Question

Someone plans to float a small, totally absorbing sphere 0.500m above an isotropic point source of light so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere’s density is 19.0 g/cm³, and its radius is 2.00mm. (a) What power would be required of the light source? (b) Even if such a source were made, why would the support of the sphere be unstable?

Short answer

1. The power of the light source is 4.68x10¹¹ w.
2. The support of the sphere is unstable because any small amount of force on the sphere in any direction will unbalance the upward radiation force and downward gravitational force from equilibrium.

Step-by-step solution

Step 1: Given

Sphere density, $\rho =19gm/c{m}^3$

The radius of the sphere, r=2mm

Distance, d=0.500m

Determining the concept

The radiation pressure formula and radiation force formula used to calculate the power of the light source.The radiation pressure formula is of total absorption of radiation. First, find the force on the surface due to radiation, and substituting this value and area of the surface, calculate the radiation pressure of the light bulb.When electromagnetic radiation of an intensityIis incident on a perfectly reflective surface, the radiation pressure P exerted on the surface is given by, P=2Ic, where c is the speed of light.

Formulae are as follows:

$\text{F =}\frac{\text{IA}}{\text{c}}$

${\text{p}}_{\text{r}}\text{=}\frac{\text{I}}{\text{c}}$

$\text{I =}\frac{\text{P}}{\text{A}}$

Here, F is the force, I is the intensity, A is the area, P_r is the radiation pressure, and P is the pressure.

(a) Determine the power of the light source.

The power of light source:

Upward radiation force from the light matches the downward gravitational force on the sphere.

Mg=F_r

The radiation pressure P_rfor the totally absorbing surface is as follows:

${\text{p}}_{\text{r}}\text{=}\frac{\text{I}}{\text{C}}$

$\text{I =}\frac{\text{P}}{{\text{A}}_{\text{r}}}$

${\text{p}}_{\text{r}}\text{=}\frac{\text{P}}{{\text{cA}}_{\text{r}}}$ …… (1)

A_r=Area of the radiation by the source,

$A_r=4{\mathrm{\pi d}}^2$ ,

d-Distance from the source to the sphere.

Radiation pressure on the sphere is,

${\text{p}}_{\text{r}}\text{=}\frac{{\text{F}}_{\text{r}}}{{\text{A}}_{\text{s}}}$

${\text{F}}_{\text{r}}{\text{=p}}_{\text{r}}{\text{A}}_{\text{s}}$

Substituting from 1),

${\text{F}}_{\text{r}}\text{=}\frac{{\text{PA}}_{\text{s}}}{{\text{cA}}_{\text{r}}}$

Here, A_s is the area of the sphere.

The only projected area of the sphere is the circle to the radiation. Hence, the area of the circle is,

$A_s={\mathrm{\pi r}}^2$

${\text{mg =F}}_{\text{r}}$

$\text{mg =}\frac{{\text{PA}}_{\text{s}}}{{\text{cA}}_{\text{r}}}$

$\text{P =}\frac{{\text{mgA}}_{\text{r}}\text{c}}{{\text{A}}_{\text{s}}}$

Resolve further as:

$\rho =\frac{m}{V_s}$

Substitute the values and solve as:

role="math" localid="1663046688980" $$\begin{gathered} m=\rho \left(\frac{4}{3}{\mathrm{\pi r}}^3\right) \\ P=\frac{\rho \left(\frac{4}{3}{\mathrm{\pi r}}^3\right)g\left(4{\mathrm{\pi d}}^2\right)c}{{\mathrm{\pi r}}^2} \\ =\frac{\left(19\times {10}^{3}kg/m^3\right)\left({\displaystyle \frac{4}{3}}\mathrm{\pi }{\left(2\times {10}^{-3}\right)}^3\right)\left(9.8m/s^2\right)\left(4\mathrm{\pi }{\left(0.5\mathrm{m}\right)}^2\right)\left(3\times {10}^{8}m/s\right)}{\mathrm{\pi }{\left(2\times {10}^{-3}\right)}^2} \\ =4.68\times {10}^{11}w \end{gathered}$$

Therefore, the power of the light source is 4.68x10¹¹ w.

(b) Determining why is the support of the sphere unstable

The movement of the sphere is unstable because any amount of minimum force in any direction will unbalance the radiation pressure upward and gravitational force downward.

Therefore, the support of the sphere is unstable because any small amount of force on the sphere in any direction will unbalance the upward radiation force and downward gravitational force from equilibrium.

Use the radiation pressure formula and radiation force formula to calculate the power of the light source.