Relationship between wavelength and refraction

In summary, the relationship between angle of refraction, wavelength, and index of refraction of a medium is dependent on the material and can be described by the Sellmeier equation or other more complex models. It is not possible to predict the angle of refraction at a specific wavelength with little information, as the material properties and wavelength range must be considered.
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While playing around with some laser diodes I have at home ~(405, 550, 650 nm) I have noticed that the refracted angles through some mediums (all?) is different.

That is, if I fire my 405nm laser through some water at [itex]\theta_{1}[/itex]=80°, the angle of refraction is ~[itex]\theta_{2}[/itex]=47.01±0.05°.

Now, if I fire my 550nm laser through the same water, at [itex]\theta_{1}[/itex], the angle of refraction is ~[itex]\theta_{3}[/itex]=47.30±0.05°.

And, finally, if I fire my 650nm laser through the same water, at [itex]\theta_{1}[/itex], the angle of refraction is ~[itex]\theta_{4}[/itex]=47.50±0.05°.

So, basically, all I know about refraction is snells law: [itex]n_{1}/n_{2}=Sin\theta_{2}/Sin\theta_{1}[/itex]. I don't really know how to mathematically find the relationship between wavelength and refraction.

I googled a bit and didn't see anything that popped out immediately to me. Aside from v=c/n => n = c/v => n = c/(fλ), and that what I'm dealing with here may be "dispersion."

So, is there a relationship here? Is there a relatively simple way for me to relate the angle refracted, wavelength, and the index of refraction of a medium?

How would I predict the angle refracted through a medium at a specific wavelength of light? Is it possible with such little information?

Could I say n = c/(λf) where c and f are fixed (what value do I use for frequency? Or is this specified on my diode?)
(My physics experienced ended with 2nd year physics, and we didn't spend too much time of refraction or optics.)
 
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  • #3
Drakkith said:
Does this help?
http://en.wikipedia.org/wiki/Refractive_index

I've been there before, and actually http://en.wikipedia.org/wiki/Dispersion_(optics ) is more helpful.

Both do have sections on what I'm asking about, but I wasn't able to construct a relationship that would or show the values I am measuring based on the information there and in other places. That's why I'm here!
 
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  • #4
In a large tank experiment, water waves are generated with straight, parallel wave fronts, 2 m apart. The wave fronts pass through two openings 5 m apart in a long board. the end of the tank is 3 m beyond the board. Where would you stand, ralative to the perpendicular bisector of the line between the openings, if you want to receive little or no wave action?
 
  • #5
How on Earth did you measure those angles in the water to that accuracy??
 
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  • #6
I reiterate:

Is there a way for me to relate the angle refracted, wavelength, and the index of refraction of a medium?

How would I predict the angle refracted through a medium at a specific wavelength of light?

Anyone have any ideas?
 
  • #7
You need to know that refractive index = speed in vacuum(air)/speed in medium
which can be written ref index = wavelength in vacuum(air)/ wavelength in medium
 
  • #8
The way that index of refraction depends on frequency is very complicated and material dependent. There is not a single simple equation to describe it. It has to do with resonant frequencies of the material which depends on the material's atomic composition as well as lattice structure. The effect is called dispersion. A simple model for dispersion is the http://faculty.uml.edu/cbaird/95.658%282012%29/Lecture2.pdf.
 
  • #9
Hey, I looked at this page cause i was looking up
Question: why wavelength of the incident wave changes the angle of bending observed as water waves in a ripple tank travel from deep to shallow water?
and this looks related.
Please & Thank you =]
 
  • #10
truesearch said:
How on Earth did you measure those angles in the water to that accuracy??

I have the same question. What apparatus did you use to obtain those angular measurements?
 
  • #11
The dependence of the angle of refraction on wavelength is a material property. Just like stiffness, or other material properties you need to look them up. If you really want to get into numerical material science simulations, you can, but you should look that info up for your material of concern in a handbook.
 
  • #12
I have the same question, how could I get refractive indices of a material for a set of wavelength? what a relation between then so it leads to calculate n for each lambda. Because I want to measure thin film thickness and it depends on table of a set of (n and lambda). Any information could help.
 
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I have another question, can we use (ps/km nm) as a refractive index unit?
 
  • #17
791980 said:
I think this link is more benefit to find n for each lambda for any material. http://www.calctool.org/CALC/phys/optics/sellmeier. it just depends on the sellmeier coefficients.
I would be more careful in using that link as a reference to calculate refractive indices, this is because Sellmeier equation is typically accurate only within certain wavelength range and this range depends on the material. Sellmeier equation doesn't describe the behavior of refractive index for any arbitrary wavelength, and that link doesn't seem to specify the range of validity of Sellmeier equation. The 2nd link shared in post #13 is more reliable.
 
  • #18
We can use this link (in post # 16) with lambda in the range of (1-2 um) and I think its enough for n.
 
  • #19
Can anyone help me finding sellmeier coefficients for CuS? I searched but no result.
I need many n with many lambda, to measure CuS thin film thickness.
 
  • #22
I need to calculate the wavelength of light knowing the refractive index but not using frequency. Can anyone help me?
 
  • #23
I don't think so. Unless you give more context.
 

1. What is the relationship between wavelength and refraction?

The relationship between wavelength and refraction is that as the wavelength of light decreases, the amount of refraction increases. This means that shorter wavelengths, such as blue and violet light, are more strongly refracted than longer wavelengths, such as red and orange light.

2. How does the medium through which light travels affect the relationship between wavelength and refraction?

The medium through which light travels can affect the relationship between wavelength and refraction. Different mediums, such as air, water, and glass, have different refractive indices, which determine how much a light ray will bend when passing through them. This means that the same wavelength of light can have different amounts of refraction depending on the medium it is passing through.

3. What is Snell's law and how does it relate to the relationship between wavelength and refraction?

Snell's law is a mathematical equation that describes the relationship between the angle of incidence and the angle of refraction when light passes through a boundary between two different mediums. It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speeds of light in the two mediums. This law helps explain why different wavelengths of light are refracted at different angles, as the speed of light is affected by the medium it is traveling through.

4. Can the relationship between wavelength and refraction be observed in everyday life?

Yes, the relationship between wavelength and refraction can be observed in everyday life. One common example is the way a pencil appears to bend when placed in a glass of water. This is due to the refraction of light as it passes from the air into the water, with shorter wavelengths being bent more than longer wavelengths.

5. Are there any real-world applications for understanding the relationship between wavelength and refraction?

Yes, understanding the relationship between wavelength and refraction has many real-world applications. This knowledge is used in the design of lenses for glasses, cameras, and other optical devices. It is also important in fields such as astronomy, where the refraction of light passing through Earth's atmosphere can affect the observation of distant objects. This relationship is also important in the study of light and its properties in various mediums, which has numerous practical applications in fields such as telecommunications and fiber optics.

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