Question

Light of intensity$=3W/m^2$ is incident on a perfectly absorbing metal surface of area $1m^2$ making an angle of ${60}^0$ with the normal. If the force exerted by the photons on the surface is $p\times {10}^{-9}$ (in Newton), then the value of p is :

• A. 5
• B. 2
• C. 1
• D. 3

Answer 1

The correct option is A 5

We will be calculating everything per unit time.
Effective area for the light on the surface is $=1\times cos{60}^0$
Energy falling on the surface$=$ intensity x Effective area $=3\times 1\times cos{60}^0=\frac{3}{2}watt$
Momentum carried by the light per sec $=3/\left(2c\right)=5\times {10}^{-9}$
So, $p=5\times {10}^{-9}$