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Optics}Thin Films

The incident wavelength changes phase at the air-oil boundary and at the oil-glass boundary. Thus, integer wavelengths produce constructive interference.

The incident wavelength travels through 2t. Thus, 2t = \lambda \Rightarrow \lambda \approx 240 nm. The closest thickness is 200 nm, so choose choice (B). Additionally, if one wants a better approximation, one can use the equation to determine the wavelength of the beam in oil. Thus, one arrives at the exact answer 200nm. (Addition due to user E123.)

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
ee7klt
2005-11-11 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 OPL = \int_{1}^{2} n dl since this is really the distance that the light 'sees'?

Thus for first order constructive interefence at the air/oil boundary, we require \lambda = OPL = 2nt giving t = \lambda / 2n = 480 nm / 2.4 = 200 nm.

Alternate Solution - Unverified
Comments
CaspianXI
2009-03-01 19:22:35
This is minor... but the solution said that \lambda \approx 240nm. I think the author meant t \approx 240nm.NEC
Poop Loops
2008-11-01 20:29:01
Okay, so what does the glass have to do with it? Do we just ignore it or what?NEC
etano
2007-11-02 10:15:13
http://physics.bu.edu/~duffy/PY106/Diffraction.htmlrnthis helped me.
etano
2007-11-02 10:15:49
http://physics.bu.edu/~duffy/PY106/Diffraction.html
NEC
irishroogie
2007-10-02 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
2007-10-02 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!
NEC
ee7klt
2005-11-11 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 OPL = \int_{1}^{2} n dl since this is really the distance that the light 'sees'?

Thus for first order constructive interefence at the air/oil boundary, we require \lambda = OPL = 2nt giving t = \lambda / 2n = 480 nm / 2.4 = 200 nm.

Alternate Solution - Unverified
E123
2005-11-09 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
2005-11-09 15:29:14
E123: thanks for the better approximation; it has been added.
f4hy
2009-04-02 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 \pi
Fixed Typos!

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