Your particle becomes more "Optically soft", because the relative index |np-nm| is smaller in EVA. Have you calculated the Anomalous diffraction approximation of van de Hulst?

Shehroz Razzaq

This is basic stuff.

There are 3 major things that control the scattering intensity:

• Size (and in certain circumstances as the size increases, the scattering can decrease. Mie theory deals with this from a theoretical basis)
• Optical Concentration (generally, until the onset of multiple scattering, if you double the concentration then the scattering intensity will double
• Optical contrast: Relative refractive index. As Gerben Vrijburg correctly explains the relative refractive index will always be lower in any medium in relation to air

BTW, Mie is named after Gustav Mie and should have an upper-case 'M'. Take a look at (needs registration):

The life of Gustav Mie and the development of the Lorenz-Mie solution to Maxwell's equations

https://www.malvernpanalytical.com/en/learn/events-and-training/webinars/W140225Gustav-mie-maxwell.html

Alan F Rawle interesting point (Gustav Mie)! Shehroz Razzaq: For more information about Gustav Mie, check out the attached article.

Frank T. Edelmann The article on Mie is a good one and I used it in my webinar with link above.

Alan F Rawle thank you for bringing this physicist to my attention. It's interesting to read that he did his PhD on the topic "Über die optischen Eigenschaften kolloidaler Goldlösungen (On the optical properties of colloidal gold suspensions)." Thus he seems to be a pioneer of gold nanoparticles.

Shehroz Razzaq

To add to the relative refractive index discussion the scattering intensity will always be greater in air (RI = 1.00; usually only 2 decimal places required) than in any fluid. There are a number of liquids with RI less than that of water (RI = 1.33 @ 589 nm) but very few with RI < 1.30). With a particle of RI of say quartz at 1.54 approx then in air the RRI is 1.54/1.00 or 1.54 and in water the RRI is 1.54/1.33 or ~ 1.16.

A corollary is that the Mie solution is more robust to optical property variation in air rather than in a liquid.

Another corollary is that the scattering intensity of a particle (very few this low) with RI 1.30 in air would be identical to that of a particle of RI 1.729 in water (RRI = 1.729/1.33 = 1.3)

Frank T. Edelmann Slight correction. Mie's paper in 1908 was his only contribution to light scattering and correctly explained the colors of gold (nano)particles. His Ph.D thesis (1892) was on partial differential equations:

'Zum Fundamentalsatz über die Existenz von Integralen partieller Differentialgleichungen'

For what it's worth, I have manage to acquire, over the years, 2 copies of his Ph.D thesis, one of which I shared a scan with Thomas Wriedt, University of Bremen and was published on his website, I believe.

Alan F Rawle agreed. 👍

Bundle of thanks for your valuable suggestions.

When particle size (5um) is greater than the incident wavelength (0.55um) the Mie scattering occurs.The forward scattered light is stronger than the back-scattered light. In literature, it is mentioned that the forward scattered light is exactly in-phase, regardless of separation of dipoles. this is not true for any other direction. The back-scattered light is out of phase. So, my question is, why forward scattered light is in-phase and back-scattered light is out of phase? If we consider scattering by dipoles, in general the scattering by dipole is coherent. The forward and back-scattered light should be same. Can anyone explain according to physics (without mathematics) ? Your cooperation will be highly appreciated. Thanks in advance.

Shehroz Razzaq

This is an exact copy of the other question you posted and contains inaccuracies. You have your replies in the other question (https://www.researchgate.net/post/Why_forward_scattered_light_is_in-phase_and_back-scattered_light_is_out_of_phase_Can_anyone_give_explanation_according_to_physics)

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