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

The Correct Option is C

Given parameters:

• Energy flux \( \phi = 20 \, \text{W/cm}^2 \)
• Surface area \( A = 30 \, \text{cm}^2 \)
• Time duration \( t = 30 \, \text{min} = 1800 \, \text{s} \)
• Complete absorption condition

 

Total energy absorbed: \[ E = \phi \times A \times t = 20 \times 30 \times 1800 = 1.08 \times 10^6 \, \text{J} \]

Momentum of light: \[ p = \frac{E}{c} = \frac{1.08 \times 10^6}{3 \times 10^8} = 3.6 \times 10^{-3} \, \text{kg m/s} \]

Thus, the correct option is (C): \( 3.6 \times 10^{-3} \, \text{kg m/s} \).

Answer 2

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): \(3.6 \times 10^{-3} \, \text{kg}\ \text{m/s}\)