Determining index of refraction and wavelength of a visible spectrum?

In summary, the problem involves a forensic pathologist finding an unknown liquid in a murder investigation. To identify the liquid, the pathologist uses constructive interference between light of wavelengths 500 nm and 700 nm. By setting up equations using the conditions for constructive interference, the pathologist can determine the index of refraction for the unknown liquid.
  • #1
Violagirl
114
0

Homework Statement


You decide to become a forensic pathologist because you enjoy working with dead people-they don't talk back-more than with the living. In one murder investigation, you find an unknown liquid in the victim's stomach. To identify this liquid, you pour a known amount of it onto a glass slide, such that it forms a film of thickness 500 nm on the side (with air above it). After doing this, you notice that, when illuminated with white light at normal incidence, light of wavelengths 500 nm and 700 nm is predominant in the reflected light (constructive interference). A) Determine the index of refraction for this liquid assuming the index of refraction of the glass slide is less than that of the liquid. B) What wavelengths of light in the visible spectrum range (400 - 700 nm) would have been observed if the index of refraction for the glass slide had been greater than the index found in part a for the unknown liquid? Hint: The observation of different wavelengths may correspond to different orders (differences in m) of the interference pattern for the different wavelengths.

Homework Equations


constructive interference:

δtot = 2m∏

λn2 = λ/n2

See attached diagram of situation.


The Attempt at a Solution



I definitely found this problem tricky. I would love to work through it with someone to better understand for my final this Saturday. Any help with it is appreciated. I have what I've been able to work out so far shown below and on the diagram attached.

δ1 = ∏

δ2 = (2d/λn2)2∏ <----- (Does this 2∏ reflect that it travels a full wavelength distance? Meaning that it shifts half a wavelength (pi) upon hitting the second medium. And since we're told that n3 is less than n2, it does not undergo a second inversion. So this would make sense for why constructive interference takes place but does the 2pi just correspond to the full wave length that is traveled?)

δtot = δ2 - δ1 = (2d/λn2) 2∏ - ∏

= 4d∏/λn2 -∏ = 2m∏

= 4d/λn2 -1 = 2m

= 4dn2/λ = 2m + 1

This is about as far as I could work out the problem. I'm not sure I understand what to do with the wavelength values provided. I know we have to solve for n2 but the hint about differences in m with wavelength did not make sense to me. Any help is appreciated.
 

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  • #2
Violagirl said:


Homework Equations


constructive interference:

δtot = 2m∏

λn2 = λ/n2

See attached diagram of situation.


The Attempt at a Solution



I definitely found this problem tricky. I would love to work through it with someone to better understand for my final this Saturday. Any help with it is appreciated. I have what I've been able to work out so far shown below and on the diagram attached.

δ1 = ∏

δ2 = (2d/λn2)2∏ <----- (Does this 2∏ reflect that it travels a full wavelength distance? Meaning that it shifts half a wavelength (pi) upon hitting the second medium. And since we're told that n3 is less than n2, it does not undergo a second inversion. So this would make sense for why constructive interference takes place but does the 2pi just correspond to the full wave length that is traveled?)

δtot = δ2 - δ1 = (2d/λn2) 2∏ - ∏

= 4d∏/λn2 -∏ = 2m∏

= 4d/λn2 -1 = 2m

= 4dn2/λ = 2m + 1

This is about as far as I could work out the problem. I'm not sure I understand what to do with the wavelength values provided. I know we have to solve for n2 but the hint about differences in m with wavelength did not make sense to me. Any help is appreciated.


There is interference between the ray reflected directly from the front surface of the film (and undergoing phase change δ1=∏) and the one reflected from the interface between film and glass. There is no phase change at the interface, but the wave traveling through the film twice undergoes a phase change δ2=4∏nd/λ. The difference of the phases is integer multiple of 2∏ in case of constructive interference:
4∏nd/λ-∏=2m∏. Simplifying by ∏, you get the condition of constructive interference: 4nd/λ= 2m+1, that is what you got. Yes, if the distance traveled is λ it corresponds to phase change 2∏.

