Interference question, thin film

[georgeh]

Here is the question I have from the textbook and what I have done so far.

A Thin film of oil ( n=1.25) is located on a smooth wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640nm and reflect no blue light at 512nm. How thick is the oil film?

$$\lambda_C=640nm$$ Constructive interference
$$\lambda_D=512nm$$ Destructive interference
We know that the light has to travel to three different mediums,
air,oil,and water.
So..
$$n_1=1.00$$ which is air.
$$n_2=1.25$$ which is oil.
$$n_3=1.33$$ which is water.

Since we know the second medium has a higher index of refraction then water.
$$n_2 > n_1$$
so our change in phase is $$\DELTA a = \lambda/2$$
Our second refraction is caused by the light hitting the oil and reaching the surface of water and going back up.
So
[tex] n_2 < n_3<br /> [\tex]<br /> so our change there is<br /> $$\DELTA b = 2t+ \lamda/2$$ <br /> Therefore, the relative shift is<br /> $$\delta=\delta b - \delta a=2t+\lamda/2-lambda/2=2t$$<br /> So we se tthat equal to constructive interference<br /> and we get<br /> $$2t=m\lamda_c$$<br /> $$t=m\lambda_c/2$$<br /> so t=320nm.<br /> So in the similar fashion, i can find the thickness t for destructive interference.<br /> instead of setting $$2t=m\lambda_c, I can set 2t=(m+1/2)\lamda_c)$$<br /> and by setting m = 1<br /> I find t= 128nm..<br /> so my question is? I have two different thicknesses, how do i 'combine them' or manipulate them to get my total thickness?[/tex]

 

[andrevdh]

I am not completely happy with this approach, but since nobody is attempting it here is my thoughts on your problem. The refractive index of a medium is somewhat dependent on the wavelength of the light. For larger wavelengths it usually increases - let's take it to be $n_o+\Delta$ for the red wavelength and $n_o-\Delta$ for the blue (the given one assumed to be for green in the middle of the spectrum).
For constructive interference we have
$$t=\frac{\lambda _r}{2(n_o+\Delta)}=\frac{a}{n_o+\Delta}$$
and for destructive
$$t=\frac{\lambda _b}{4(n_o-\Delta)}=\frac{b}{n_o-\Delta}$$
the ticknesses should be equal therfore
$$\frac{n_o+\Delta}{a}=\frac{n_o-\Delta}{b}$$
which enables you to solve for $\Delta$
The problem with this approach is to assume that the two refractive indices will be the same amount from the given one. If one can find another equation relating the quantities then this assumption is not needed.

 

[topsquark]

Here is the question I have from the textbook and what I have done so far.

A Thin film of oil ( n=1.25) is located on a smooth wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640nm and reflect no blue light at 512nm. How thick is the oil film?

$$\lambda_C=640nm$$ Constructive interference
$$\lambda_D=512nm$$ Destructive interference
We know that the light has to travel to three different mediums,
air,oil,and water.
So..
$$n_1=1.00$$ which is air.
$$n_2=1.25$$ which is oil.
$$n_3=1.33$$ which is water.

Since we know the second medium has a higher index of refraction then water.
$$n_2 > n_1$$
so our change in phase is $$\DELTA a = \lambda/2$$
Our second refraction is caused by the light hitting the oil and reaching the surface of water and going back up.
So
[tex] n_2 < n_3<br /> [\tex]<br /> so our change there is<br /> $$\DELTA b = 2t+ \lamda/2$$ <br /> Therefore, the relative shift is<br /> $$\delta=\delta b - \delta a=2t+\lamda/2-lambda/2=2t$$<br /> So we se tthat equal to constructive interference<br /> and we get<br /> $$2t=m\lamda_c$$<br /> $$t=m\lambda_c/2$$<br /> so t=320nm.<br /> So in the similar fashion, i can find the thickness t for destructive interference.<br /> instead of setting $$2t=m\lambda_c, I can set 2t=(m+1/2)\lamda_c)$$<br /> and by setting m = 1<br /> I find t= 128nm..<br /> so my question is? I have two different thicknesses, how do i 'combine them' or manipulate them to get my total thickness?[/tex]

$$<br /> Another possibility: Your constructive and destructive interference distances are periodic. So perhaps you are to look for a least common multiple of them? (This isn't so much a hint as a suggestion. I really don't know.)<br /> <br /> -Dan$$

 

[Andrew Mason]

You have the correct approach but I am a little confused by your analysis.

Since it has minimum reflection of blue at 512 nm, the thickness of oil is some number of wavelengths + a quarter wavelength of blue:$t = n\lambda + \lambda_{b}/4$. This results in the light reflecting from the oil/water boundary having a relative phase shift of $2*2\pi/4 = \pi$ relative to the light reflecting from the air/oil boundary producing destructive interference with the latter.

Since it has maximum reflection of red at 640 nm, the thickness of oil is some number of half wavelengths of the red light: [itex]t = m\lambda_{r}/2[/tex] sp that the relative phase shift is $2*2\pi/2 = 2\pi$, thereby producing constructive interference upon reflection from both surfaces.<br /> <br /> So:<br /> <br /> $$m\lambda_{r}/2 = n\lambda_{b} +\lambda_{b}/4$$<br /> <br /> Work out the minimum value for n from that, keeping in mind that m and n must be whole numbers. You also have to keep in mind that the wavelengths are affected by the index of refraction of the oil so you have to use the wavelength of the light in the oil medium in the above equation.<br /> <br /> AM<br /> <br /> [Correction: I think the above equation should be $$m\lambda_{r}/2 = n\lambda_{b}/2 +\lambda_{b}/4$$ since the thickness of oil is some number of half wavelengths + a quarter wavelength of blue:$t = n\lambda/2 + \lambda_{b}/4$. ][/itex]

 

[BobG]

Here is the question I have from the textbook and what I have done so far.

A Thin film of oil ( n=1.25) is located on a smooth wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640nm and reflect no blue light at 512nm. How thick is the oil film?

$$\lambda_C=640nm$$ Constructive interference
$$\lambda_D=512nm$$ Destructive interference
We know that the light has to travel to three different mediums,
air,oil,and water.
So..
$$n_1=1.00$$ which is air.
$$n_2=1.25$$ which is oil.
$$n_3=1.33$$ which is water.

Since we know the second medium has a higher index of refraction then water.
$$n_2 > n_1$$
so our change in phase is $$\DELTA a = \lambda/2$$
Our second refraction is caused by the light hitting the oil and reaching the surface of water and going back up.
So
[tex] n_2 < n_3<br /> [\tex]<br /> so our change there is<br /> $$\DELTA b = 2t+ \lamda/2$$ <br /> Therefore, the relative shift is<br /> $$\delta=\delta b - \delta a=2t+\lamda/2-lambda/2=2t$$<br /> So we se tthat equal to constructive interference<br /> and we get<br /> $$2t=m\lamda_c$$<br /> $$t=m\lambda_c/2$$<br /> so t=320nm.<br /> So in the similar fashion, i can find the thickness t for destructive interference.<br /> instead of setting $$2t=m\lambda_c, I can set 2t=(m+1/2)\lamda_c)$$<br /> and by setting m = 1<br /> I find t= 128nm..<br /> so my question is? I have two different thicknesses, how do i 'combine them' or manipulate them to get my total thickness?[/tex]

$$To answer the original post, the question was the thickness of the oil film (which rides on the top of the water). It is true that the light will travel through three mediums, but I think it's assumed that the pavement won't reflect much light (or at least be a very diffuse reflection), making the third medium, water, irrelevant.$$