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Question: A beam of wavelength 500nm is known to be incident normally on a plane surface. The incident beam is...

A beam of wavelength 500nm is known to be incident normally on a plane surface. The incident beam is known to carry a power of 10W. Find the force exerted by the light beam on the surface.
A. 6.66×108N6.66\times {{10}^{8}}N
B. 6.66×10−8N6.66\times {{10}^{-8}}N
C. 3.33×108N3.33\times {{10}^{8}}N
D. 3.33×10−8N3.33\times {{10}^{-8}}N

Explanation

Solution

As a first step, one could read the question well and hence note down all the given quantities from the question. You could then recall the expression as per requirement and hence substitute these given values into this expression and hence find the answer.

Formula used:
Energy of photon,
E=hcλE=\dfrac{hc}{\lambda }
Number of photons absorbed,
N=PEN=\dfrac{P}{E}
Force exerted on the surface,
F=NhλF=\dfrac{Nh}{\lambda }

Complete step-by-step solution:
In the question we are given a beam of light wavelength λ=500nm=500×10−9m\lambda =500nm=500\times {{10}^{-9}}mand the power of this incident beam is also known to be P = 10W. WE are supposed to find the force exerted by the beam on the surface.
You may recall that the energy of a photon could be given by,
E=hcλE=\dfrac{hc}{\lambda }……………………………………….. (1)
We know that the number of photons that are absorbed every second is given by,
N=PEN=\dfrac{P}{E}
Substituting the equation (1) here, we get,
N=PλhcN=\dfrac{P\lambda }{hc}
Now on substituting the given values from the question into the above expression we get,
N=10×500×10−96.62×10−34×3×108=2.52×1019N=\dfrac{10\times 500\times {{10}^{-9}}}{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}=2.52\times {{10}^{19}}……………………………… (2)
Now let us recall the expression for force exerted by light beam on the surface.
F=NhλF=\dfrac{Nh}{\lambda }
Now we could simply substitute the value from equation (2) here to get,
F=2.52×1019×6.62×10−34500×10−9F=\dfrac{2.52\times {{10}^{19}}\times 6.62\times {{10}^{-34}}}{500\times {{10}^{-9}}}
∴F=3.33×10−8N\therefore F=3.33\times {{10}^{-8}}N
Therefore, we found the force exerted by the given beam on the surface to be 3.33×10−8N3.33\times {{10}^{-8}}N. Hence, option D is found to be the answer.

Note: It is actually not necessary that the required quantities would be directly given in the question. That is, most of the time we will have to go through other equations and substitute them so as to work accordingly with the quantities that we are directly in the question.