Interference of light in thin films

  • #1
jerry222
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Homework Statement:
Suppose we have a thin oil film (n=1.5) in air. The light strikes the film vertically. We use a spectrometer to see that an intensity maximum at wavelength of 600nm and a minimum at 300nm. And there is no minima inbetween. How thick is the film?
Relevant Equations:
\delta=(2pi/lamda)
Phase difference is $\phi=\frac{2pi}{\lambda}* \Delta+\pi$

Phase difference, max: $\Delta \phi=2pim=\frac{2pi}{\lamda_{max}}*2nd$

Phase difference, max: $\Delta \phi=2pim=\frac{2pi}{\lamda_{min}}*2nd+pi$

Flim thickness: $d=100nm$

Set the equations equal to each other i got a d=-100nm which is strange. Can someone help?
 

Answers and Replies

  • #2
haruspex
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Homework Statement:: Suppose we have a thin oil film (n=1.5) in air. The light strikes the film vertically. We use a spectrometer to see that an intensity maximum at wavelength of 600nm and a minimum at 300nm. And there is no minima inbetween. How thick is the film?
Relevant Equations:: ##\delta=(2\pi/\lambda)##

Phase difference is ##\phi=\frac{2pi}{\lambda}* \Delta+\pi##
Phase difference, max: ##\Delta \phi=2pim=\frac{2pi}{\lamda_{max}}*2nd##
Phase difference, max: ##\Delta \phi=2pim=\frac{2pi}{\lamda_{min}}*2nd+pi##
Flim thickness: ##d=100nm##

Set the equations equal to each other i got a d=-100nm which is strange. Can someone help?
I tried fixing up your LaTeX but it still isn’t making much sense:
Phase difference is ##\phi=\frac{2\pi}{\lambda}* \Delta+\pi##
Phase difference, max: ##\Delta \phi=2\pi m=\frac{2\pi}{\lambda_{max}}*2nd##
Phase difference, max: ##\Delta \phi=2\pi m=\frac{2\pi}{\lambda_{min}}*2nd+\pi##
Flim thickness: ##d=100nm##

Don't you mean Phase difference is ##\pi##?
Anyway, a difference is unsigned, so in the equation it's ##\pm\pi##.
 
  • #3
jerry222
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I tried fixing up your LaTeX but it still isn’t making much sense:
Phase difference is ##\phi=\frac{2\pi}{\lambda}* \Delta+\pi##
Phase difference, max: ##\Delta \phi=2\pi m=\frac{2\pi}{\lambda_{max}}*2nd##
Phase difference, max: ##\Delta \phi=2\pi m=\frac{2\pi}{\lambda_{min}}*2nd+\pi##
Flim thickness: ##d=100nm##

Don't you mean Phase difference is ##\pi##?
Anyway, a difference is unsigned, so in the equation it's ##\pm\pi##.
yeah where pi is the phase change due to refraction and reflection
 
  • #4
haruspex
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yeah where pi is the phase change due to refraction and reflection
So which sign should you select?
 
  • #5
jerry222
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So which sign should you select?
ah i got -100nm, the correct answer is 100...i just set those two eaquations equal to each other, that s what i wanted to ask...
 
  • #6
haruspex
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ah i got -100nm, the correct answer is 100...i just set those two eaquations equal to each other, that s what i wanted to ask...
I think you are not getting the point I am making.
if the difference is ##\pi##, your equation should be ##\Delta \phi=2\pi m=\frac{2\pi}{\lambda_{min}}*2nd\pm\pi##. I.e., you do not at this stage know which way the difference is.
So your answer becomes ##\pm 100nm##, and you have to choose the sign that makes sense.
 
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  • #7
jerry222
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ahhhhhhhhh Thank You !
 

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