There is constructive interference at λ1=700 nm and at λ2=500 nm, and no other in between. So the order of interference differs by 1 between them. Use it to get the refractive index of the film.

ehild
 
  • #3
ehild said:
There is interference between the ray reflected directly from the front surface of the film (and undergoing phase change δ1=∏) and the one reflected from the interface between film and glass. There is no phase change at the interface, but the wave traveling through the film twice undergoes a phase change δ2=4∏nd/λ. The difference of the phases is integer multiple of 2∏ in case of constructive interference:
4∏nd/λ-∏=2m∏. Simplifying by ∏, you get the condition of constructive interference: 4nd/λ= 2m+1, that is what you got. Yes, if the distance traveled is λ it corresponds to phase change 2∏.

There is constructive interference at λ1=700 nm and at λ2=500 nm, and no other in between. So the order of interference differs by 1 between them. Use it to get the refractive index of the film.

ehild


Thanks a lot for your help! I got a little bit farther with the problem. Based on what you said about each wavelength having its own maxima, this is what I got:

For 700 nm:

4d (n2)/λ700 = 2m+1

For 500 nm:

4d (n2)/λ500 = 2m+1

So I suppose then you can combine these. Based on the above equation, the 500 nm wavelength will have a greater m than the 700 nm wavelength as a smaller wavelength corresponds to a greater m.

So we'll have:

4d (n2)/λ500 - 4d (n2)/λ700 = 2m+1

So from here, I can then solve for n2 if m corresponds to 1 to show for a difference of 1 for m to for between the wavelengths, right?
 
  • #4
Violagirl said:
So I suppose then you can combine these. Based on the above equation, the 500 nm wavelength will have a greater m than the 700 nm wavelength as a smaller wavelength corresponds to a greater m.

That is right but you do not know the value of m. But it is clear that the interference maxima are next to each other: the 700 nm one corresponds to some order m=M and the 500 nm one corresponds to the next order, which is M+1.

4d (n2)/700 = 2M+1,
4d (n2)/500 = 2(M+1)+1.

Subtract the equation: M cancels and you can isolate n(2).

ehild
 
  • #5
Violagirl said:
Thanks a lot for your help! I got a little bit farther with the problem. Based on what you said about each wavelength having its own maxima, this is what I got:

For 700 nm:

4d (n2)/λ700 = 2m+1

For 500 nm:

4d (n2)/λ500 = 2m+1

So I suppose then you can combine these. Based on the above equation, the 500 nm wavelength will have a greater m than the 700 nm wavelength as a smaller wavelength corresponds to a greater m.

So we'll have:

4d (n2)/λ500 - 4d (n2)/λ700 = 2m+1

So from here, I can then solve for n2 if m corresponds to 1 to show for a difference of 1 for m to for between the wavelengths, right?

That equation is not quite correct: you have to subtract the right hand sides too.

4d (n2)/λ700 = 2m1+1
4d (n2)/λ500 = 2m2+1

The difference is

4d (n2)/λ500 - 4d (n2)/λ700 = 2(m2-m1).

The maxima are next to each other, so m2-m1=1.

ehild
 

Related to Determining index of refraction and wavelength of a visible spectrum?

1. How do you determine the index of refraction of a material?

The index of refraction of a material can be determined by measuring the ratio of the speed of light in a vacuum to the speed of light in the material. This can be done using a device called a spectrometer, which measures the angles of refraction for different wavelengths of light passing through the material.

2. What is the visible spectrum?

The visible spectrum is the range of electromagnetic radiation that is visible to the human eye. It includes all the colors of the rainbow, from red to violet, and is typically measured in nanometers (nm).

3. How is the wavelength of light in the visible spectrum determined?

The wavelength of light in the visible spectrum can be determined by using a spectrometer to measure the angles of refraction for different wavelengths of light passing through a material. The wavelength can then be calculated using the index of refraction of the material and the known speed of light in a vacuum.

4. What factors can affect the index of refraction of a material?

The index of refraction of a material can be affected by various factors such as temperature, pressure, and the chemical composition of the material. It can also vary for different wavelengths of light, which is known as dispersion.

5. Why is determining the index of refraction and wavelength of the visible spectrum important?

Determining the index of refraction and wavelength of the visible spectrum is important in many scientific fields, including optics, materials science, and astronomy. It can help us understand the properties of different materials, how they interact with light, and how light is dispersed through a medium. This information is also crucial for developing technologies such as lenses, fiber optics, and spectroscopy devices.

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