Light with an energy flux of 25 × 104 W m -2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm 2, the average force (in micro newton) exerted on the surface is

Question

Light with an energy flux of 25 × 104 W m -2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm 2, the average force (in micro newton) exerted on the surface is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Average force, F av =(Δ p/Δ t)=(2 I A/c) =(2 × 25 × 104 × 15 × 10-4/3 × 108) =2.50 × 10-6 N

Q. Light with an energy flux of $25 \times 1 0^{4} W m^{- 2}$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15 c m^{2}$ , the average force (in micro newton) exerted on the surface is

Average force, $F_{a v} = \frac{Δ p}{Δ t} = \frac{2 I A}{c}$
$= \frac{2 \times 25 \times 1 0 ^{4} \times 15 \times 1 0 ^{- 4}}{3 \times 1 0 ^{8}}$
$= 2.50 \times 1 0^{- 6} N$

Q. Light with an energy flux of 25×104Wm−2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15cm2, the average force (in micro newton) exerted on the surface is

Average force, Fav​=ΔtΔp​=c2IA​ =3×1082×25×104×15×10−4​ =2.50×10−6N

Q. Light with an energy flux of $25 \times 1 0^{4} W m^{- 2}$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15 c m^{2}$ , the average force (in micro newton) exerted on the surface is

Average force, $F_{a v} = \frac{Δ p}{Δ t} = \frac{2 I A}{c}$
$= \frac{2 \times 25 \times 1 0 ^{4} \times 15 \times 1 0 ^{- 4}}{3 \times 1 0 ^{8}}$
$= 2.50 \times 1 0^{- 6} N$