Interference of light in thin films

[jerry222]

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?

 

[haruspex]

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 $\pm\pi$.

 

[jerry222]

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

 

[haruspex]

yeah where pi is the phase change due to refraction and reflection

So which sign should you select?

 

[jerry222]

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...

 

[haruspex]

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.

 

[jerry222]

ahhhhhhhhh Thank You !