Question

A laser beam with an energy flux of 20 W/cm2 is incident on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm2,the total momentum delivered by the laser in 30 minutes for complete absorption will be:

• A. 2.8x10-3 kg m/s
• B. 4.2x10-3 kg m/s
• C. 3.6x10-3 kg m/s
• D. 3.3x10-3 kg m/s
• E. 2.4x10-3 kg m/s

Answer 1

Answer: (C)

3.6x10-3 kg m/s

Answer 2

The correct option is (C): 3.6x10-3 kg m/s.
Let's calculate the total energy delivered by the laser:
Given:
The energy flux of the laser beam is $20 \, \text{W/cm}^2$.
The area of the surface is $30 \, \text{cm}^2$.
The total power P incident on the surface is:
$P = (\text{energy flux}) \times (\text{area}) = 20 \, \text{W/cm}^2 \times 30 \, \text{cm}^2 = 600 \, \text{W}$
Now, calculate the total energy absorbed by the surface over 30 minutes:
The duration t is 30 minutes, which is equal to $30 \times 60 = 1800$ seconds.
The total energy E absorbed by the surface is:
$E = P \times t = 600 \, \text{W} \times 1800 \, \text{s} = 1{,}080{,}000 \, \text{J}$
The relationship between energy E and momentum p for a photon is given by E = pc, where c is the speed of light in a vacuum.
Rearranging for momentum, we get:
$p = \frac{E}{c}$
Now, calculate the total momentum delivered:
Using $c \approx 3 \times 10^8 \, \text{m/s}$:
$p = \frac{1{,}080{,}000 \, \text{J}}{3 \times 10^8 \, \text{m/s}} = \frac{1{,}080{,}000}{300{,}000{,}000} = 3.6 \times 10^{-3} \, \text{kg}\ \text{m/s}$
So, the correct option is (C): $3.6 \times 10^{-3} \, \text{kg}\ \text{m/s}$