411 21:21:35  I don't know why we can assume that it is air/vacuum above the oil.   risyou 20121107 21:01:01  I thought with the light distance in oil is l'= n d (n, d: index and thinness of the oil)
And for the reflected point, which is the lightest (m)λ=2l' where m = 1. You could got the exactly 200   johnhero2010 20121020 04:03:08  why did u use the destructive interference formula, while here we have a constructive one because the 1st ray reflected from the oil surface and also the 2nd ray reflected from the glass surface,both have a 180 degree change in phase so they actually are in phase so constructive interference formula should be used, but the problem is that the answer is 100 nm which is not in the answer choices..so what is wrong?rn
cz 20121104 22:10:18 
I believe all your issues have been corrected in the solution. The formula used is for constructive interference. and the answer should be 200nm.

  atomike238 20110310 12:50:58  Assuming the boundaries are vacuumoil, oilglass, then there will be a phase shift at the vacuumoil boundary since < and another phase shift at the oilglass boundary because < .
That tell us that constructive interference happens for integer wavelengths (m).
Light crosses a distance d in t=d/c. but since its speed is lower in oil, . Then t=d/v=nd/c. Its as if light had to travel nd instead of d.
Therefore the condition for constructive interference becomes
The next minimum thickness other than zero is when m=1.
So = 200nm   Kabuto Yakushi 20100911 09:27:02  I think it's important to note that the question only asks about the reflection off the thin film of oil. This makes for constructive interference because:
the phase change of the reflected light entering the oil from the air is 180 degrees
<
the phase change for the reflected light entering the glass from the oil is 180 degrees.
< ,
both wave are out of phase by 180 degrees constructive interference occurs. We use the constructive interference formula for thin lenses:
We have to be careful because if the question had asked about the entire oil/lens system than we would have had partially constructive interference. The light reflected off of the back of the glass (if the back of the glass is exposed to air that is) would have a 0 degree phase change and would destructively interfere with the reflected light from the front of the glass.
For more on this topic refer to Halliday and Resnick Fundamentals of Physics chapter 357 which has a fairly concise treatment of thin film interference.   CaspianXI 20090301 19:22:35  This is minor... but the solution said that . I think the author meant .   Poop Loops 20081101 20:29:01  Okay, so what does the glass have to do with it? Do we just ignore it or what?
gamercora 20100907 09:15:25 
the index of refraction of glass is bigger than that of oil. it means there is a halfwave loss. but there is also a halfwave loss between air and oil so no loss of course.

  etano 20071102 10:15:13  http://physics.bu.edu/~duffy/PY106/Diffraction.htmlrnthis helped me.
etano 20071102 10:15:49 
http://physics.bu.edu/~duffy/PY106/Diffraction.html

  irishroogie 20071002 14:15:17  But I thought the formula for constructive interference in thin films was: 2nt = (m+1/2)*lambda, m=0,1,2... and m=0 for minimum thickness.
Using that formula you get 100nm which is clearly none of the answers, but I am confused as to where I am going wrong?
irishroogie 20071002 20:44:06 
Kk, So my question was flawed. 2nt=m*lambda is correct since the phase changes again by 180 at the oil glass interface. Woops!

  ee7klt 20051111 04:25:27  Hi,
I think E123 has got it right and just wanted to add my 2 cents to this to see if this reasoning is right.
I think the quantity of interest is the optical path length (OPL) where since this is really the distance that the light 'sees'?
Thus for first order constructive interefence at the air/oil boundary, we require giving .
  E123 20051109 14:52:08  i think the 200nm instead of 240nm is because of the 1.2 refractive index > lambda(oil) = lambda(air) / 1.2
yosun 20051109 15:29:14 
E123: thanks for the better approximation; it has been added.

f4hy 20090402 17:33:39 
also it is important to note that the index of refraction of glass is > than that of oil. Otherwise on the reflection there would have been a phase shift of

 

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I don't know if I'm completely misunderstanding this problem or what, but it seems to me that when the wave which reflects off the glass reemerges, it should be 180 degrees out of phase with the wave which reflected off the oil.
We have a wave incident on a surface of higher index of refraction than air. Part of it reflects with a phase change of 180 degree, part of it is transmitted, also with a phase change of 180 degrees.
The portion which is transmitted then reflects off of the glass, again undergoing a 180 degree phase change, making it back in phase with the initial incoming wave, and now 180 degrees or out of phase with the first reflected wave.
In which case shouldn't we want the path length difference to be equal to half a wavelength (in medium) so that overall the second reflected wave is 180 degrees out of phase with the incident wave, and in phase with the first reflected wave?
for odd

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