# Power required to move a glass bead in a viscous fluid

1. Dec 11, 2014

### rwooduk

1. The problem statement, all variables and given/known data

A bead of radius R(=5 μm) is trapped by an optical beam and moved through a
viscous fluid at a speed vd of 20 μms-1. If the viscous drag is given by Stokes
law:

$$F_{d}=6\pi \eta Rv_{d}$$

obtain an expression for the laser power (intensity). If the process only has an efficiency of 25% what laser power
is required to drag the bead?

(Assume that the viscosity of the medium is η=10-3Pa s, and its refractive index, n=1.3)

2. Relevant equations

To keep the particle still, the scattering force from the laser must be equal to the viscous drag from the fluid:

Fs=Fd

3. The attempt at a solution

I can obtain an expression for keeping the particle still, as described in the Relevant equations above. However if we want to move the particle it must (?) involve the gradient force. The force that when the laser is moved will shift the particle towards it's centre, hence the particle will move. But we are not given required info for this derivation, such as the refractive index of the bead.

I'm really lost with this question, has anyone any experience with optics type questions such as these? I've looked online and cant find a similar question anywhere.

Any point in the right direction would really be appreciated.

2. Dec 11, 2014

### Bystander

"Still?"
Fdrag(v=0,still) = 0

3. Dec 12, 2014

### rwooduk

hm, indeed i misinterpretd the question wrong. In that the laser moves the bead in the fluid, the bead isnt already moving in the fluid. Still, would the drag be due to the gradient force as described in the OP?

4. Dec 12, 2014

### rwooduk

5. Dec 12, 2014

### Bystander

You have two forces working, viscous and scattering. Don't get too sidetracked pursuing a "gradient" force until you've defined it.

6. Dec 13, 2014

### rwooduk

Ahhh i see what you are saying, why define the gradient force when it must be equal (slightly greater than) the viscous drag. Will work on this some more, thanks for the replies!