A radiation of 200 W is incident on a surface which is 60% reflecting and 40% absorbing. The total force on the surface is
Answer & explanation
Correct answer: option 1
$F_{total} + F_{ref} + F_{abs}$
$=\frac{1.2P}{c}+\frac{0.4P}{c}=\frac{1.6P}{c}$
$=\frac{1.6×200}{3×10^8}=1.07 × 10^{-6} N$