Question

Light is incident normally on a completely absorbing surface with an energy flux of 25 Wcm^-2. If the surface has an area of 25 cm², the momentum transferred to the surface in 40 min time duration will be:
1. 6.3 × 10^-4 Ns
2. 1.4 × 10^-6 Ns
3. 5.0 × 10^-3 Ns
4. 3.5 × 10^-6 Ns

Answer 1

Correct Answer - Option 3 : 5.0 × 10^-3 Ns

Concept:

As Total momentum of the incident light:

We know that, the momentum of incident light = U(total energy)/c

The total momentum of the incident light delivered (for complete absorption) is

P = U/C

Where, U is the intensity and c is the velocity of the light

Calculation:

Given,

Surface Area, A = 25 cm²

Energy flux, ϕ = 25 Wcm^-2

Time, t = 40 mins = 40 × 60 sec

Total energy falling on the surface in time t is

U = ϕAt    

U = 25 × 25 × 40 × 60

U = 625 × 2400

U = 1500000= 15 × 10⁵ J

Therefore, the total momentum of the incident light delivered (for complete absorption) is

\({\rm{P}} = \frac{{\rm{U}}}{{\rm{C}}}\) 

\({\rm{P}} = \frac{{15 \times {{10}^5}}}{{3 \times {{10}^8}}}\) 

P = 5 × 10^-3 Ns

Therefore, the momentum transferred to the surface in 40 min time duration will be 5 × 10^-3 Ns.