Answer
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Hint: In order to get the solution of the given question, we need to find the change in momentum of the laser beam. After that, we need to relate it with conservation of momentum and conservation of energy. After comparing, we will get an equation and solving the equation will give us the required solution of the given question.
Complete step by step solution:
The energy of the laser that falls on the mirror is given as $E$.
If we assume the laser beam to be perfectly reflected, then the change in momentum of laser beam is known as $\dfrac{{2E}}{c}.$
Now, if we apply conservation of momentum, we can write it as,
$mv = \dfrac{{2E}}{c}$
$ \Rightarrow v = \dfrac{{2E}}{{mc}}$
As, we know that from conservation of energy, Kinetic energy = Potential energy
Therefore, we can write, $\dfrac{1}{2}m{v^2} = mg(1 - \cos \theta )$………… (i)
Also, from trigonometry, we know that,$1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
Now, since the angle is very small, we can write the above relation as, $1 - \cos \theta = 2{\left( {\dfrac{\theta }{2}} \right)^2}$=$\dfrac{{{\theta ^2}}}{2}$
Now, using the above relation in equation (i), we can write it as,
$\dfrac{1}{2}m{\left( {\dfrac{{2E}}{{mc}}} \right)^2} = mgl\dfrac{{{\theta ^2}}}{2}$
$ \Rightarrow {\theta ^2} = \dfrac{{2 \times 2{E^2}}}{{c{m^2}gl}}$
$\therefore \theta = \dfrac{{2E}}{{mc\sqrt {gl} }}$
As we can notice, the required value of the angle is not given in the options of the given question.
Hence, option (D), i.e. none of the above is the correct choice of the given question.
Note: We define a laser beam as a ray of light with high intensity that is generated through the emission of photons (also known as light particles). Photons do not have mass but they are capable of redirecting particles with which they collide and transfer their momentum to them.
Complete step by step solution:
The energy of the laser that falls on the mirror is given as $E$.
If we assume the laser beam to be perfectly reflected, then the change in momentum of laser beam is known as $\dfrac{{2E}}{c}.$
Now, if we apply conservation of momentum, we can write it as,
$mv = \dfrac{{2E}}{c}$
$ \Rightarrow v = \dfrac{{2E}}{{mc}}$
As, we know that from conservation of energy, Kinetic energy = Potential energy
Therefore, we can write, $\dfrac{1}{2}m{v^2} = mg(1 - \cos \theta )$………… (i)
Also, from trigonometry, we know that,$1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
Now, since the angle is very small, we can write the above relation as, $1 - \cos \theta = 2{\left( {\dfrac{\theta }{2}} \right)^2}$=$\dfrac{{{\theta ^2}}}{2}$
Now, using the above relation in equation (i), we can write it as,
$\dfrac{1}{2}m{\left( {\dfrac{{2E}}{{mc}}} \right)^2} = mgl\dfrac{{{\theta ^2}}}{2}$
$ \Rightarrow {\theta ^2} = \dfrac{{2 \times 2{E^2}}}{{c{m^2}gl}}$
$\therefore \theta = \dfrac{{2E}}{{mc\sqrt {gl} }}$
As we can notice, the required value of the angle is not given in the options of the given question.
Hence, option (D), i.e. none of the above is the correct choice of the given question.
Note: We define a laser beam as a ray of light with high intensity that is generated through the emission of photons (also known as light particles). Photons do not have mass but they are capable of redirecting particles with which they collide and transfer their momentum to them.
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