Relationship between wavelength and refraction

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SUMMARY

The discussion centers on the relationship between wavelength and refraction, specifically examining how different wavelengths (405 nm, 550 nm, and 650 nm) affect the angle of refraction in water. The angles measured were approximately 47.01°, 47.30°, and 47.50° respectively for each wavelength at an incident angle of 80°. Participants highlighted that the refractive index varies with wavelength due to dispersion, and emphasized the complexity of deriving a simple formula for predicting refracted angles based on wavelength and the index of refraction. The Sellmeier equation was mentioned as a potential resource for calculating refractive indices across different wavelengths.

PREREQUISITES
  • Understanding of Snell's Law and its application in optics
  • Familiarity with the concept of dispersion in materials
  • Basic knowledge of refractive index and its dependence on wavelength
  • Experience with measuring angles accurately in experimental setups
NEXT STEPS
  • Research the Sellmeier equation for calculating refractive indices at various wavelengths
  • Explore the concept of dispersion and its implications in optical materials
  • Investigate methods for accurately measuring angles of refraction in laboratory settings
  • Look into material-specific refractive index data for various wavelengths
USEFUL FOR

Optics researchers, physics students, and engineers involved in material science or optical design who seek to understand the relationship between wavelength and refraction in different media.

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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 \theta_{1}=80°, the angle of refraction is ~\theta_{2}=47.01±0.05°.

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

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

So, basically, all I know about refraction is snells law: n_{1}/n_{2}=Sin\theta_{2}/Sin\theta_{1}. 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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Drakkith said:

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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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?
 
How on Earth did you measure those angles in the water to that accuracy??
 
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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?
 
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
 
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.
 
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.
 
  • #13
  • #14
  • #15
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.
 

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