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Relationship between wavelength and refraction 
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#1
Apr2612, 08:48 PM

P: 3

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.) 


#2
Apr2712, 12:41 AM

Mentor
P: 11,878

Does this help?
http://en.wikipedia.org/wiki/Refractive_index 


#3
Apr2712, 01:14 AM

P: 3

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! 


#4
Apr2812, 04:03 PM

P: 1

Relationship between wavelength and refraction
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
Apr2812, 04:07 PM

P: 349

How on earth did you measure those angles in the water to that accuracy??



#6
Apr2812, 05:35 PM

P: 3

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
Apr2912, 11:40 AM

P: 349

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
Apr3012, 02:59 PM

P: 617

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 classical harmonic oscillator model.



#9
Jul1812, 05:32 PM

P: 1

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
Jul1812, 06:15 PM

PF Gold
P: 1,909




#11
Apr2114, 11:44 AM

P: 46

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 in to numerical material science simulations, you can, but you should look that info up for your material of concern in a handbook.



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