In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam ofintensity$\mathbf{6}\mathbf{.0}\mathbf{\text{\hspace{0.17em}mW}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$.The sphere is totally absorbing and has a radius of$\mathbf{2}\mathbf{.0}\mathbf{}\mathit{\mu }\mathbf{\text{m}}$and a uniform density of$\mathbf{5}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}{\mathbf{\text{\hspace{0.17em}kg/m}}}^{\mathbf{\text{3}}}$. What is the magnitude of the sphere’s acceleration due to the light?

• Intensity of light,$I=6.0\text{\hspace{0.17em}mW}/{\text{m}}^{\text{2}}\text{or\hspace{0.17em}}6.0\times {10}^{-3}{\text{W/m}}^{\text{2}}$
• The radius of sphere,$r=2.0\mu \text{mor\hspace{0.17em}}2.0\times {10}^{-6}\text{m}$
• The density of sphere,$\rho =5.0\times {10}^3{\text{kg/m}}^{\text{3}}$

Here, we need to use the concept of radiation pressure and the equation of force on area A due to the total absorption of light. Then, to get the acceleration, we can use the equation of Newton’s second law of motion.

Formulae:

The force due to Newton’s second law,$F=ma\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

The force acting on a reflective surface due to intensity of the incident beam,

$F=\frac{IA}{c}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Area of a circle,$A=\pi r^2\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

Volume of a sphere,$V=\frac{4}{3}\pi r^3\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)$

The mass of a body due to its density and volume,$m=\rho V\cdot \cdot \cdot \cdot \cdot \cdot \left(5\right)$

Equation equations (1) and (2), we can get the acceleration equation as follows:

$a=\frac{IA}{mc}\cdot \cdot \cdot \cdot \cdot \cdot \left(6\right)$

Now, using equation (4) in equation (5), we can get the mass of the sphere as follows:

$m=\rho \times \frac{4}{3}\pi r^3$

Substituting the above value and equation (3) in equation (6), we get the magnitude of acceleration of the sphere as follows:

$\begin{array}{c}a=\frac{3I}{4\rho cr}\\ =\frac{3\times 6.0\times {10}^{-3}}{4\times 5.0\times {10}^3\times 3.0\times {10}^8\times 2.0\times {10}^{-6}}\\ =1.5\times {10}^{-9}{\text{m/s}}^{\text{2}}\end{array}$

Hence, the value of acceleration is.$1.5\times {10}^{-9}{\text{m/s}}^{\text{2}}$