Question

Light of intensity 1.5 kW/m2 is incident normally on a perfectly absorbing surface. The pressure exerted by the light on the surface is

• A. 5 × ${10}^{-6}$ Pa
• B. 2 × ${10}^{-6}$ Pa
• C. 4 × ${10}^{-5}$ Pa
• D. 1.0 × ${10}^{-5}$ Pa

Answer 1

The correct option is A 5 × ${10}^{-6}$ Pa

$P=\frac{I}{C}=\left(\frac{1.5\times {10}^3}{3\times {10}^8}\right)$ = 0.5 × ${10}^{-5}$ = 5 × ${10}^{-6}$