Question

Light of intensity $3W{{m}^{-2}}$ is incident on a perfectly absorbing metal surface of area $1{{m}^{2}}$ making an angle of ${{60}^{0}}$ with the normal. If the force exerted by the photons on the surface is $p\times {{10}^{-9}}N$, then the value of ‘p’ is:
(A) 5
(B) 2
(C) 1
(D) 3

Answer 1

The light incident on the perfectly absorbing material is hitting it at a certain angle. This angle is made by the normal to the surface. Hence, the effective area for the light on the surface will be written as the cosine component of area. Once we get the effective area, we can calculate the total energy falling on the surface per second, that is the power delivered to the surface.

Complete answer:
Let the area of a perfectly absorbing metal surface be A.
Then,
$\Rightarrow A=1{{m}^{2}}$
Also, the effective area over which power is delivered by light will be:
$\begin{aligned} & \Rightarrow {{A}_{eff}}=A\cos {{60}^{0}} \\\ & \Rightarrow {{A}_{eff}}=1\times \dfrac{1}{2}{{m}^{2}} \\\ & \therefore {{A}_{eff}}=0.5{{m}^{2}} \\\ \end{aligned}$
Now, the power delivered by the incident light on the metal surface can be calculated as follows:
$\Rightarrow Power=Intensity\times {{A}_{eff}}$
Here, intensity and effective area is known to us. Putting their values in the above equation, we get:
$\begin{aligned} & \Rightarrow P=3\times 0.5Watt \\\ & \Rightarrow P=1.5Watt \\\ \end{aligned}$
Now, the force exerted by these photons on the metal surface will be given by the following equation:
$\Rightarrow F=\dfrac{P}{c}$
Where, ‘c’ is the speed of light. Putting the values of Power delivered and speed of light in the above equation, we get:
$\begin{aligned} & \Rightarrow F=\dfrac{1.5}{3\times {{10}^{8}}}N \\\ & \Rightarrow F=5\times {{10}^{-9}}N \\\ \end{aligned}$
This value of Force exerted by the photons on the metal is given as:
$\Rightarrow F=p\times {{10}^{-9}}N$
Therefore, comparing the force calculated with the force given in the problem, we get:
$\begin{aligned} & \Rightarrow 5\times {{10}^{-9}}N=p\times {{10}^{-9}}N \\\ & \therefore p=5 \\\ \end{aligned}$
Hence, the value of ‘p’ is calculated to be 5.

Hence, option (A) is the correct option.

Note:
Even though as impossible as it sounds but photons do exert force on a substance. This means light incident on any substance exerts force on it. This is not observed by us or one can say felt by us because this force is of the order of nano-Newtons or even smaller under normal conditions.