A perfectly reflecting mirror has an area of 1 cm 2 . Light energy is allowed to fall on it for 1 h at the rate of 10 W cm -2 . The force that acts on the mirror is:

Question

A perfectly reflecting mirror has an area of 1 cm 2 . Light energy is allowed to fall on it for 1 h at the rate of 10 W cm -2 . The force that acts on the mirror is: (A) 3.35 × 10-8 N (B) 6.7 × 10-8 N (C) 1.34 × 10-7 N  (D) 2.4 × 10-4 N

Tardigrade Admin · Accepted Answer

Let, E= energy falling on the surface per second =10 J Momentum of photons p=(h/λ)=(h/(c / v))=(h v/c)=(E/c) On reflection, change in momentum per second =2 p=(2 E/c)=(2 × 10/3 × 108) =6.7 × 10-8 N

Q. A perfectly reflecting mirror has an area of $1 c m^{2} .$ Light energy is allowed to fall on it for $1 h$ at the rate of $10 W$ $c m^{- 2} .$ The force that acts on the mirror is:

$3.35 \times 1 0^{- 8} N$

$6.7 \times 1 0^{- 8} N$

$1.34 \times 1 0^{- 7} N$

$2.4 \times 1 0^{- 4} N$

Let, $E =$ energy falling on the surface per second
$= 10 J$
Momentum of photons $p = \frac{h}{λ} = \frac{h}{( c / v )} = \frac{h v}{c} = \frac{E}{c}$
On reflection, change in momentum per second
$= 2 p = \frac{2 E}{c} = \frac{2 \times 10}{3 \times 1 0 ^{8}}$
$= 6.7 \times 1 0^{- 8} N$

Q. A perfectly reflecting mirror has an area of 1cm2. Light energy is allowed to fall on it for 1h at the rate of 10W cm−2. The force that acts on the mirror is:

3.35×10−8N

6.7×10−8N

1.34×10−7N

2.4×10−4N