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# Optical Levitation-An Unexpected Finding

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Waly M.Z Karim Author ✔

### Waly M.Z Karim

#### Review

In the previous two discussions , we derived mathematical condition of optical levitation of hemisphere and a sphere. In case of the hemisphere(I will post the link below for those who are new comers) , given that laser intensity is flexible and refractive index , mass ,and the radius of the transparent hemisphere is known, we can observe levitation if the following condition is met:

\displaystyle{I_0\frac{\pi}{4c}\mu^2(1-\frac{1}{\mu})^2\frac{a^4}{R^2}\geq mg }

Here, $I_0$ is the intensity of the laser beam , $\mu$ denotes refractive index of the hemisphere ; “a” and “R” denotes the radius of the beam and hemisphere accordingly.

Now , what this equation tells us? At the very first glance, we can easily understand that we can levitate different hemispheres with different mass and radius simply by varying the intensity of the beam.

A similar equation holds for sphere that we derived in part-3:

\displaystyle{I_0\frac{\pi}{c}(1-\frac{1}{\mu})^2\frac{a^4}{R^2}\geq mg }

Here, all the notations are same as the case of sphere.

Before we explore further, let’s spend some minute in guessing. Hemisphere and sphere have quite similarity in fact the former is a part of the latter. The very nature of this similarity should enable us to relate two conditions of levitation by a simple equation. We can say this with greater confidence since all those conditions and related equations sprang from geometric configurations of those objects. This relation is pretty simple to derive, but before that let’s play a guessing game to enhance our power of imagination.

#### Let’s Have An Educated Guess!

What is the difference between a sphere and a hemisphere? Given that their radius is same, a sphere is simply double of a hemisphere. That means, a photon has to traverse more path in the former than the latter which will cause the photon to remain a longer period of time in sphere compared to hemisphere. Now, the longer the photon travels in one medium, the more energy it transfers to the medium. More energy and corresponding “more” momentum transfer will give a “greater” impulse. Consequently, we can infer that the sphere will feel a larger force compared to the hemisphere. Let’s derive the relation and see if that’s the case!

#### Mathematical Derivation

From previous two discussion , we can write:

Force due to laser beam on a hemisphere is given by,

\displaystyle{F_{hemisphere} =I_0\frac{\pi}{4c}\mu^2(1-\frac{1}{\mu})^2\frac{a^4}{R^2}}

Force due to laser beam on a sphere is given by,

\displaystyle{F_{sphere}=I_0\frac{\pi}{c}(1-\frac{1}{\mu})^2\frac{a^4}{R^2}}

Dividing the latter by the former, we get:

\displaystyle{\frac{F_{sphere}}{F_{hemisphere}}= \frac{4}{\mu^2}}

Aha! For how many materials , you can find $\mu^2$ is less than 4?If we take things like glass , our assumption that the sphere will feel a larger force will hold. But what if we take a material having refractive index greater than “2” like diamond? Obviously $\mu$ greater than 2 will end up with

$F_{sphere} > F_{hemisphere}$ which wasn’t our assumption in the preceding section. Then, what is making the difference or why refractive index is moving the needle to a considerable context? I do not have answer to this question.That’s why I called this an unexpected one!{ I’m leaving this as an open challenge: whoever may answer it with an appropriate explanation , will get a considerable reward from me}.

#### Main Take Home Message

In the preceding section we derived a relation between two forces that depends only on the refractive index and fundamental constants. It is yet another evidence that geometry of an object plays a greater role in optical levitation. It also concludes that ,it isn’t always necessary that the more time photon spends in a particular medium, the more impulse it transfers . In our next discussions, we will come up with more fun topics such as, kinematics of this levitation phenomena.