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 cm-2, the momentum transferred to the surface in 40 min time duration will be :

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 cm-2, the momentum transferred to the surface in 40 min time duration will be : (A) 5.0 × 10-3 Ns (B) 3.5 × 10-6 Ns (C) 1.4 × 10-6 Ns (D) 6.3 × 10-4 Ns

Tardigrade Admin · Accepted Answer

Pressure = (I/C) Force = Pressure × Area = (I/C) . Area Momentum transferred = Force . Δ t = (I/C) . Area . Δ t = (25 × 104/3 × 108) × 25 × 10-4 × 40 × 60 = 5 × 10-3 N - s

Q. Light is incident normally on a completely absorbing surface with an energy flux of $25 W c m^{- 2}$ . if the surface has an area of $25 c m^{- 2}$ , the momentum transferred to the surface in $40$ min time duration will be :

$5.0 \times 1 0^{- 3} N s$

$3.5 \times 1 0^{- 6} N s$

$1.4 \times 1 0^{- 6} N s$

$6.3 \times 1 0^{- 4} N s$

Pressure = $\frac{I}{C}$
Force = Pressure $\times$ Area = $\frac{I}{C}$ . Area
Momentum transferred = Force .
$Δ t = \frac{I}{C}$ . Area . $Δ t$
$= \frac{25 \times 1 0 ^{4}}{3 \times 1 0 ^{8}} \times 25 \times 1 0^{- 4} \times 40 \times 60$
$= 5 \times 1 0^{- 3} N - s$

Q. Light is incident normally on a completely absorbing surface with an energy flux of 25Wcm−2. if the surface has an area of 25cm−2, the momentum transferred to the surface in 40 min time duration will be :

5.0×10−3Ns

3.5×10−6Ns

1.4×10−6Ns

6.3×10−4Ns