Q. A radiation of 200 W is incident on a surface which is 60% reflecting and 40% absorbing. The total force on the surface is
A) $1.07 \times 10^{- 6}$N
B) $1.3 \times 10^{- 6}$N
C) $1.07 \times 10^{- 7}$N
D) $1.03 \times 10^{- 7}$N
Explanation
(a) $F_{total} + F_{ref} + F_{abs}$
$= \frac{1.2P}{c} + \frac{0.4P}{c} = \frac{1.6P}{c}$
$= \frac{1.6 \times 200}{3 \times 10^{8}} = 1.07 \times 10^{- 6}\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.