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 c m2 , the average force 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 c m2 , the average force exerted on the surface is (A) 1.25 × 10-6 N  (B) 2.50×10-6 N (C) 1.20×10-6N (D) 3.0×10-6N

Tardigrade Admin · Accepted Answer

Here, Energy flux, I=25 × 104 W m-2 Area, A=15 cm 2=15 × 10-4 m 2 Speed of light, c=3 × 108 ms -1 For a perfectly reflecting surface, the average force exerted on the surface is F=(2 I A/c)=(2 × 25 × 104 Wm -2 × 15 × 10-4 m 2/3 × 108 m s -1) =250 × 10-8 N=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 exerted on the surface is

$1.25 \times 1 0^{- 6} N$

$2.50 \times 1 0^{- 6} N$

$1.20 \times 1 0^{- 6} N$

$3.0 \times 1 0^{- 6} N$

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 exerted on the surface is

1.25×10−6N

2.50×10−6N

1.20×10−6N

3.0×10−6N

Here, Energy flux, I=25×104Wm−2 Area, A=15cm2=15×10−4m2 Speed of light, c=3×108ms−1 For a perfectly reflecting surface, the average force exerted on the surface is F=c2IA​=3×108ms−12×25×104Wm−2×15×10−4m2​ =250×10−8N=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 exerted on the surface is

$1.25 \times 1 0^{- 6} N$

$2.50 \times 1 0^{- 6} N$

$1.20 \times 1 0^{- 6} N$

$3.0 \times 1 0^{- 6} N$