A small laser emits light at power 5.00m/w and wavelength 633nm. The laser beam is focused (narrowed) until its diameter matches the 1266nm diameter of a sphere placed in its path. The sphere is perfectly absorbing and has density . What are (a) the beam intensity at the sphere’s location (b) the radiation pressure on the sphere? (c) the magnitude of the corresponding force? And (d) the magnitude of the acceleration that force alone would give the sphere?

Power is, $I=5.0\times {10}^{-3}\text{W}$.

Diameter is, $d=1266\times {10}^{-9}\text{m}$.

Density is, $\rho =5000{\text{kg/m}}^{\text{3}}$.

Intensity is the power transferred per unit area, where the area is measured on the plane perpendicular to the direction of the propagation of the energy. The intensity or flux of radiant energy is the power transferred per unit area, where the area is measured on the plane perpendicular to the direction of propagation of the energy. In the SI system, it has a unit of watts per square meter or kg⋅s⁻³ in base units.

Intensity (I),

$I=\frac{\text{Power}}{\text{Area}}$

Radiation pressure (P_r),

$P_r=\frac{I}{c}$

Where, I is the intensity, P_r is the radiation pressure, and c is the speed of light.

Intensity (I)can be calculated as,

$I=\frac{P}{\frac{\pi d^2}{4}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} I=\frac{5.0\times {10}^{-3}}{\frac{\pi {\left(1266\times {10}^{-9}\right)}^2}{4}} \\ I=3.974\times {10}^9{\text{W/m}}^{\text{2}} \end{gathered}$$

Therefore, the beam intensity at the sphere’s location is $\text{3.974}\times {\text{10}}^{\text{9}}{\text{W/m}}^{\text{2}}$.

Radiation pressure(P_r)can be calculated as,

$P_r=\frac{I}{c}$

$\begin{array}{l}{P}_r=\frac{(3.974\times {10}^9)}{3\times {10}^8}\\ P_{\text{r}}=13.2\text{\hspace{0.17em}Pa}\end{array}$

Therefore, the radiation pressure on the sphere is 13.25Pa.

The magnitude of the force (F_r) can be calculated as,

$F_r=P_r\times \frac{\pi d^2}{4}$

$$\begin{gathered} F_r=13.25\times \pi \times \frac{{\left(1266\times {10}^{-9}\right)}^2}{4} \\ {F}_r=1.667\times {10}^{-11}\text{N} \end{gathered}$$

Therefore, the magnitude of the corresponding force is$\text{1.667}\times {\text{10}}^{\text{-11}}\text{\hspace{0.17em}N}$

Mass can be calculated as,

$\left(m\right)=\frac{\rho \times \pi d^3}{6}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} m=\frac{5000\times \pi \times {\left(1266\times {10}^{-9}\right)}^2}{6} \\ m=5.31\times {10}^{-15}\text{kg} \end{gathered}$$

To find the acceleration of the sphere (a), we can use the equation as,

$a=\frac{F_r}{m}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} a=\frac{1.667\times {10}^{-11}}{5.31\times {10}^{-15}} \\ a=3.14\times {10}^3{\text{m/s}}^{\text{2}} \end{gathered}$$

Therefore, the magnitude of the acceleration that force would give the sphere is $\text{3.14}\times {\text{10}}^{\text{3}}{\text{m/s}}^{\text{2}}$.