Why does the wave of color that has a greater index of refraction bend more?

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Discussion Overview

The discussion centers around the behavior of light as it passes through a prism, specifically addressing why colors with a greater index of refraction appear to bend more despite Snell's law suggesting otherwise. The scope includes conceptual understanding and technical reasoning related to optics.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the apparent contradiction between the bending of light in a prism and Snell's law, noting that shorter wavelengths (higher index of refraction) bend more.
  • Another participant points out that light is bent twice when passing through a prism, once entering and once exiting, suggesting that the second bending contributes significantly to the overall deviation.
  • A later reply asserts that according to Snell's law, while a higher index of refraction results in a smaller angle of refraction, this leads to a greater overall change in angle, implying that blue light, with a higher index, deviates more than red light.

Areas of Agreement / Disagreement

Participants express differing interpretations of Snell's law and its implications for light bending in a prism. There is no consensus on the resolution of the apparent contradiction regarding the bending of different colors of light.

Contextual Notes

Participants reference Snell's law and the behavior of light in a prism, but there are unresolved aspects regarding the interpretation of the law and its application to the bending of light of different wavelengths.

Who May Find This Useful

This discussion may be useful for individuals interested in optics, particularly those exploring the behavior of light in prisms and the implications of refractive indices on light bending.

101nancyma
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Hi guys, I just have a short question. A prism splits white light into it component color depending on the wavelength of the color. As the wavelength decreases, the index of refraction increases, so the wave bends more. However according to Snell's law, a greater index of refraction results in smaller degree of bend of the light. So why does the wave of color that has a greater index of refraction bend more?

thanks so much
 
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keep in mind the light gets bent twice, once when entering the prism, and once when leaving the prism. The second is where most of the bending takes place.
 
jpreed said:
keep in mind the light gets bent twice, once when entering the prism, and once when leaving the prism. The second is where most of the bending takes place.

Thanks a lot for the response. Blue has a greater index refraction than red, so should red bend more the second time than blue if snell's law holds true?
 
101nancyma said:
A prism splits white light into it component color depending on the wavelength of the color.
Good.
As the wavelength decreases, the index of refraction increases, so the wave bends more.
Good.
However according to Snell's law, a greater index of refraction results in smaller degree of bend of the light. So why does the wave of color that has a greater index of refraction bend more?
You are misunderstanding Snell's law, which says n1sinθ1 = n2sinθ2. It's true that as the index of refraction (n2) increases, the angle of refraction (θ2) decreases. But that leads to a greater change in angle, which is what counts. (If θ1 = θ2, that would mean the light didn't bend at all.)

So blue, which has the greatest index of refraction, deviates from its original path the most; red, the least.
 

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