Question

A light beam of intensity $20\, W/cm^2$ is incident normally on a perfectly reflecting surface of sides $25 \,cm \times 15 \,cm$. The momentum imparted to the surface by the light per second is

• A. $2\times10^{-5} kg \,ms^{-1}$
• B. $1\times10^{-5} kg\, ms^{-1}$
• C. $5\times10^{-5} kg \,ms^{-1}$
• D. $1.2\times10^{-5} kg\, ms^{-1}$

Answer 1

The Correct Option is C

Intensity $I = E / A$
where $E$ is energy of radiation and $A$ is incident area
$\Rightarrow E = IA$
Momentum of radiation is given by
$P =2 E / c =2 IA / c$
Where $c$ is speed of light.
$P=\frac{2 \times 20 \times 10^{4} \frac{ W }{ m ^{2}} \times 375 \times 10^{-4} m ^{2}}{3 \times 10^{8} m / s }$
$=5 \times 10^{-5} kg\,ms ^{-1}$

Answer 2

The momentum imparted to a surface by a light beam can be calculated using the relation: \[ p = \frac{I \cdot A}{c} \] Where:
\(I\) is the intensity of the light,
\(A\) is the area of the surface, 
\(c\) is the speed of light in vacuum (\(3 \times 10^8 \, \text{m/s}\)).
Given:
\(I = 20 \, \text{W/cm}^2 = 20 \times 10^4 \, \text{W/m}^2\),
\(A = 25 \, \text{cm} \times 15 \, \text{cm} = 0.25 \, \text{m} \times 0.15 \, \text{m} = 0.0375 \, \text{m}^2\),
\(c = 3 \times 10^8 \, \text{m/s}\). Substitute the values into the formula: \[ p = \frac{(20 \times 10^4) \times 0.0375}{3 \times 10^8} = \frac{7500}{3 \times 10^8} = 2.5 \times 10^{-5} \, \text{kg ms}^{-1} \]
Since the light is perfectly reflected, the momentum doubles, so the total momentum imparted is: \[ 2 \times 2.5 \times 10^{-5} = 5 \times 10^{-5} \, \text{kg ms}^{-1} \] Thus, the momentum imparted to the surface by the light per second is \(1 \times 10^{-5} \, \text{kg ms}^{-1}\).