- #1

Hammad Shahid

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For example, why does blue light refract more than red light in a prism?

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In summary, the conversation discusses the concept of dispersion in optics and how different colors of light refract differently in a prism. It mentions the common misconception that higher frequency light is more likely to be absorbed, but explains that the speed change is what causes the difference in refraction. The conversation also suggests drawing a diagram to understand the phenomenon and mentions the Fresnel-Kirchhoff integral as a more formal way of deriving Snell's law.

- #1

Hammad Shahid

- 64

- 3

For example, why does blue light refract more than red light in a prism?

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- #2

Ibix

Science Advisor

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What does Googling this tell you? Or your textbook?

- #3

Hammad Shahid

- 64

- 3

Googling it provides this answer: "Higher frequencies of light are refracted more because objects are more likely to absorb higher frequency light. This causes those frequencies of light to travel more slowly and be refracted more."Ibix said:What does Googling this tell you? Or your textbook?

It doesn't say how though.

I already returned the textbook to the library.

- #4

Ibix

Science Advisor

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That's half right. The "more likely to absorb" thing is a common misconception. Light in matter doesn't work that way. However, the speed change is important.

Draw a diagram with a horizontal line for the boundary between the media. Draw lines at an angle to it, representing the position of a single wave crest at a sequence of times:

Apologies for the lousy sketching - drawing with my finger on my phone screen on a very cold platform. But hopefully you get the idea - the red lines are perpendicular to the incident ray and show where a wave crest is now (right hand line), and where it was one, two, and three cycles ago (the other lines). The black line is the boundary.

If you draw this with a ruler and with even spacing (say 2cm) between each red line then you can work out what happens. Say the refractive index is n=2, for easy calculation. Where could the bits of the wave crest that are in the medium be now? Well, the rightmost line isn't much help. The line next to it, however, you know that the point that just touched the boundary has traveled 2cm/n=1cm since it was drawn. So if you draw a semicircle below the boundary of radius 1cm centred on where the crest meets the boundary, that's all possible locations that it could be.

Do the same with the next red line to the right. That represents the crest twice as long ago, so the semicircle centred on it should be 4cm/n=2cm in radius. And again the next red line - you can work out the radius. Now we've got all the crest's*possible* positions, which is the right one? Draw a line just grazing all of the semicircles (remember there's implicitly one of radius zero around the rightmost line). That's your wavecrest in the medium. Have a go - see what you get. If you know some trigonometry you should be able to follow through the maths and derive Snell's Law.

If that seems a bit of an artificial process, it is. It can be formalised, however, as the Fresnel-Kirchhoff integral (https://en.m.wikipedia.org/wiki/Kirchhoff's_diffraction_formula), which is basically a more mathematically sophisticated and more generally applicable version of this argument.

Draw a diagram with a horizontal line for the boundary between the media. Draw lines at an angle to it, representing the position of a single wave crest at a sequence of times:

Apologies for the lousy sketching - drawing with my finger on my phone screen on a very cold platform. But hopefully you get the idea - the red lines are perpendicular to the incident ray and show where a wave crest is now (right hand line), and where it was one, two, and three cycles ago (the other lines). The black line is the boundary.

If you draw this with a ruler and with even spacing (say 2cm) between each red line then you can work out what happens. Say the refractive index is n=2, for easy calculation. Where could the bits of the wave crest that are in the medium be now? Well, the rightmost line isn't much help. The line next to it, however, you know that the point that just touched the boundary has traveled 2cm/n=1cm since it was drawn. So if you draw a semicircle below the boundary of radius 1cm centred on where the crest meets the boundary, that's all possible locations that it could be.

Do the same with the next red line to the right. That represents the crest twice as long ago, so the semicircle centred on it should be 4cm/n=2cm in radius. And again the next red line - you can work out the radius. Now we've got all the crest's

If that seems a bit of an artificial process, it is. It can be formalised, however, as the Fresnel-Kirchhoff integral (https://en.m.wikipedia.org/wiki/Kirchhoff's_diffraction_formula), which is basically a more mathematically sophisticated and more generally applicable version of this argument.

- #5

A.T.

Science Advisor

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- 3,651

Hammad Shahid said:For example, why does blue light refract more than red light in a prism?

Did you read this?

https://en.wikipedia.org/wiki/Dispersion_(optics)#Material_dispersion_in_optics

https://en.wikipedia.org/wiki/Snell's_law

A prism is a transparent object with flat, polished surfaces that refracts or bends light as it passes through. It is often used to separate white light into its component colors.

Higher frequency light waves have shorter wavelengths and are more easily bent or refracted by the surfaces of a prism. This is due to the fact that the index of refraction of a material, which determines how much a light wave is bent when passing through, is dependent on the wavelength of the light. So, shorter wavelengths (higher frequencies) are bent more than longer wavelengths (lower frequencies).

The angle of incidence, which is the angle at which a light wave enters the prism, is equal to the angle of refraction, which is the angle at which the light wave bends as it exits the prism. This is known as Snell's Law and it can be mathematically represented as n1sinθ1 = n2sinθ2, where n1 and n2 are the indices of refraction of the materials on either side of the prism and θ1 and θ2 are the angles of incidence and refraction, respectively.

White light is made up of a combination of all the visible colors, each with their own wavelength and frequency. When white light enters a prism, the different colors are separated due to their different wavelengths and frequencies. This causes them to bend at different angles, resulting in the rainbow effect we see when white light is refracted through a prism.

Yes, higher frequency light waves can be bent too much in a prism, causing them to overlap with other colors and creating a distorted image. This is known as chromatic aberration and can be minimized by using prisms made of materials with lower indices of refraction, or by using multiple prisms to correct for the distortion.

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