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

[101nancyma]

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

 

[jpreed]

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.

 

[101nancyma]

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?

 

[Doc Al]

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.